# Why the torque exerted by the moon upon the Earth makes the Moon to increase his orbit?

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For what I know, this torque exists because of the misalignment of the tidal bulge with the apsidal line Earth-Moon a certain angle $$alpha$$, which makes the earth rotation to slow down a bit. But, how that pulls away the moon? I thought this "loss" of rotational momentum transferred into tidal heating. I fail to see how this energy is transferred into more orbital momentum, because I thought the conservation of angular momentum didn't have to do with the rotation of the objects. I mean, isn't true that $$c$$ remains constant in the two-body problem (being $$c = vec{r} imes dot{vec{r}}$$)? How the rotation plays into that?

because I thought the conservation of angular momentum didn't have to do with the rotation of the objects.

The angular momentum of the system is conserved. As rotation can contribute to angular momentum, it does affect the equation. You can ignore it for point-like objects. In the case of the earth-moon system, the earth's rotation is a significant portion of the total momentum.

A not-too-wrong version would be to assume an axis centered on the earth. Then the total angular momentum of the system could be found by summing the individual momenta for:

• the earth's rotation
• the moon's rotation
• the moon's orbit

Assuming the moon is tidally locked to the earth and has a nearly circular orbit, this gives you only two variables. You can then calculate that, if you slow the earth's rotation, the only way to keep angular momentum of the system constant is to increase the angular momentum of the moon by an equal amount. This would correspond to a larger orbit.

Further if you calculate the KE from both configurations, you find the second one has less energy. So there must be an energy loss to transition between them.

## When the Moon orbits Earth, what is the centripetal force?

Any two objects will attract one another gravitationally with a force given by Newton's equation:

#G# is the universal gravitational constant, usually referred to as "big G".

#m_1# and #m_2# are the masses of the objects.

#r# is the distance between the centres of mass of the two objects.

Notice that the gravitational force decreases in inverse proportion to the square of the distance between the objects.

If there were no force of gravity then the moon would tend to travel in a straight line. With gravity providing a centripetal force, that path is curved towards the Earth, resulting in a roughly circular orbit. In essence the moon is in continuous freefall towards the Earth.

Actually the moon and the Earth orbit around a centre which lies between the centre of the Earth and that of the moon. However, since the Earth is much more massive than the moon, that centre lies within the Earth (at a distance of #4670# km or so from the centre of the Earth).

There is no centripetal force between the Earth and Moon according to General Relativity.

#### Explanation:

Newton's laws of gravity and motion are a good approximation as long as the masses of the bodies are not too large and they are travelling at speeds significantly slower than the speed of light.

Newton describes gravity as a force when in fact gravity is not a force. In fact it is inaccurate to use the term force of gravity.

Anything with mass causes curvature of 4 dimensional spacetime. The three dimensional analogy is placing a ball on a stretched rubber sheet.

A body moving in the absence of an external force travels in a straight line. In 4 dimensional spacetime this is called a geodesic which is the shortest possible line between two points on a curved surface.

So, gravity is not a force. It is the curvature of spacetime. A geodesic in curved spacetime is not a straight line, it is a curve.

So, there is no force acting between the Earth and the Moon. The shape of the Moon's orbit is the shape of the geodesic of the Moon travelling through the curved spacetime caused by the Earth's mass.

## Hall of Fame

### Interaction of Gravitational Mass and Photons

Einstein extended his special theory of relativity to phenomena involving acceleration into his general theory of relativity, which he wrote in 1915 ( Einstein, 1916b ). He proposed that as mass was equivalent to energy, as described above, the same principle of equivalence would require that gravitational mass would interact with the mass of photons of visible light (electromagnetic radiation). From this reasoning, Einstein predicted the deflection of the light from stars as the light would pass near a massive body such as the sun ( Einstein, 1911 ). The gravitational pull of the sun would attract and bend the light from a distant star as that light passed near the body of the sun. Einstein concluded that this deflection of light toward the body of the sun could be observed from earth when the light of the sun would be blocked by a total eclipse. In 1913, prior to the total eclipse of May, 1919, Einstein diagrammed a sketch illustrating how the gravity of the sun would deflect light near the sun making stars to appear to observers on earth that they have shifted their position in space. Einstein's prediction was found to be true when British astronomers in May of 1919 took photographs of the total eclipse of the sun. The British astronomer Arthur Eddington demonstrated Einstein's prediction to be true. He spotted a star that should have been hidden behind the sun. Photographs of the total eclipse illustrated how the positions of some stars deviated from their positions when the stars were photographed on other occasions with the sun in a different location in the sky. This finding made Einstein an instant celebrity. The London Times on November 7, 1919 ran the headline “Revolution in Science, New Theory of the Universe, Newtonian Ideas Overthrown.” The demonstration of Einstein's theory is commemorated in the stamp issued by Serbia in 2004. The stamp illustrates how the bending of the light near the sun gave the appearance that the distant star was located to the side rather than directly behind the sun. Eddington commented that it was the greatest moment of his life when he measured the image of a star and found that the sun's gravity warped the space through which the light had traveled. The effect was reconfirmed in photographs taken during the solar eclipse of 1922. As noted by the American Institute of Physics:

The eclipse experiments, like most important new science, were done at the very limit of available techniques. It was not until the 1960s, with vastly improved methods, that the gravitational bending of light could be demonstrated beyond reasonable doubt. Until then one could almost say that the logic and beauty of Einstein's theory did as much to confirm the observations as the observations did to confirm his theory.

Eddingtons findings during the solar eclipse of 1919 confirming Einstein's predictions were commemorated with the postage stamp issued in 2009 by São Tomé and Príncipe illustrated here.

In addition to the deflection of light by the sun's gravity described above, Einstein made several predictions from general relativity theory including the perihelic orbit of the planet Mercury ( Einstein, 1915 ). In his Nobel lecture presented on July 11, 1923 (see clarification below for the lateness and subject matter of the lecture), Einstein underscored some of the predictions of his theory when he stated:

The considerations mentioned led to the theory of gravity which yields the Newtonian theory as a first approximation and furthermore it yields the motion of the perihelion of Mercury, the deflection of light by the sun, and the red shift of spectral lines [due to an expanding universe]…

Einstein (1923) .

## Problems

Andrea, a 63.0-kg sprinter, starts a race with an acceleration of 4.200 m/s 2 4.200 m/s 2 . What is the net external force on her?

If the sprinter from the previous problem accelerates at that rate for 20.00 m and then maintains that velocity for the remainder of a 100.00-m dash, what will her time be for the race?

A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate the magnitude of his cart’s acceleration.

Astronauts in orbit are apparently weightless. This means that a clever method of measuring the mass of astronauts is needed to monitor their mass gains or losses, and adjust their diet. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of 50.0 N is exerted, and an astronaut’s acceleration is measured to be 0.893 m/s 2 0.893 m/s 2 . (a) Calculate her mass. (b) By exerting a force on the astronaut, the vehicle in which she orbits experiences an equal and opposite force. Use this knowledge to find an equation for the acceleration of the system (astronaut and spaceship) that would be measured by a nearby observer. (c) Discuss how this would affect the measurement of the astronaut’s acceleration. Propose a method by which recoil of the vehicle is avoided.

In Figure 5.12, the net external force on the 24-kg mower is given as 51 N. If the force of friction opposing the motion is 24 N, what force F (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m/s when the force F is removed. How far will the mower go before stopping?

If the rocket sled shown in the previous problem starts with only one rocket burning, what is the magnitude of this acceleration? Assume that the mass of the system is 2.10 × 10 3 2.10 × 10 3 kg, the thrust T is 2.40 × 10 4 N , 2.40 × 10 4 N , and the force of friction opposing the motion is 650.0 N. (b) Why is the acceleration not one-fourth of what it is with all rockets burning?

What is the acceleration opposite to the motion of the rocket sled if it comes to rest in 1.10 s from a speed of 1000.0 km/h? (Such acceleration opposite to the motion caused one test subject to black out and have temporary blindness.)

Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second exerts a force of 90.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 23.0 kg. (a) What is the system of interest if the acceleration of the child in the wagon is to be calculated? (See the free-body diagram.) (b) Calculate the acceleration. (c) What would the acceleration be if friction were 15.0 N?

A car with a mass of 1000.0 kg accelerates from 0 to 90.0 km/h in 10.0 s. (a) What is its acceleration? (b) What is the net force on the car?

The driver in the previous problem applies the brakes when the car is moving at 90.0 km/h, and the car comes to rest after traveling 40.0 m. What is the net force on the car during its acceleration opposite to the motion?

Suppose that the particle of the previous problem also experiences forces F → 2 = −15 i ^ N F → 2 = −15 i ^ N and F → 3 = 6.0 j ^ N . F → 3 = 6.0 j ^ N . What is its acceleration in this case?

Find the acceleration of the body of mass 5.0 kg shown below.

In the following figure, the horizontal surface on which this block slides is frictionless. If the two forces acting on it each have magnitude F = 30.0 N F = 30.0 N and M = 10.0 kg M = 10.0 kg , what is the magnitude of the resulting acceleration of the block?

### 5.4 Mass and Weight

The weight of an astronaut plus his space suit on the Moon is only 250 N. (a) How much does the suited astronaut weigh on Earth? (b) What is the mass on the Moon? On Earth?

Repeat the previous problem for a situation in which the rocket sled accelerates opposite to the motion at a rate of 201 m/s 2 201 m/s 2 . In this problem, the forces are exerted by the seat and the seat belt.

A body of mass 2.00 kg is pushed straight upward by a 25.0 N vertical force. What is its acceleration?

A car weighing 12,500 N starts from rest and accelerates to 83.0 km/h in 5.00 s. The friction force is 1350 N. Find the applied force produced by the engine.

A body with a mass of 10.0 kg is assumed to be in Earth’s gravitational field with g = 9.80 m/s 2 g = 9.80 m/s 2 . What is the net force on the body if there are no other external forces acting on the object?

A fireman has mass m he hears the fire alarm and slides down the pole with acceleration a (which is less than g in magnitude). (a) Write an equation giving the vertical force he must apply to the pole. (b) If his mass is 90.0 kg and he accelerates at 5.00 m/s 2 , 5.00 m/s 2 , what is the magnitude of his applied force?

A baseball catcher is performing a stunt for a television commercial. He will catch a baseball (mass 145 g) dropped from a height of 60.0 m above his glove. His glove stops the ball in 0.0100 s. What is the force exerted by his glove on the ball?

### 5.5 Newton’s Third Law

(a) What net external force is exerted on a 1100.0-kg artillery shell fired from a battleship if the shell is accelerated at 2.40 × 10 4 m/s 2 ? 2.40 × 10 4 m/s 2 ? (b) What is the magnitude of the force exerted on the ship by the artillery shell, and why?

A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800.0 N on him. The mass of the losing player plus equipment is 90.0 kg, and he is accelerating backward at 1.20 m/s 2 1.20 m/s 2 . (a) What is the force of friction between the losing player’s feet and the grass? (b) What force does the winning player exert on the ground to move forward if his mass plus equipment is 110.0 kg?

A history book is lying on top of a physics book on a desk, as shown below a free-body diagram is also shown. The history and physics books weigh 14 N and 18 N, respectively. Identify each force on each book with a double subscript notation (for instance, the contact force of the history book pressing against physics book can be described as F → HP F → HP ), and determine the value of each of these forces, explaining the process used.

A truck collides with a car, and during the collision, the net force on each vehicle is essentially the force exerted by the other. Suppose the mass of the car is 550 kg, the mass of the truck is 2200 kg, and the magnitude of the truck’s acceleration is 10 m/s 2 10 m/s 2 . Find the magnitude of the car’s acceleration.

### 5.6 Common Forces

A leg is suspended in a traction system, as shown below. (a) Which pulley in the figure is used to calculate the force exerted on the foot? (b) What is the tension in the rope? Here T → T → is the tension, w → leg w → leg is the weight of the leg, and w → w → is the weight of the load that provides the tension.

Suppose the shinbone in the preceding image was a femur in a traction setup for a broken bone, with pulleys and rope available. How might we be able to increase the force along the femur using the same weight?

Two teams of nine members each engage in a tug-of-war, pulling in opposite directions on a horizontal rope. Each of the first team’s members has an average mass of 68 kg and exerts an average force of 1350 N horizontally on the ground as they pull on the rope. Each of the second team’s members has an average mass of 73 kg and exerts an average force of 1365 N horizontally on the ground as they pull on the rope in the opposite direction. (a) What is the magnitude of the acceleration of the two teams, and which team wins? (b) What is the tension in the section of rope between the teams?

What force does a trampoline have to apply to Jennifer, a 45.0-kg gymnast, to accelerate her straight up at 7.50 m/s 2 7.50 m/s 2 ? The answer is independent of the velocity of the gymnast—she can be moving up or down or can be instantly stationary.

## 39 Newton’s Universal Law of Gravitation

What do aching feet, a falling apple, and the orbit of the Moon have in common? Each is caused by the gravitational force. Our feet are strained by supporting our weight—the force of Earth’s gravity on us. An apple falls from a tree because of the same force acting a few meters above Earth’s surface. And the Moon orbits Earth because gravity is able to supply the necessary centripetal force at a distance of hundreds of millions of meters. In fact, the same force causes planets to orbit the Sun, stars to orbit the center of the galaxy, and galaxies to cluster together. Gravity is another example of underlying simplicity in nature. It is the weakest of the four basic forces found in nature, and in some ways the least understood. It is a force that acts at a distance, without physical contact, and is expressed by a formula that is valid everywhere in the universe, for masses and distances that vary from the tiny to the immense.

Sir Isaac Newton was the first scientist to precisely define the gravitational force, and to show that it could explain both falling bodies and astronomical motions. See (Figure). But Newton was not the first to suspect that the same force caused both our weight and the motion of planets. His forerunner Galileo Galilei had contended that falling bodies and planetary motions had the same cause. Some of Newton’s contemporaries, such as Robert Hooke, Christopher Wren, and Edmund Halley, had also made some progress toward understanding gravitation. But Newton was the first to propose an exact mathematical form and to use that form to show that the motion of heavenly bodies should be conic sections—circles, ellipses, parabolas, and hyperbolas. This theoretical prediction was a major triumph—it had been known for some time that moons, planets, and comets follow such paths, but no one had been able to propose a mechanism that caused them to follow these paths and not others.

The gravitational force is relatively simple. It is always attractive, and it depends only on the masses involved and the distance between them. Stated in modern language, Newton’s universal law of gravitation states that every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

The magnitude of the force on each object (one has larger mass than the other) is the same, consistent with Newton’s third law.

The bodies we are dealing with tend to be large. To simplify the situation we assume that the body acts as if its entire mass is concentrated at one specific point called the center of mass (CM), which will be further explored in Linear Momentum and Collisions. For two bodies having masses and with a distance between their centers of mass, the equation for Newton’s universal law of gravitation is

where is the magnitude of the gravitational force and is a proportionality factor called the gravitational constant . is a universal gravitational constant—that is, it is thought to be the same everywhere in the universe. It has been measured experimentally to be

in SI units. Note that the units of are such that a force in newtons is obtained from , when considering masses in kilograms and distance in meters. For example, two 1.000 kg masses separated by 1.000 m will experience a gravitational attraction of . This is an extraordinarily small force. The small magnitude of the gravitational force is consistent with everyday experience. We are unaware that even large objects like mountains exert gravitational forces on us. In fact, our body weight is the force of attraction of the entire Earth on us with a mass of .

Recall that the acceleration due to gravity is about on Earth. We can now determine why this is so. The weight of an object mg is the gravitational force between it and Earth. Substituting mg for in Newton’s universal law of gravitation gives

where is the mass of the object, is the mass of Earth, and is the distance to the center of Earth (the distance between the centers of mass of the object and Earth). See (Figure). The mass of the object cancels, leaving an equation for :

Substituting known values for Earth’s mass and radius (to three significant figures),

and we obtain a value for the acceleration of a falling body:

This is the expected value and is independent of the body’s mass. Newton’s law of gravitation takes Galileo’s observation that all masses fall with the same acceleration a step further, explaining the observation in terms of a force that causes objects to fall—in fact, in terms of a universally existing force of attraction between masses.

Take a marble, a ball, and a spoon and drop them from the same height. Do they hit the floor at the same time? If you drop a piece of paper as well, does it behave like the other objects? Explain your observations.

Attempts are still being made to understand the gravitational force. As we shall see in Particle Physics, modern physics is exploring the connections of gravity to other forces, space, and time. General relativity alters our view of gravitation, leading us to think of gravitation as bending space and time.

In the following example, we make a comparison similar to one made by Newton himself. He noted that if the gravitational force caused the Moon to orbit Earth, then the acceleration due to gravity should equal the centripetal acceleration of the Moon in its orbit. Newton found that the two accelerations agreed “pretty nearly.”

(a) Find the acceleration due to Earth’s gravity at the distance of the Moon.

(b) Calculate the centripetal acceleration needed to keep the Moon in its orbit (assuming a circular orbit about a fixed Earth), and compare it with the value of the acceleration due to Earth’s gravity that you have just found.

This calculation is the same as the one finding the acceleration due to gravity at Earth’s surface, except that is the distance from the center of Earth to the center of the Moon. The radius of the Moon’s nearly circular orbit is .

Solution for (a)

Substituting known values into the expression for found above, remembering that is the mass of Earth not the Moon, yields

Centripetal acceleration can be calculated using either form of

We choose to use the second form:

where is the angular velocity of the Moon about Earth.

Solution for (b)

Given that the period (the time it takes to make one complete rotation) of the Moon’s orbit is 27.3 days, (d) and using

The centripetal acceleration is

The direction of the acceleration is toward the center of the Earth.

The centripetal acceleration of the Moon found in (b) differs by less than 1% from the acceleration due to Earth’s gravity found in (a). This agreement is approximate because the Moon’s orbit is slightly elliptical, and Earth is not stationary (rather the Earth-Moon system rotates about its center of mass, which is located some 1700 km below Earth’s surface). The clear implication is that Earth’s gravitational force causes the Moon to orbit Earth.

Why does Earth not remain stationary as the Moon orbits it? This is because, as expected from Newton’s third law, if Earth exerts a force on the Moon, then the Moon should exert an equal and opposite force on Earth (see (Figure)). We do not sense the Moon’s effect on Earth’s motion, because the Moon’s gravity moves our bodies right along with Earth but there are other signs on Earth that clearly show the effect of the Moon’s gravitational force as discussed in Satellites and Kepler’s Laws: An Argument for Simplicity.

### Tides

Ocean tides are one very observable result of the Moon’s gravity acting on Earth. (Figure) is a simplified drawing of the Moon’s position relative to the tides. Because water easily flows on Earth’s surface, a high tide is created on the side of Earth nearest to the Moon, where the Moon’s gravitational pull is strongest. Why is there also a high tide on the opposite side of Earth? The answer is that Earth is pulled toward the Moon more than the water on the far side, because Earth is closer to the Moon. So the water on the side of Earth closest to the Moon is pulled away from Earth, and Earth is pulled away from water on the far side. As Earth rotates, the tidal bulge (an effect of the tidal forces between an orbiting natural satellite and the primary planet that it orbits) keeps its orientation with the Moon. Thus there are two tides per day (the actual tidal period is about 12 hours and 25.2 minutes), because the Moon moves in its orbit each day as well).

The Sun also affects tides, although it has about half the effect of the Moon. However, the largest tides, called spring tides, occur when Earth, the Moon, and the Sun are aligned. The smallest tides, called neap tides, occur when the Sun is at a angle to the Earth-Moon alignment.

(a, b) Spring tides: The highest tides occur when Earth, the Moon, and the Sun are aligned. (c) Neap tide: The lowest tides occur when the Sun lies at to the Earth-Moon alignment. Note that this figure is not drawn to scale.

Tides are not unique to Earth but occur in many astronomical systems. The most extreme tides occur where the gravitational force is the strongest and varies most rapidly, such as near black holes (see (Figure)). A few likely candidates for black holes have been observed in our galaxy. These have masses greater than the Sun but have diameters only a few kilometers across. The tidal forces near them are so great that they can actually tear matter from a companion star.

### ”Weightlessness” and Microgravity

In contrast to the tremendous gravitational force near black holes is the apparent gravitational field experienced by astronauts orbiting Earth. What is the effect of “weightlessness” upon an astronaut who is in orbit for months? Or what about the effect of weightlessness upon plant growth? Weightlessness doesn’t mean that an astronaut is not being acted upon by the gravitational force. There is no “zero gravity” in an astronaut’s orbit. The term just means that the astronaut is in free-fall, accelerating with the acceleration due to gravity. If an elevator cable breaks, the passengers inside will be in free fall and will experience weightlessness. You can experience short periods of weightlessness in some rides in amusement parks.

Microgravity refers to an environment in which the apparent net acceleration of a body is small compared with that produced by Earth at its surface. Many interesting biology and physics topics have been studied over the past three decades in the presence of microgravity. Of immediate concern is the effect on astronauts of extended times in outer space, such as at the International Space Station. Researchers have observed that muscles will atrophy (waste away) in this environment. There is also a corresponding loss of bone mass. Study continues on cardiovascular adaptation to space flight. On Earth, blood pressure is usually higher in the feet than in the head, because the higher column of blood exerts a downward force on it, due to gravity. When standing, 70% of your blood is below the level of the heart, while in a horizontal position, just the opposite occurs. What difference does the absence of this pressure differential have upon the heart?

Some findings in human physiology in space can be clinically important to the management of diseases back on Earth. On a somewhat negative note, spaceflight is known to affect the human immune system, possibly making the crew members more vulnerable to infectious diseases. Experiments flown in space also have shown that some bacteria grow faster in microgravity than they do on Earth. However, on a positive note, studies indicate that microbial antibiotic production can increase by a factor of two in space-grown cultures. One hopes to be able to understand these mechanisms so that similar successes can be achieved on the ground. In another area of physics space research, inorganic crystals and protein crystals have been grown in outer space that have much higher quality than any grown on Earth, so crystallography studies on their structure can yield much better results.

Plants have evolved with the stimulus of gravity and with gravity sensors. Roots grow downward and shoots grow upward. Plants might be able to provide a life support system for long duration space missions by regenerating the atmosphere, purifying water, and producing food. Some studies have indicated that plant growth and development are not affected by gravity, but there is still uncertainty about structural changes in plants grown in a microgravity environment.

### The Cavendish Experiment: Then and Now

As previously noted, the universal gravitational constant is determined experimentally. This definition was first done accurately by Henry Cavendish (1731–1810), an English scientist, in 1798, more than 100 years after Newton published his universal law of gravitation. The measurement of is very basic and important because it determines the strength of one of the four forces in nature. Cavendish’s experiment was very difficult because he measured the tiny gravitational attraction between two ordinary-sized masses (tens of kilograms at most), using apparatus like that in (Figure). Remarkably, his value for differs by less than 1% from the best modern value.

One important consequence of knowing was that an accurate value for Earth’s mass could finally be obtained. This was done by measuring the acceleration due to gravity as accurately as possible and then calculating the mass of Earth from the relationship Newton’s universal law of gravitation gives

where is the mass of the object, is the mass of Earth, and is the distance to the center of Earth (the distance between the centers of mass of the object and Earth). See (Figure). The mass of the object cancels, leaving an equation for :

Rearranging to solve for yields

So can be calculated because all quantities on the right, including the radius of Earth , are known from direct measurements. We shall see in Satellites and Kepler’s Laws: An Argument for Simplicity that knowing also allows for the determination of astronomical masses. Interestingly, of all the fundamental constants in physics, is by far the least well determined.

The Cavendish experiment is also used to explore other aspects of gravity. One of the most interesting questions is whether the gravitational force depends on substance as well as mass—for example, whether one kilogram of lead exerts the same gravitational pull as one kilogram of water. A Hungarian scientist named Roland von Eötvös pioneered this inquiry early in the 20th century. He found, with an accuracy of five parts per billion, that the gravitational force does not depend on the substance. Such experiments continue today, and have improved upon Eötvös’ measurements. Cavendish-type experiments such as those of Eric Adelberger and others at the University of Washington, have also put severe limits on the possibility of a fifth force and have verified a major prediction of general relativity—that gravitational energy contributes to rest mass. Ongoing measurements there use a torsion balance and a parallel plate (not spheres, as Cavendish used) to examine how Newton’s law of gravitation works over sub-millimeter distances. On this small-scale, do gravitational effects depart from the inverse square law? So far, no deviation has been observed.

Cavendish used an apparatus like this to measure the gravitational attraction between the two suspended spheres () and the two on the stand () by observing the amount of torsion (twisting) created in the fiber. Distance between the masses can be varied to check the dependence of the force on distance. Modern experiments of this type continue to explore gravity.

### Section Summary

• Newton’s universal law of gravitation: Every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In equation form, this is

where F is the magnitude of the gravitational force. is the gravitational constant, given by .

### Conceptual Questions

Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the ultimate determinant of the truth in physics, and why was this action ultimately accepted?

Two friends are having a conversation. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward Earth. Tom says a satellite in orbit is not in freefall because the acceleration due to gravity is not . Who do you agree with and why?

Draw a free body diagram for a satellite in an elliptical orbit showing why its speed increases as it approaches its parent body and decreases as it moves away.

Newton’s laws of motion and gravity were among the first to convincingly demonstrate the underlying simplicity and unity in nature. Many other examples have since been discovered, and we now expect to find such underlying order in complex situations. Is there proof that such order will always be found in new explorations?

### Problem Exercises

(a) Calculate Earth’s mass given the acceleration due to gravity at the North Pole is and the radius of the Earth is 6371 km from center to pole.

(b) Compare this with the accepted value of .

a)

b) This is identical to the best value to three significant figures.

(a) Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.

(b) Calculate the magnitude of the acceleration due to gravity at Earth due to the Sun.

(c) Take the ratio of the Moon’s acceleration to the Sun’s and comment on why the tides are predominantly due to the Moon in spite of this number.

(a) What is the acceleration due to gravity on the surface of the Moon?

(b) On the surface of Mars? The mass of Mars is and its radius is .

a)

b)

(a) Calculate the acceleration due to gravity on the surface of the Sun.

(b) By what factor would your weight increase if you could stand on the Sun? (Never mind that you cannot.)

The Moon and Earth rotate about their common center of mass, which is located about 4700 km from the center of Earth. (This is 1690 km below the surface.)

(a) Calculate the magnitude of the acceleration due to the Moon’s gravity at that point.

(b) Calculate the magnitude of the centripetal acceleration of the center of Earth as it rotates about that point once each lunar month (about 27.3 d) and compare it with the acceleration found in part (a). Comment on whether or not they are equal and why they should or should not be.

a)

b)

The values are nearly identical. One would expect the gravitational force to be the same as the centripetal force at the core of the system.

Solve part (b) of (Figure) using .

Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of one’s birth. The only known force a planet exerts on Earth is gravitational.

(a) Calculate the magnitude of the gravitational force exerted on a 4.20 kg baby by a 100 kg father 0.200 m away at birth (he is assisting, so he is close to the child).

(b) Calculate the magnitude of the force on the baby due to Jupiter if it is at its closest distance to Earth, some away. How does the force of Jupiter on the baby compare to the force of the father on the baby? Other objects in the room and the hospital building also exert similar gravitational forces. (Of course, there could be an unknown force acting, but scientists first need to be convinced that there is even an effect, much less that an unknown force causes it.)

a)

b) ,

The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune’s orbit. Pluto was subsequently discovered near its predicted position. But it now appears that the discovery was fortuitous, because Pluto is small and the irregularities in Neptune’s orbit were not well known. To illustrate that Pluto has a minor effect on the orbit of Neptune compared with the closest planet to Neptune:

(a) Calculate the acceleration due to gravity at Neptune due to Pluto when they are apart, as they are at present. The mass of Pluto is .

(b) Calculate the acceleration due to gravity at Neptune due to Uranus, presently about apart, and compare it with that due to Pluto. The mass of Uranus is .

(a) The Sun orbits the Milky Way galaxy once each , with a roughly circular orbit averaging light years in radius. (A light year is the distance traveled by light in 1 y.) Calculate the centripetal acceleration of the Sun in its galactic orbit. Does your result support the contention that a nearly inertial frame of reference can be located at the Sun?

(b) Calculate the average speed of the Sun in its galactic orbit. Does the answer surprise you?

a)

b)

Unreasonable Result

A mountain 10.0 km from a person exerts a gravitational force on him equal to 2.00% of his weight.

(a) Calculate the mass of the mountain.

(b) Compare the mountain’s mass with that of Earth.

(c) What is unreasonable about these results?

(d) Which premises are unreasonable or inconsistent? (Note that accurate gravitational measurements can easily detect the effect of nearby mountains and variations in local geology.)

a)

b)

c) The mass of the mountain and its fraction of the Earth’s mass are too great.

d) The gravitational force assumed to be exerted by the mountain is too great.

## 13.4 Satellite Orbits and Energy

The Moon orbits Earth. In turn, Earth and the other planets orbit the Sun. The space directly above our atmosphere is filled with artificial satellites in orbit. We examine the simplest of these orbits, the circular orbit, to understand the relationship between the speed and period of planets and satellites in relation to their positions and the bodies that they orbit.

### Circular Orbits

As noted at the beginning of this chapter, Nicolaus Copernicus first suggested that Earth and all other planets orbit the Sun in circles. He further noted that orbital periods increased with distance from the Sun. Later analysis by Kepler showed that these orbits are actually ellipses, but the orbits of most planets in the solar system are nearly circular. Earth’s orbital distance from the Sun varies a mere 2%. The exception is the eccentric orbit of Mercury, whose orbital distance varies nearly 40%.

Determining the orbital speed and orbital period of a satellite is much easier for circular orbits, so we make that assumption in the derivation that follows. As we described in the previous section, an object with negative total energy is gravitationally bound and therefore is in orbit. Our computation for the special case of circular orbits will confirm this. We focus on objects orbiting Earth, but our results can be generalized for other cases.

Consider a satellite of mass m in a circular orbit about Earth at distance r from the center of Earth ((Figure)). It has centripetal acceleration directed toward the center of Earth. Earth’s gravity is the only force acting, so Newton’s second law gives

Figure 13.12 A satellite of mass m orbiting at radius r from the center of Earth. The gravitational force supplies the centripetal acceleration.

We solve for the speed of the orbit, noting that m cancels, to get the orbital speed

Consistent with what we saw in (Figure) and (Figure), m does not appear in (Figure). The value of g, the escape velocity, and orbital velocity depend only upon the distance from the center of the planet, and not upon the mass of the object being acted upon. Notice the similarity in the equations for

. The escape velocity is exactly

times greater, about 40%, than the orbital velocity. This comparison was noted in (Figure), and it is true for a satellite at any radius.

To find the period of a circular orbit, we note that the satellite travels the circumference of the orbit

in one period T. Using the definition of speed, we have

. We substitute this into (Figure) and rearrange to get

We see in the next section that this represents Kepler’s third law for the case of circular orbits. It also confirms Copernicus’s observation that the period of a planet increases with increasing distance from the Sun. We need only replace

We conclude this section by returning to our earlier discussion about astronauts in orbit appearing to be weightless, as if they were free-falling towards Earth. In fact, they are in free fall. Consider the trajectories shown in (Figure). (This figure is based on a drawing by Newton in his Principia and also appeared earlier in Motion in Two and Three Dimensions.) All the trajectories shown that hit the surface of Earth have less than orbital velocity. The astronauts would accelerate toward Earth along the noncircular paths shown and feel weightless. (Astronauts actually train for life in orbit by riding in airplanes that free fall for 30 seconds at a time.) But with the correct orbital velocity, Earth’s surface curves away from them at exactly the same rate as they fall toward Earth. Of course, staying the same distance from the surface is the point of a circular orbit.

Figure 13.13 A circular orbit is the result of choosing a tangential velocity such that Earth’s surface curves away at the same rate as the object falls toward Earth.

We can summarize our discussion of orbiting satellites in the following Problem-Solving Strategy.

#### Problem-Solving Strategy: Orbits and Conservation of Energy

1. Determine whether the equations for speed, energy, or period are valid for the problem at hand. If not, start with the first principles we used to derive those equations.
2. To start from first principles, draw a free-body diagram and apply Newton’s law of gravitation and Newton’s second law.
3. Along with the definitions for speed and energy, apply Newton’s second law of motion to the bodies of interest.

### Example

#### The International Space Station

Determine the orbital speed and period for the International Space Station (ISS).

#### Strategy

above Earth’s surface, the radius at which it orbits is

. We use (Figure) and (Figure) to find the orbital speed and period, respectively.

#### Solution

Using (Figure), the orbital velocity is

which is about 17,000 mph. Using (Figure), the period is

which is just over 90 minutes.

#### Significance

The ISS is considered to be in low Earth orbit (LEO). Nearly all satellites are in LEO, including most weather satellites. GPS satellites, at about 20,000 km, are considered medium Earth orbit. The higher the orbit, the more energy is required to put it there and the more energy is needed to reach it for repairs. Of particular interest are the satellites in geosynchronous orbit. All fixed satellite dishes on the ground pointing toward the sky, such as TV reception dishes, are pointed toward geosynchronous satellites. These satellites are placed at the exact distance, and just above the equator, such that their period of orbit is 1 day. They remain in a fixed position relative to Earth’s surface.

By what factor must the radius change to reduce the orbital velocity of a satellite by one-half? By what factor would this change the period?

In (Figure), the radius appears in the denominator inside the square root. So the radius must increase by a factor of 4, to decrease the orbital velocity by a factor of 2. The circumference of the orbit has also increased by this factor of 4, and so with half the orbital velocity, the period must be 8 times longer. That can also be seen directly from (Figure).

### Example

#### Determining the Mass of Earth

Determine the mass of Earth from the orbit of the Moon.

#### Strategy

, and substitute for the period and radius of the orbit. The radius and period of the Moon’s orbit was measured with reasonable accuracy thousands of years ago. From the astronomical data in Appendix D, the period of the Moon is 27.3 days

, and the average distance between the centers of Earth and the Moon is 384,000 km.

#### Significance

Compare this to the value of

that we obtained in (Figure), using the value of g at the surface of Earth. Although these values are very close (

0.8%), both calculations use average values. The value of g varies from the equator to the poles by approximately 0.5%. But the Moon has an elliptical orbit in which the value of r varies just over 10%. (The apparent size of the full Moon actually varies by about this amount, but it is difficult to notice through casual observation as the time from one extreme to the other is many months.)

There is another consideration to this last calculation of

. We derived (Figure) assuming that the satellite orbits around the center of the astronomical body at the same radius used in the expression for the gravitational force between them. What assumption is made to justify this? Earth is about 81 times more massive than the Moon. Does the Moon orbit about the exact center of Earth?

The assumption is that orbiting object is much less massive than the body it is orbiting. This is not really justified in the case of the Moon and Earth. Both Earth and the Moon orbit about their common center of mass. We tackle this issue in the next example.

### Example

#### Galactic Speed and Period

Let’s revisit (Figure). Assume that the Milky Way and Andromeda galaxies are in a circular orbit about each other. What would be the velocity of each and how long would their orbital period be? Assume the mass of each is 800 billion solar masses and their centers are separated by 2.5 million light years.

#### Strategy

We cannot use (Figure) and (Figure) directly because they were derived assuming that the object of mass m orbited about the center of a much larger planet of mass M. We determined the gravitational force in (Figure) using Newton’s law of universal gravitation. We can use Newton’s second law, applied to the centripetal acceleration of either galaxy, to determine their tangential speed. From that result we can determine the period of the orbit.

#### Solution

In (Figure), we found the force between the galaxies to be

and that the acceleration of each galaxy is

Since the galaxies are in a circular orbit, they have centripetal acceleration. If we ignore the effect of other galaxies, then, as we learned in Linear Momentum and Collisions and Fixed-Axis Rotation, the centers of mass of the two galaxies remain fixed. Hence, the galaxies must orbit about this common center of mass. For equal masses, the center of mass is exactly half way between them. So the radius of the orbit,

, is not the same as the distance between the galaxies, but one-half that value, or 1.25 million light-years. These two different values are shown in (Figure).

Figure 13.14 The distance between two galaxies, which determines the gravitational force between them, is r, and is different from

, which is the radius of orbit for each. For equal masses,

. (credit: modification of work by Marc Van Norden)

Using the expression for centripetal acceleration, we have

Solving for the orbit velocity, we have

. Finally, we can determine the period of the orbit directly from

, to find that the period is

#### Significance

The orbital speed of 47 km/s might seem high at first. But this speed is comparable to the escape speed from the Sun, which we calculated in an earlier example. To give even more perspective, this period is nearly four times longer than the time that the Universe has been in existence.

In fact, the present relative motion of these two galaxies is such that they are expected to collide in about 4 billion years. Although the density of stars in each galaxy makes a direct collision of any two stars unlikely, such a collision will have a dramatic effect on the shape of the galaxies. Examples of such collisions are well known in astronomy.

Galaxies are not single objects. How does the gravitational force of one galaxy exerted on the “closer” stars of the other galaxy compare to those farther away? What effect would this have on the shape of the galaxies themselves?

The stars on the “inside” of each galaxy will be closer to the other galaxy and hence will feel a greater gravitational force than those on the outside. Consequently, they will have a greater acceleration. Even without this force difference, the inside stars would be orbiting at a smaller radius, and, hence, there would develop an elongation or stretching of each galaxy. The force difference only increases this effect.

See the Sloan Digital Sky Survey page for more information on colliding galaxies.

### Energy in Circular Orbits

In Gravitational Potential Energy and Total Energy, we argued that objects are gravitationally bound if their total energy is negative. The argument was based on the simple case where the velocity was directly away or toward the planet. We now examine the total energy for a circular orbit and show that indeed, the total energy is negative. As we did earlier, we start with Newton’s second law applied to a circular orbit,

In the last step, we multiplied by r on each side. The right side is just twice the kinetic energy, so we have

The total energy is the sum of the kinetic and potential energies, so our final result is

We can see that the total energy is negative, with the same magnitude as the kinetic energy. For circular orbits, the magnitude of the kinetic energy is exactly one-half the magnitude of the potential energy. Remarkably, this result applies to any two masses in circular orbits about their common center of mass, at a distance r from each other. The proof of this is left as an exercise. We will see in the next section that a very similar expression applies in the case of elliptical orbits.

### Example

#### Energy Required to Orbit

In (Figure), we calculated the energy required to simply lift the 9000-kg Soyuz vehicle from Earth’s surface to the height of the ISS, 400 km above the surface. In other words, we found its change in potential energy. We now ask, what total energy change in the Soyuz vehicle is required to take it from Earth’s surface and put it in orbit with the ISS for a rendezvous ((Figure))? How much of that total energy is kinetic energy?

Figure 13.15 The Soyuz in a rendezvous with the ISS. Note that this diagram is not to scale the Soyuz is very small compared to the ISS and its orbit is much closer to Earth. (credit: modification of works by NASA)

#### Strategy

The energy required is the difference in the Soyuz’s total energy in orbit and that at Earth’s surface. We can use (Figure) to find the total energy of the Soyuz at the ISS orbit. But the total energy at the surface is simply the potential energy, since it starts from rest. [Note that we do not use (Figure) at the surface, since we are not in orbit at the surface.] The kinetic energy can then be found from the difference in the total energy change and the change in potential energy found in (Figure). Alternatively, we can use (Figure) to find

and calculate the kinetic energy directly from that. The total energy required is then the kinetic energy plus the change in potential energy found in (Figure).

#### Solution

From (Figure), the total energy of the Soyuz in the same orbit as the ISS is

The total energy at Earth’s surface is

. To get the kinetic energy, we subtract the change in potential energy from (Figure),

. As stated earlier, the kinetic energy of a circular orbit is always one-half the magnitude of the potential energy, and the same as the magnitude of the total energy. Our result confirms this.

The second approach is to use (Figure) to find the orbital speed of the Soyuz, which we did for the ISS in (Figure).

So the kinetic energy of the Soyuz in orbit is

the same as in the previous method. The total energy is just

#### Significance

The kinetic energy of the Soyuz is nearly eight times the change in its potential energy, or 90% of the total energy needed for the rendezvous with the ISS. And it is important to remember that this energy represents only the energy that must be given to the Soyuz. With our present rocket technology, the mass of the propulsion system (the rocket fuel, its container and combustion system) far exceeds that of the payload, and a tremendous amount of kinetic energy must be given to that mass. So the actual cost in energy is many times that of the change in energy of the payload itself.

### Summary

• Orbital velocities are determined by the mass of the body being orbited and the distance from the center of that body, and not by the mass of a much smaller orbiting object.
• The period of the orbit is likewise independent of the orbiting object’s mass.
• Bodies of comparable masses orbit about their common center of mass and their velocities and periods should be determined from Newton’s second law and law of gravitation.

### Conceptual Questions

One student argues that a satellite in orbit is in free fall because the satellite keeps falling toward Earth. Another says a satellite in orbit is not in free fall because the acceleration due to gravity is not

. With whom do you agree with and why?

Many satellites are placed in geosynchronous orbits. What is special about these orbits? For a global communication network, how many of these satellites would be needed?

The period of the orbit must be 24 hours. But in addition, the satellite must be located in an equatorial orbit and orbiting in the same direction as Earth’s rotation. All three criteria must be met for the satellite to remain in one position relative to Earth’s surface. At least three satellites are needed, as two on opposite sides of Earth cannot communicate with each other. (This is not technically true, as a wavelength could be chosen that provides sufficient diffraction. But it would be totally impractical.)

### Problems

If a planet with 1.5 times the mass of Earth was traveling in Earth’s orbit, what would its period be?

Two planets in circular orbits around a star have speeds of v and 2v. (a) What is the ratio of the orbital radii of the planets? (b) What is the ratio of their periods?

Using the average distance of Earth from the Sun, and the orbital period of Earth, (a) ﬁnd the centripetal acceleration of Earth in its motion about the Sun. (b) Compare this value to that of the centripetal acceleration at the equator due to Earth’s rotation.

Calculate the mass of the Sun based on data for Earth’s orbit and compare the value obtained with the Sun’s actual mass.

Find the mass of Jupiter based on the fact that Io, its innermost moon, has an average orbital radius of 421,700 km and a period of 1.77 days.

Astronomical observations of our Milky Way galaxy indicate that it has a mass of about

solar masses. A star orbiting on the galaxy’s periphery is about

light-years from its center. (a) What should the orbital period of that star be? (b) If its period is

years instead, what is the mass of the galaxy? Such calculations are used to imply the existence of other matter, such as a very massive black hole at the center of the Milky Way.

(a) In order to keep a small satellite from drifting into a nearby asteroid, it is placed in orbit with a period of 3.02 hours and radius of 2.0 km. What is the mass of the asteroid? (b) Does this mass seem reasonable for the size of the orbit?

b. The satellite must be outside the radius of the asteroid, so it can’t be larger than this. If it were this size, then its density would be about

. This is just above that of water, so this seems quite reasonable.

The Moon and Earth rotate about their common center of mass, which is located about 4700 km from the center of Earth. (This is 1690 km below the surface.) (a) Calculate the acceleration due to the Moon’s gravity at that point. (b) Calculate the centripetal acceleration of the center of Earth as it rotates about that point once each lunar month (about 27.3 d) and compare it with the acceleration found in part (a). Comment on whether or not they are equal and why they should or should not be.

The Sun orbits the Milky Way galaxy once each

, with a roughly circular orbit averaging a radius of

light-years. (A light-year is the distance traveled by light in 1 year.) Calculate the centripetal acceleration of the Sun in its galactic orbit. Does your result support the contention that a nearly inertial frame of reference can be located at the Sun? (b) Calculate the average speed of the Sun in its galactic orbit. Does the answer surprise you?

Yes, the centripetal acceleration is so small it supports the contention that a nearly inertial frame of reference can be located at the Sun. b.

A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth’s rotation). Calculate the radius of such an orbit based on the data for Earth in Astronomical Data.

### Glossary

orbital period

time required for a satellite to complete one orbit orbital speed speed of a satellite in a circular orbit it can be also be used for the instantaneous speed for noncircular orbits in which the speed is not constant

## The Cavendish Experiment: Then and Now

As previously noted, the universal gravitational constant is determined experimentally. This definition was first done accurately by Henry Cavendish (1731–1810), an English scientist, in 1798, more than 100 years after Newton published his universal law of gravitation. The measurement of is very basic and important because it determines the strength of one of the four forces in nature. Cavendish’s experiment was very difficult because he measured the tiny gravitational attraction between two ordinary-sized masses (tens of kilograms at most), using apparatus like that in Figure 9. Remarkably, his value for differs by less than 1% from the best modern value.

One important consequence of knowing was that an accurate value for Earth’s mass could finally be obtained. This was done by measuring the acceleration due to gravity as accurately as possible and then calculating the mass of Earth from the relationship Newton’s universal law of gravitation gives

where is the mass of the object, is the mass of Earth, and is the distance to the center of Earth (the distance between the centers of mass of the object and Earth). See Figure 2. The mass of the object cancels, leaving an equation for

Rearranging to solve for yields

So can be calculated because all quantities on the right, including the radius of Earth are known from direct measurements. We shall see in Chapter 6.6 Satellites and Kepler’s Laws: An Argument for Simplicity that knowing also allows for the determination of astronomical masses. Interestingly, of all the fundamental constants in physics, is by far the least well determined.

The Cavendish experiment is also used to explore other aspects of gravity. One of the most interesting questions is whether the gravitational force depends on substance as well as mass—for example, whether one kilogram of lead exerts the same gravitational pull as one kilogram of water. A Hungarian scientist named Roland von Eötvös pioneered this inquiry early in the 20th century. He found, with an accuracy of five parts per billion, that the gravitational force does not depend on the substance. Such experiments continue today, and have improved upon Eötvös’ measurements. Cavendish-type experiments such as those of Eric Adelberger and others at the University of Washington, have also put severe limits on the possibility of a fifth force and have verified a major prediction of general relativity—that gravitational energy contributes to rest mass. Ongoing measurements there use a torsion balance and a parallel plate (not spheres, as Cavendish used) to examine how Newton’s law of gravitation works over sub-millimeter distances. On this small-scale, do gravitational effects depart from the inverse square law? So far, no deviation has been observed.

. Figure 9. Cavendish used an apparatus like this to measure the gravitational attraction between the two suspended spheres (m) and the two on the stand (M) by observing the amount of torsion (twisting) created in the fiber. Distance between the masses can be varied to check the dependence of the force on distance. Modern experiments of this type continue to explore gravity.

## Why the torque exerted by the moon upon the Earth makes the Moon to increase his orbit? - Astronomy

This simple picture is unfortunately complicated by the difficulty of defining a suitable equator and equinox. One problem is that the Sun's apparent motion is not completely regular, due to the ellipticity of the Earth's orbit and its continuous disturbance by the Moon and planets. This is dealt with by separating the motion into (i) a smooth and steady mean Sun and (ii) a set of periodic corrections and perturbations only the former is involved in establishing reference frames and timescales. A second, far larger problem, is that the celestial equator and the ecliptic are both moving with respect to the stars. These motions arise because of the gravitational interactions between the Earth and the other solar-system bodies.

By far the largest effect is the so-called precession of the equinoxes'', where the Earth's rotation axis sweeps out a cone centred on the ecliptic pole, completing one revolution in about 26,000 years. The cause of the motion is the torque exerted on the distorted and spinning Earth by the Sun and the Moon. Consider the effect of the Sun alone, at or near the northern summer solstice. The Sun sees' the top (north pole) of the Earth tilted towards it (by about , the obliquity of the ecliptic ), and sees the nearer part of the Earth's equatorial bulge below centre and the further part above centre. Although the Earth is in free fall, the gravitational force on the nearer part of the equatorial bulge is greater than that on the further part, and so there is a net torque acting as if to eliminate the tilt. Six months later the same thing is happening in reverse, except that the torque is still trying to eliminate the tilt. In between (at the equinoxes) the torque shrinks to zero. A torque acting on a spinning body is gyroscopically translated into a precessional motion of the spin axis at right-angles to the torque, and this happens to the Earth. The motion varies during the year, going through two maxima, but always acts in the same direction. The Moon produces the same effect, adding a contribution to the precession which peaks twice per month. The Moon's proximity to the Earth more than compensates for its smaller mass and gravitational attraction, so that it in fact contributes most of the precessional effect.

The complex interactions between the three bodies produce a precessional motion that is wobbly rather than completely smooth. However, the main 26,000-year component is on such a grand scale that it dwarfs the remaining terms, the biggest of which has an amplitude of only and a period of about 18.6 years. This difference of scale makes it convenient to treat these two components of the motion separately. The main 26,000-year effect is called luni-solar precession the smaller, faster, periodic terms are called the nutation .

Note that precession and nutation are simply different frequency components of the same physical effect. It is a common misconception that precession is caused by the Sun and nutation is caused by the Moon. In fact the Moon is responsible for two-thirds of the precession, and, while it is true that much of the complex detail of the nutation is a reflection of the intricacies of the lunar orbit, there are nonetheless important solar terms in the nutation.

In addition to and quite separate from the precession-nutation effect, the orbit of the Earth-Moon system is not fixed in orientation, a result of the attractions of the planets. This slow (about per year) secular rotation of the ecliptic about a slowly-moving diameter is called, confusingly, planetary precession and, along with the luni-solar precession is included in the general precession . The equator and ecliptic as affected by general precession are what define the various `mean'' reference frames.

The models for precession and nutation come from a combination of observation and theory, and are subject to continuous refinement. Nutation models in particular have reached a high degree of sophistication, taking into account such things as the non-rigidity of the Earth and the effects of the planets SLALIB's nutation model (IAU 1980) involves 106 terms in each of (longitude) and (obliquity), some as small as .

Copyright © 2003 Council for the Central Laboratory of the Research Councils

## MINUTES

OGDEN ASTRONOMICAL SOCIETY

April 10, 1997

The regular meeting opened at 7:30 p.m., President Steve Peterson directing.

The Barnes and Noble star party washed out twice. No plans to reschedule.

Because of poor weather but high interest, another Antelope Island event is scheduled for April 26. Follow the signs to White Rock Bay.

John needs help for this quarter's final Wed. Night star party at W.S.U. Arrive by 8:30 p.m.

Elgie Mills and Jim Seargeant showed the latest results of their CCD and film photography. Some incredible images of the comet were shown.

Planetarium assistant Jarett Bartholomew ran "Voyage to the Planets".

Meeting adjourned at 8:40 p.m. August 29, 1997"

## 13 Chapter Review

Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the ultimate determinant of the truth in science, and why was this action at a distance ultimately accepted?

In the law of universal gravitation, Newton assumed that the force was proportional to the product of the two masses (

m1m2 ). While all scientific conjectures must be experimentally verified, can you provide arguments as to why this must be? (You may wish to consider simple examples in which any other form would lead to contradictory results.)

#### 13.2 Gravitation Near Earth’s Surface

Must engineers take Earth’s rotation into account when constructing very tall buildings at any location other than the equator or very near the poles?

#### 13.3 Gravitational Potential Energy and Total Energy

It was stated that a satellite with negative total energy is in a bound orbit, whereas one with zero or positive total energy is in an unbounded orbit. Why is this true? What choice for gravitational potential energy was made such that this is true?

It was shown that the energy required to lift a satellite into a low Earth orbit (the change in potential energy) is only a small fraction of the kinetic energy needed to keep it in orbit. Is this true for larger orbits? Is there a trend to the ratio of kinetic energy to change in potential energy as the size of the orbit increases?

#### 13.4 Satellite Orbits and Energy

One student argues that a satellite in orbit is in free fall because the satellite keeps falling toward Earth. Another says a satellite in orbit is not in free fall because the acceleration due to gravity is not 9.80 m/s 2 9.80m/s2 . With whom do you agree with and why?

Many satellites are placed in geosynchronous orbits. What is special about these orbits? For a global communication network, how many of these satellites would be needed?

#### 13.5 Kepler’s Laws of Planetary Motion

Are Kepler’s laws purely descriptive, or do they contain causal information?

In the diagram below for a satellite in an elliptical orbit about a much larger mass, indicate where its speed is the greatest and where it is the least. What conservation law dictates this behavior? Indicate the directions of the force, acceleration, and velocity at these points. Draw vectors for these same three quantities at the two points where the y-axis intersects (along the semi-minor axis) and from this determine whether the speed is increasing decreasing, or at a max/min.

#### 13.6 Tidal Forces

As an object falls into a black hole, tidal forces increase. Will these tidal forces always tear the object apart as it approaches the Schwarzschild radius? How does the mass of the black hole and size of the object affect your answer?

#### 13.7 Einstein’s Theory of Gravity

The principle of equivalence states that all experiments done in a lab in a uniform gravitational field cannot be distinguished from those done in a lab that is not in a gravitational field but is uniformly accelerating. For the latter case, consider what happens to a laser beam at some height shot perfectly horizontally to the floor, across the accelerating lab. (View this from a nonaccelerating frame outside the lab.) Relative to the height of the laser, where will the laser beam hit the far wall? What does this say about the effect of a gravitational field on light? Does the fact that light has no mass make any difference to the argument?

As a person approaches the Schwarzschild radius of a black hole, outside observers see all the processes of that person (their clocks, their heart rate, etc.) slowing down, and coming to a halt as they reach the Schwarzschild radius. (The person falling into the black hole sees their own processes unaffected.) But the speed of light is the same everywhere for all observers. What does this say about space as you approach the black hole?

### Problems

#### 13.1 Newton’s Law of Universal Gravitation

Evaluate the magnitude of gravitational force between two 5-kg spherical steel balls separated by a center-to-center distance of 15 cm.

Estimate the gravitational force between two sumo wrestlers, with masses 220 kg and 240 kg, when they are embraced and their centers are 1.2 m apart.

Astrology makes much of the position of the planets at the moment of one’s birth. The only known force a planet exerts on Earth is gravitational. (a) Calculate the gravitational force exerted on a 4.20-kg baby by a 100-kg father 0.200 m away at birth (he is assisting, so he is close to the child). (b) Calculate the force on the baby due to Jupiter if it is at its closest distance to Earth, some 6.29 × 10 11 m 6.29×1011m away. How does the force of Jupiter on the baby compare to the force of the father on the baby? Other objects in the room and the hospital building also exert similar gravitational forces. (Of course, there could be an unknown force acting, but scientists first need to be convinced that there is even an effect, much less that an unknown force causes it.)

A mountain 10.0 km from a person exerts a gravitational force on him equal to 2.00% of his weight. (a) Calculate the mass of the mountain. (b) Compare the mountain’s mass with that of Earth. (c) What is unreasonable about these results? (d) Which premises are unreasonable or inconsistent? (Note that accurate gravitational measurements can easily detect the effect of nearby mountains and variations in local geology.)

The International Space Station has a mass of approximately 370,000 kg. (a) What is the force on a 150-kg suited astronaut if she is 20 m from the center of mass of the station? (b) How accurate do you think your answer would be?

Asteroid Toutatis passed near Earth in 2006 at four times the distance to our Moon. This was the closest approach we will have until 2060. If it has mass of 5.0 × 10 13 kg 5.0×1013kg , what force did it exert on Earth at its closest approach?

(a) What was the acceleration of Earth caused by asteroid Toutatis (see previous problem) at its closest approach? (b) What was the acceleration of Toutatis at this point?

#### 13.2 Gravitation Near Earth’s Surface

(a) Calculate Earth’s mass given the acceleration due to gravity at the North Pole is measured to be 9.832 m/s 2 9.832m/s2 and the radius of the Earth at the pole is 6356 km. (b) Compare this with the NASA’s Earth Fact Sheet value of 5.9726 × 10 24 kg 5.9726×1024kg .

(a) What is the acceleration due to gravity on the surface of the Moon? (b) On the surface of Mars? The mass of Mars is 6.418 × 10 23 kg 6.418×1023kg and its radius is 3.38 × 10 6 m 3.38×106m .

(a) Calculate the acceleration due to gravity on the surface of the Sun. (b) By what factor would your weight increase if you could stand on the Sun? (Never mind that you cannot.)

The mass of a particle is 15 kg. (a) What is its weight on Earth? (b) What is its weight on the Moon? (c) What is its mass on the Moon? (d) What is its weight in outer space far from any celestial body? (e) What is its mass at this point?

On a planet whose radius is 1.2 × 10 7 m 1.2×107m , the acceleration due to gravity is 18 m/s 2 18m/s2 . What is the mass of the planet?

The mean diameter of the planet Saturn is 1.2 × 10 8 m 1.2×108m , and its mean mass density is 0.69 g/cm 3 0.69g/cm3 . Find the acceleration due to gravity at Saturn’s surface.

The mean diameter of the planet Mercury is 4.88 × 10 6 m 4.88×106m , and the acceleration due to gravity at its surface is 3.78 m/s 2 3.78m/s2 . Estimate the mass of this planet.

The acceleration due to gravity on the surface of a planet is three times as large as it is on the surface of Earth. The mass density of the planet is known to be twice that of Earth. What is the radius of this planet in terms of Earth’s radius?

A body on the surface of a planet with the same radius as Earth’s weighs 10 times more than it does on Earth. What is the mass of this planet in terms of Earth’s mass?

#### 13.3 Gravitational Potential Energy and Total Energy

Find the escape speed of a projectile from the surface of Mars.

Find the escape speed of a projectile from the surface of Jupiter.

What is the escape speed of a satellite located at the Moon’s orbit about Earth? Assume the Moon is not nearby.

(a) Evaluate the gravitational potential energy between two 5.00-kg spherical steel balls separated by a center-to-center distance of 15.0 cm. (b) Assuming that they are both initially at rest relative to each other in deep space, use conservation of energy to find how fast will they be traveling upon impact. Each sphere has a radius of 5.10 cm.

An average-sized asteroid located 5.0 × 10 7 km 5.0×107km from Earth with mass 2.0 × 10 13 kg 2.0×1013kg is detected headed directly toward Earth with speed of 2.0 km/s. What will its speed be just before it hits our atmosphere? (You may ignore the size of the asteroid.)

(a) What will be the kinetic energy of the asteroid in the previous problem just before it hits Earth? b) Compare this energy to the output of the largest fission bomb, 2100 TJ. What impact would this have on Earth?

(a) What is the change in energy of a 1000-kg payload taken from rest at the surface of Earth and placed at rest on the surface of the Moon? (b) What would be the answer if the payload were taken from the Moon’s surface to Earth? Is this a reasonable calculation of the energy needed to move a payload back and forth?

#### 13.4 Satellite Orbits and Energy

If a planet with 1.5 times the mass of Earth was traveling in Earth’s orbit, what would its period be?

Two planets in circular orbits around a star have speeds of v and 2v. (a) What is the ratio of the orbital radii of the planets? (b) What is the ratio of their periods?

Using the average distance of Earth from the Sun, and the orbital period of Earth, (a) ﬁnd the centripetal acceleration of Earth in its motion about the Sun. (b) Compare this value to that of the centripetal acceleration at the equator due to Earth’s rotation.

Calculate the mass of the Sun based on data for Earth’s orbit and compare the value obtained with the Sun’s actual mass.

Find the mass of Jupiter based on the fact that Io, its innermost moon, has an average orbital radius of 421,700 km and a period of 1.77 days.

Astronomical observations of our Milky Way galaxy indicate that it has a mass of about 8.0 × 10 11 8.0×1011 solar masses. A star orbiting on the galaxy’s periphery is about 6.0 × 10 4 6.0×104 light-years from its center. (a) What should the orbital period of that star be? (b) If its period is 6.0 × 10 7 6.0×107 years instead, what is the mass of the galaxy? Such calculations are used to imply the existence of other matter, such as a very massive black hole at the center of the Milky Way.

(a) In order to keep a small satellite from drifting into a nearby asteroid, it is placed in orbit with a period of 3.02 hours and radius of 2.0 km. What is the mass of the asteroid? (b) Does this mass seem reasonable for the size of the orbit?

The Moon and Earth rotate about their common center of mass, which is located about 4700 km from the center of Earth. (This is 1690 km below the surface.) (a) Calculate the acceleration due to the Moon’s gravity at that point. (b) Calculate the centripetal acceleration of the center of Earth as it rotates about that point once each lunar month (about 27.3 d) and compare it with the acceleration found in part (a). Comment on whether or not they are equal and why they should or should not be.

The Sun orbits the Milky Way galaxy once each 2.60 × 10 8 years 2.60×108years , with a roughly circular orbit averaging a radius of 3.00 × 10 4 3.00×104 light-years. (A light-year is the distance traveled by light in 1 year.) Calculate the centripetal acceleration of the Sun in its galactic orbit. Does your result support the contention that a nearly inertial frame of reference can be located at the Sun? (b) Calculate the average speed of the Sun in its galactic orbit. Does the answer surprise you?

A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth’s rotation). Calculate the radius of such an orbit based on the data for Earth inAppendix D.

#### 13.5 Kepler’s Laws of Planetary Motion

Calculate the mass of the Sun based on data for average Earth’s orbit and compare the value obtained with the Sun’s commonly listed value of 1.989 × 10 30 kg 1.989×1030kg .

Io orbits Jupiter with an average radius of 421,700 km and a period of 1.769 days. Based upon these data, what is the mass of Jupiter?

The “mean” orbital radius listed for astronomical objects orbiting the Sun is typically not an integrated average but is calculated such that it gives the correct period when applied to the equation for circular orbits. Given that, what is the mean orbital radius in terms of aphelion and perihelion?

The perihelion of Halley’s comet is 0.586 AU and the aphelion is 17.8 AU. Given that its speed at perihelion is 55 km/s, what is the speed at aphelion ( 1 AU = 1.496 × 10 11 m 1AU=1.496×1011m )? (Hint: You may use either conservation of energy or angular momentum, but the latter is much easier.)

The perihelion of the comet Lagerkvist is 2.61 AU and it has a period of 7.36 years. Show that the aphelion for this comet is 4.95 AU.

What is the ratio of the speed at perihelion to that at aphelion for the comet Lagerkvist in the previous problem?

Eros has an elliptical orbit about the Sun, with a perihelion distance of 1.13 AU and aphelion distance of 1.78 AU. What is the period of its orbit?

#### 13.6 Tidal Forces

(a) What is the difference between the forces on a 1.0-kg mass on the near side of Io and far side due to Jupiter? Io has a mean radius of 1821 km and a mean orbital radius about Jupiter of 421,700 km. (b) Compare this difference to that calculated for the difference for Earth due to the Moon calculated in Example 13.14. Tidal forces are the cause of Io’s volcanic activity.

If the Sun were to collapse into a black hole, the point of no return for an investigator would be approximately 3 km from the center singularity. Would the investigator be able to survive visiting even 300 km from the center? Answer this by finding the difference in the gravitational attraction the black holes exerts on a 1.0-kg mass at the head and at the feet of the investigator.

Consider Figure 13.23 in Tidal Forces. This diagram represents the tidal forces for spring tides. Sketch a similar diagram for neap tides. (Hint: For simplicity, imagine that the Sun and the Moon contribute equally. Your diagram would be the vector sum of two force fields (as in Figure 13.23), reduced by a factor of two, and superimposed at right angles.)

#### 13.7 Einstein’s Theory of Gravity

What is the Schwarzschild radius for the black hole at the center of our galaxy if it has the mass of 4 million solar masses?

What would be the Schwarzschild radius, in light years, if our Milky Way galaxy of 100 billion stars collapsed into a black hole? Compare this to our distance from the center, about 13,000 light years.

A neutron star is a cold, collapsed star with nuclear density. A particular neutron star has a mass twice that of our Sun with a radius of 12.0 km. (a) What would be the weight of a 100-kg astronaut on standing on its surface? (b) What does this tell us about landing on a neutron star?

(a) How far from the center of Earth would the net gravitational force of Earth and the Moon on an object be zero? (b) Setting the magnitudes of the forces equal should result in two answers from the quadratic. Do you understand why there are two positions, but only one where the net force is zero?

How far from the center of the Sun would the net gravitational force of Earth and the Sun on a spaceship be zero?

Calculate the values of g at Earth’s surface for the following changes in Earth’s properties: (a) its mass is doubled and its radius is halved (b) its mass density is doubled and its radius is unchanged (c) its mass density is halved and its mass is unchanged.

Suppose you can communicate with the inhabitants of a planet in another solar system. They tell you that on their planet, whose diameter and mass are 5.0 × 10 3 km 5.0×103km and 3.6 × 10 23 kg 3.6×1023kg , respectively, the record for the high jump is 2.0 m. Given that this record is close to 2.4 m on Earth, what would you conclude about your extraterrestrial friends’ jumping ability?

(a) Suppose that your measured weight at the equator is one-half your measured weight at the pole on a planet whose mass and diameter are equal to those of Earth. What is the rotational period of the planet? (b) Would you need to take the shape of this planet into account?

A body of mass 100 kg is weighed at the North Pole and at the equator with a spring scale. What is the scale reading at these two points? Assume that g = 9.83 m/s 2 g=9.83m/s2 at the pole.

Find the speed needed to escape from the solar system starting from the surface of Earth. Assume there are no other bodies involved and do not account for the fact that Earth is moving in its orbit. [Hint: Equation 13.6 does not apply. UseEquation 13.5 and include the potential energy of both Earth and the Sun.

Consider the previous problem and include the fact that Earth has an orbital speed about the Sun of 29.8 km/s. (a) What speed relative to Earth would be needed and in what direction should you leave Earth? (b) What will be the shape of the trajectory?

A comet is observed 1.50 AU from the Sun with a speed of 24.3 km/s. Is this comet in a bound or unbound orbit?

An asteroid has speed 15.5 km/s when it is located 2.00 AU from the sun. At its closest approach, it is 0.400 AU from the Sun. What is its speed at that point?

Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a satellite in an orbit 900 km above Earth’s surface. (b) Suppose a loose rivet is in an orbit of the same radius that intersects the satellite’s orbit at an angle of 90 ° 90° . What is the velocity of the rivet relative to the satellite just before striking it? (c) If its mass is 0.500 g, and it comes to rest inside the satellite, how much energy in joules is generated by the collision? (Assume the satellite’s velocity does not change appreciably, because its mass is much greater than the rivet’s.)

A satellite of mass 1000 kg is in circular orbit about Earth. The radius of the orbit of the satellite is equal to two times the radius of Earth. (a) How far away is the satellite? (b) Find the kinetic, potential, and total energies of the satellite.

After Ceres was promoted to a dwarf planet, we now recognize the largest known asteroid to be Vesta, with a mass of 2.67 × 10 20 kg 2.67×1020kg and a diameter ranging from 578 km to 458 km. Assuming that Vesta is spherical with radius 520 km, find the approximate escape velocity from its surface.

(a) Using the data in the previous problem for the asteroid Vesta, what would be the orbital period for a space probe in a circular orbit of 10.0 km from its surface? (b) Why is this calculation marginally useful at best?

What is the orbital velocity of our solar system about the center of the Milky Way? Assume that the mass within a sphere of radius equal to our distance away from the center is about a 100 billion solar masses. Our distance from the center is 27,000 light years.

(a) Using the information in the previous problem, what velocity do you need to escape the Milky Way galaxy from our present position? (b) Would you need to accelerate a spaceship to this speed relative to Earth?

Circular orbits in Equation 13.10 for conic sections must have eccentricity zero. From this, and using Newton’s second law applied to centripetal acceleration, show that the value of α α in Equation 13.10 is given by α = L 2 G M m 2 α=L2GMm2 where L is the angular momentum of the orbiting body. The value of α α is constant and given by this expression regardless of the type of orbit.

Show that for eccentricity equal to one in Equation 13.10 for conic sections, the path is a parabola. Do this by substituting Cartesian coordinates, x and y, for the polar coordinates, r and θ θ , and showing that it has the general form for a parabola, x = a y 2 + b y + c x=ay2+by+c .

Using the technique shown in Satellite Orbits and Energy, show that two masses m 1 m1 and m 2 m2 in circular orbits about their common center of mass, will have total energy E = K + E = K 1 + K 2 − G m 1 m 2 r = − G m 1 m 2 2 r E=K+E=K1+K2−Gm1m2r=−Gm1m22r . We have shown the kinetic energy of both masses explicitly. (Hint: The masses orbit at radii r 1 r1 and r 2 r2 , respectively, where r = r 1 + r 2 r=r1+r2 . Be sure not to confuse the radius needed for centripetal acceleration with that for the gravitational force.)

Given the perihelion distance, p, and aphelion distance, q, for an elliptical orbit, show that the velocity at perihelion, v p vp , is given by v p = 2 G M Sun ( q + p ) q p ‾‾‾‾‾‾‾‾√ vp=2GMSun(q+p)qp . (Hint: Use conservation of angular momentum to relate v p vp and v q vq , and then substitute into the conservation of energy equation.)

Comet P/1999 R1 has a perihelion of 0.0570 AU and aphelion of 4.99 AU. Using the results of the previous problem, find its speed at aphelion. (Hint: The expression is for the perihelion. Use symmetry to rewrite the expression for aphelion.)

### Challenge Problems

A tunnel is dug through the center of a perfectly spherical and airless planet of radius R. Using the expression for g derived in Gravitation Near Earth’s Surface for a uniform density, show that a particle of mass m dropped in the tunnel will execute simple harmonic motion. Deduce the period of oscillation of m and show that it has the same period as an orbit at the surface.

Following the technique used in Gravitation Near Earth’s Surface, find the value of g as a function of the radius r from the center of a spherical shell planet of constant density ρ ρ with inner and outer radii R in Rin and R out Rout . Find g for both R in < r < R out Rin<r<Rout and for r < R in r<Rin . Assuming the inside of the shell is kept airless, describe travel inside the spherical shell planet.

Show that the areal velocity for a circular orbit of radius r about a mass M is Δ A Δ t = 1 2 G M r ‾‾‾‾‾√ ΔAΔt=12GMr . Does your expression give the correct value for Earth’s areal velocity about the Sun?

Show that the period of orbit for two masses, m 1 m1 and m 2 m2 , in circular orbits of radii r 1 r1 and r 2 r2 , respectively, about their common center-of-mass, is given by T = 2 π r 3 G ( m 1 + m 2 ) ‾‾‾‾‾‾‾‾√ where r = r 1 + r 2 T=2πr3G(m1+m2)wherer=r1+r2 . (Hint: The masses orbit at radii r 1 r1 and r 2 r2 , respectively where r = r 1 + r 2 r=r1+r2 . Use the expression for the center-of-mass to relate the two radii and note that the two masses must have equal but opposite momenta. Start with the relationship of the period to the circumference and speed of orbit for one of the masses. Use the result of the previous problem using momenta in the expressions for the kinetic energy.)

Show that for small changes in height h, such that h < < R E h<<RE , Equation 13.4 reduces to the expression Δ U = m g h ΔU=mgh .

Using Figure 13.9, carefully sketch a free body diagram for the case of a simple pendulum hanging at latitude lambda, labeling all forces acting on the point mass, m. Set up the equations of motion for equilibrium, setting one coordinate in the direction of the centripetal acceleration (toward P in the diagram), the other perpendicular to that. Show that the deflection angle ε ε , defined as the angle between the pendulum string and the radial direction toward the center of Earth, is given by the expression below. What is the deflection angle at latitude 45 degrees? Assume that Earth is a perfect sphere. tan ( λ + ε ) = g ( g − ω 2 R E ) tan λ tan(λ+ε)=g(g−ω2RE)tanλ , where ω ω is the angular velocity of Earth.

(a) Show that tidal force on a small object of mass m, defined as the difference in the gravitational force that would be exerted on m at a distance at the near and the far side of the object, due to the gravitation at a distance R from M, is given by F tidal = 2 G M m R 3 Δ r Ftidal=2GMmR3Δr where Δ r Δr is the distance between the near and far side and Δ r < < R Δr<<R . (b) Assume you are falling feet first into the black hole at the center of our galaxy. It has mass of 4 million solar masses. What would be the difference between the force at your head and your feet at the Schwarzschild radius (event horizon)? Assume your feet and head each have mass 5.0 kg and are 2.0 m apart. Would you survive passing through the event horizon?

Find the Hohmann transfer velocities, Δ v EllipseEarth ΔvEllipseEarth and Δ v EllipseMars ΔvEllipseMars , needed for a trip to Mars. Use Equation 13.7 to find the circular orbital velocities for Earth and Mars. Using Equation 13.4 and the total energy of the ellipse (with semi-major axisa), given by E = − G m M s 2 a E=−GmMs2a , find the velocities at Earth (perihelion) and at Mars (aphelion) required to be on the transfer ellipse. The difference, Δ v Δv , at each point is the velocity boost or transfer velocity needed.

## Watch the video: Neil deGrasse Tyson Explains the Tides (July 2022).

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