## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 5 - Problems - Page 187: 36

#### Answer

The Hubble Space Telescope's orbital period is $97.0~minutes$

#### Work Step by Step

When a satellite orbits a planet, the gravitational force provides the centripetal force to keep the satellite moving in a circle. Let $M_p$ be the mass of the planet and let $M_s$ be the mass of the satellite. We can find an expression for the angular speed of a satellite: $\frac{G~M_p~M_s}{R^2} = M_s~\omega^2~R$ $\omega = \sqrt{\frac{G~M_p}{R^3}}$ We can find an expression for a satellite's orbital period $P$: $P = \frac{2\pi}{\omega} = 2\pi~\sqrt{\frac{R^3}{G~M_p}}$ We can find the Hubble telescope's orbital period $P$: $P = 2\pi~\sqrt{\frac{R^3}{G~M_p}}$ $P = 2\pi~\sqrt{\frac{(6.38\times 10^6~m+6.13\times 10^5~m)^3}{(6.67\times 10^{-11}~m^3/kg~s^2)~(5.97\times 10^{24}~kg)}}$ $P = 5822.7~s = 97.0~minutes$ The Hubble Space Telescope's orbital period is $97.0~minutes$

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