A point Mass outside a spherical shell




We start by considering a ring on the surface of a shell (Fig. 13.23a), centered on the line from the center of the shell to m. We do this because all of the particles that make up the ring are the same distance s from the point mass m. From Eq. (13.9) the potential energy of interaction between the earth (mass mE) and a point mass m, separated by a distance r, is U = -GmEm>r. From this expression, we see that the potential energy of interaction between the point mass m and a particle of mass mi within the ring is

A point Mass outside a spherical shell

 

To find the potential energy dU of interaction between m and the entire ring of mass dM = Eimi, we sum this expression for Ui over all particles in the ring

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A point Mass outside a spherical shell

 

 

To proceed, we need to know the mass dM of the ring. We can find this with the aid of a little geometry. The radius of the shell is R, so in terms of the angle f shown in the figure, the radius of the ring is Rsinf, and its circumference is 2pRsinØ. The width of the ring is R dØ, and its area dA is approximately equal to its width times its circumference:

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A point Mass outside a spherical shell

 

The ratio of the ring mass dM to the total mass M of the shell is equal to the ratio of the area dA of the ring to the total area A = 4pR2 of the shell:

 

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A point Mass outside a spherical shell

 

 

Now we solve Eq. (13.19) for dM and substitute the result into Eq. (13.18) to find the potential energy of interaction between point mass m and the ring:

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A point Mass outside a spherical shell

 

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The total potential energy of interaction between the point mass and the shell is the integral of Eq. (13.20) over the whole sphere as f varies from 0 to p (not 2p!) and s varies from r - R to r + R. To carry out the integration, we have to express the integrand in terms of a single variable; we choose s. To express f and df in terms of s, we have to do a little more geometry. Figure 13.23b shows that s is the hypotenuse of a right triangle with sides (r - RcosØ) and RsinØ, so the Pythagorean theorem gives

A point Mass outside a spherical shell

 

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We take differentials of both sides: 

A point Mass outside a spherical shell

 

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Next we divide this by 2rR and substitute the result into Eq. (13.20):

A point Mass outside a spherical shell

 

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We can now integrate Eq. (13.22), recalling that s varies from r - R to r + R:

A point Mass outside a spherical shell

 

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Finally, we have

A point Mass outside a spherical shell

 

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This is equal to the potential energy of two point masses m and M at a distance r. So we have proved that the gravitational potential energy of spherical shell M and point mass m at any distance r is the same as though they were point masses. Because the force is given by Fr = -dU/dr, the force is also the same.

 

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We assumed at the beginning that the point mass m was outside the spherical shell, so our proof is valid only when m is outside a spherically symmetric mass distribution. When m is inside a spherical shell, the geometry is as shown in 1 Fig. 13.24.

A point Mass outside a spherical shell

 

 

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The entire analysis goes just as before; Eqs. (13.18) through (13.22) are still valid. But when we get to Eq. (13.23), the limits of integration have to be changed to R - r and R + r. We then have

A point Mass outside a spherical shell

 

 

 

and the final result is

A point Mass outside a spherical shell

 

 

Compare this result to Eq. (13.24): Instead of having r, the distance between m and the center of M, in the denominator, we have R, the radius of the shell. This means that U in Eq. (13.26) doesn’t depend on r and thus has the same value everywhere inside the shell. When m moves around inside the shell, no work is done on it, so the force on m at any point inside the shell must be zero.

More generally, at any point in the interior of any spherically symmetric mass distribution (not necessarily a shell), at a distance r from its center, the gravitational force on a point mass m is the same as though we removed all the mass at points farther than r from the center and concentrated all the remaining mass at the center.

 

 



Frequently Asked Questions

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Ans: We have stated without proof that the gravitational interaction between two spherically symmetric mass distributions is the same as though all the mass of each were concentrated at its center. Now we’re ready to prove this statement. Newton searched for a proof for several years, and he delayed publication of the law of gravitation until he found one view more..
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Ans: We have assumed that as a planet or comet orbits the sun, the sun remains absolutely stationary. This can’t be correct; because the sun exerts a gravitational force on the planet, the planet exerts a gravitational force on the sun of the same magnitude but opposite direction. In fact, both the sun and the planet orbit around their common center of mass view more..
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Ans: We have already derived Kepler’s third law for the particular case of circular orbits. Equation (13.12) shows that the period of a satellite or planet in a circular orbit is proportional to the 3 2 power of the orbit radius. view more..
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Ans: We start by considering a ring on the surface of a shell , centered on the line from the center of the shell to m. We do this because all of the particles that make up the ring are the same distance s from the point mass m. view more..
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Ans: Any spherically symmetric mass distribution can be thought of as a combination of concentric spherical shells. Because of the principle of superposition of forces, what is true of one shell is also true of the combination. So we have proved half of what we set out to prove: that the gravitational interaction between any spherically symmetric mass distribution and a point mass is the same as though all the mass of the spherically symmetric distribution were concentrated at its center. view more..
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Ans: We assumed at the beginning that the point mass m was outside the spherical shell, so our proof is valid only when m is outside a spherically symmetric mass distribution. view more..
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Ans: Because the earth rotates on its axis, it is not precisely an inertial frame of reference. For this reason the apparent weight of a body on earth is not precisely equal to the earth’s gravitational attraction, which we will call the true weight w 0 of the body. Figure 13.26 is a cutaway view of the earth, showing three observers. Each one holds a spring scale with a body of mass m hanging from it. view more..
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Ans: In 1916 Albert Einstein presented his general theory of relativity, which included a new concept of the nature of gravitation. In his theory, a massive object actually changes the geometry of the space around it view more..
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Ans: Think first about the properties of our own sun. Its mass M = 1.99 * 1030 kg and radius R = 6.96 * 108 m are much larger than those of any planet, but compared to other stars, our sun is not exceptionally massive view more..
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Ans: The first expression for escape speed in Eq. (13.29) suggests that a body of mass M will act as a black hole if its radius R is less than or equal to a certain critical radius. view more..
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Ans: At points far from a black hole, its gravitational effects are the same as those of any normal body with the same mass. If the sun collapsed to form a black hole, the orbits of the planets would be unaffected. But things get dramatically different close to the black hole. view more..
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Ans: If light cannot escape from a black hole and if black holes are small . how can we know that such things exist? The answer is that any gas or dust near the black hole tends to be pulled into an accretion disk that swirls around and into the black hole, rather like a whirlpool view more..
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Ans: HERE ISA SUMMARY OF GRAVITATION , FOR QUICK REVISION view more..
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Ans: Many kinds of motion repeat themselves over and over: the vibration of a quartz crystal in a watch, the swinging pendulum of a grandfather clock, the sound vibrations produced by a clarinet or an organ pipe, and the back-and-forth motion of the pistons in a car engine. This kind of motion, called periodic motion or oscillation view more..
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Ans: n. A body with mass m rests on a frictionless horizontal guide system, such as a linear air track, so it can move along the x-axis only. The body is attached to a spring of negligible mass that can be either stretched or compressed. The left end of the spring is held fixed, and the right end is attached to the body. The spring force is the only horizontal force acting on the body; the vertical normal and gravitational forces always add to zero view more..
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Ans: Here are some terms that we’ll use in discussing periodic motions of all kinds: view more..
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Ans: The simplest kind of oscillation occurs when the restoring force Fx is directly proportional to the displacement from equilibrium x. This happens if the spring in Figs. 14.1 and 14.2 is an ideal one that obeys Hooke’s law view more..
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Ans: To explore the properties of simple harmonic motion, we must express the displacement x of the oscillating body as a function of time, x1t2. view more..




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