## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 13 - Newton's Theory of Gravity - Exercises and Problems - Page 354: 32

#### Answer

(a) At a height of $1.38\times 10^7~m$ above the earth, the free-fall acceleration is 10% of its value at the surface. (b) The speed of a satellite orbiting at that height is 4.44 km/s

#### Work Step by Step

(a) We can write an expression for the free-fall acceleration $g$ at the earth's surface. Let $M_e$ be the mass of the earth. Let $R_e$ be the radius of the earth. $g = \frac{G~M_e}{R_e^2}$ Let's suppose the free-fall acceleration at a height $h$ above the earth is $0.1~g$. $0.1~g = \frac{G~M_e}{(R_e+h)^2}$ $(0.1)~(\frac{G~M_e}{R_e^2}) = \frac{G~M_e}{(R_e+h)^2}$ $(R_e+h)^2 = 10~R_e^2$ $R_e+h = \sqrt{10}~R_e$ $h = (\sqrt{10}-1)~R_e$ $h = (\sqrt{10}-1)(6.38\times 10^6~m)$ $h = 1.38\times 10^7~m$ At a height of $1.38\times 10^7~m$ above the earth, the free-fall acceleration is 10% of its value at the surface. (b) We can find the orbital radius $R$. $R = R_e+h$ $R = (6.38\times 10^6~m)+(1.38\times 10^7~m)$ $R = 2.02\times 10^7~m$ We can use the orbital radius to find the orbital speed. $v = \sqrt{\frac{G~M_e}{R}}$ $v = \sqrt{\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(5.98\times 10^{24}~kg)}{2.02\times 10^{7}~m}}$ $v = 4.44\times 10^3~m/s$ $v = 4.44~km/s$ The speed of a satellite orbiting at that height is 4.44 km/s.

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