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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 13 / Problem 57P

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Chapter 13, Problem 57P

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QUESTION:

Problem 57P

There are two equations from which a change in the gravitational potential energy U of the system of a mass m and the earth can be calculated. One is U = mgy (Eq. 7.2). The other is U = −GmEm/r (Eq. 13.9). As shown in Section 13.3, the first equation is correct only if the gravitational force is a constant over the change in height Δy. The second is always correct. Actually, the gravitational force is never exactly constant over any change in height, but if the variation is small, we can ignore it. Consider the difference in U between a mass at the earth’s surface and a distance h above it using both equations, and find the value of h for which Eq (7.2) is in error by 1%. Express this value of h as a fraction of the earth’s radius, and also obtain a numerical value for it.

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QUESTION:

Problem 57P

There are two equations from which a change in the gravitational potential energy U of the system of a mass m and the earth can be calculated. One is U = mgy (Eq. 7.2). The other is U = −GmEm/r (Eq. 13.9). As shown in Section 13.3, the first equation is correct only if the gravitational force is a constant over the change in height Δy. The second is always correct. Actually, the gravitational force is never exactly constant over any change in height, but if the variation is small, we can ignore it. Consider the difference in U between a mass at the earth’s surface and a distance h above it using both equations, and find the value of h for which Eq (7.2) is in error by 1%. Express this value of h as a fraction of the earth’s radius, and also obtain a numerical value for it.

ANSWER:

Solution 57P

Introduction

We have to calculate the height at which the potential energy calculated using the formula $U=mgy$ has $1\%$ error.

Step 1

The potential energy at height $h$ is given by the formula ${U}_{1}=mgh$ is actually the difference between the potential energy at the surface of the earth and at the height $h$ considering the acceleration due to gravity is constant throughout the distance.

Now the actual potential energy of the mass $m$ at the surface of the earth is

${U}_{s}=-\frac{GMm}{R}$

And the potential energy at the height $h$ is given by

${U}_{h}=-\frac{GMm}{R+h}$

Hence the difference is

$\Delta{}U=-\frac{GMm}{R+h}+\frac{GMm}{R}=\frac{GMmh}{{R}^{2}+Rh}$

Now the acceleration due to gravity is given by

$g=\frac{GM}{{R}^{2}}$

Putting this value in the form of ${U}_{1}$ , we have

${U}_{1}=\frac{GMmh}{{R}^{2}}$

Now for 1% error we have

$\frac{GMmh}{{R}^{2}}-\frac{GMmh}{{R}^{2}+Rh}=0.1\frac{GMmh}{{R}^{2}+Rh}$

$\Rightarrow{}1.01\frac{GMmh}{{R}^{2}+Rh}=\frac{GMmh}{{R}^{2}}$

$\Rightarrow{}\frac{1.01}{R+h}=\frac{1}{R}$

$\Rightarrow{}1.01R=R+h$

$\Rightarrow{}h=0.01R$

So potential energy, calculated using formula $U=mgh$ , will have 1% error at $h=0.01R$ or at a height 1% of the radius of the earth.

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