Question

A satellite has a mass of 5613 kg and is in a circular orbit 3.70 105 m above the surface of a planet. The period of the orbit is 2.20 hours. The radius of the planet is 3.85 106 m. What would be the true weight of the satellite if it were at rest on the planet's surface?

Answer 1

Force due to gravity = Force due to centripetal

Fg = Fc
GMm/r² = m4pi²r/T²
GM/r² = 4pi²r/T²
M = 4pi²(r^3)/(T²G)
= 4pi²(3.85*10^6 + 3.7*10^5)^3/((2.2 * 3600)^2 * 6.67*10^-11)
= 2.9668*10^21/4.1838*10^-3

= 7.091*10^23 [this is the mass of the planet]

The gravitational force, Fg, is also equal to m*a. To find the acceleration,

ma=GMm/r²
a=GM/r²

At the surface of the planet, a=(6.67*10^-11)(7.091*10^23)/(3.85*10^6)² = 3.19 m/s² = 3.19 N/kg

So, Fg=mg= 5613 kg * 3.19N/kg = 17910N