(adsbygoogle = window.adsbygoogle || []).push({}); [SOLVED] Period of a satellite

1. Given: G = 6.67 x 10^-11 N m^2/ kg^2

The acceleration of gravity on the surface of a planet of radius R = 4910 km is 11.8 m/s^2. What is the period T of a satellite in circular h = 8445.2 km above the surface? Answer in units of s.

2. F = GMm/r^2

F = ma = mv^2/r

v = sqrt(GM/r)

T = 2pi*r/v

3. i got r = 13355200 m; a = 11.8 m/s^2; that gave me the velocity = 12553.54m/s. i plugged it into the period equation and got T = 6684.425s but its wrong

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# Period of a satellite

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