Question

An artificial satellite is in a circular orbit d=300.0 km above the surface of a planet of radius r=4.05X10^3. The period of revolution of the satellite around the planet is T=2.15 hours. What is the average density of the planet?

Answer 1

using kepler's 3rd law of planetary motion

T^2 = 4 pi^2 (R +h) ^3 / (GM)

(2.15× 3600)^2 = 4* 3.14^2* ( 4050* 10^3 + 300* 10^3) / ( 6.67* 10^-11 M)

M = 8.124* 10^23 kg

density

rho = M / (4/3 pi (4050* 10^3)^3)

rho = 2921.1 kg/m^3

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