Question

An artificial satellite is in a circular orbit d=730.0 km above the surface of a planet of radius  r=2.75×103 km. The period of revolution of the satellite around the planet is T=3.15 hours. What is the average density of the planet?

Answer 1

here,

the height above the surface , h = 730 km

h = 0.73 * 10^6 m

radius of planet , r = 2.75 * 10^3 m

r = 2.75 * 10^6 m

let the mass of satellite be M

the period of revolution , T = 3.15 h = 3.15 * 3600 s

3.15 * 3600 s = 2*pi*sqrt((r + h)^3 /(G * M))

3.15 * 3000 = 2*pi*sqrt((2.75 * 10^6 + 0.73 * 10^6)^3 /( 6.67 * 10^-11 * M))

solving for M

M = 2.79 * 10^23 kg

the density of planet , p = M/volume of planet

p = M /(4/3 * pi * (r)^3)

p = 2.79 * 10^23 /( 1.33 * pi * (2.75 * 10^6)^3)

p = 32.1 kg/m^3

the density of planet is 32.1 kg/m^3