## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 10 - Problems - Page 401: 73

#### Answer

(a) The gravitational field strength on the other planet is less than that on Earth. (b) The gravitational field strength on the other planet is $5.57~m/s^2$

#### Work Step by Step

(a) We can write an expression for the period of a pendulum: $T = 2\pi~\sqrt{\frac{L}{g}}$ We can see that increasing the gravitational field strength decreases the period, and decreasing the gravitational field strength increases the period. Since the period on the other planet is greater than the period on Earth, the gravitational field strength on the other planet is less than that on Earth. (b) We can write an expression for the gravitational field strength on Earth $g_E$: $T_E = 2\pi~\sqrt{\frac{L}{g_E}}$ $(\frac{T_E}{2\pi})^2 = \frac{L}{g_E}$ $g_E = \frac{L~(2\pi)^2}{T_E^2}$ We can write an expression for the gravitational field strength on the other planet $g_o$: $T_o = 2\pi~\sqrt{\frac{L}{g_o}}$ $(\frac{T_o}{2\pi})^2 = \frac{L}{g_o}$ $g_o = \frac{L~(2\pi)^2}{T_o^2}$ We can find $g_o$: $\frac{g_o}{g_E} = \frac{\frac{L~(2\pi)^2}{T_o^2}}{\frac{L~(2\pi)^2}{T_E^2}}$ $g_o = \frac{T_E^2}{T_o^2}~g_E$ $g_o = \frac{(0.650~s)^2}{(0.862~s)^2}~(9.80~m/s^2)$ $g_o = 5.57~m/s^2$ The gravitational field strength on the other planet is $5.57~m/s^2$

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