Question

A 1320 kg geosynchronous satellite orbits a planet similar to Earth at a radius 193000 km from the planets center. Its angular speed at this radius is the same as the rotational speed of the Earth, and so they appear stationary in the sky. That is, the period of the satellite is 24 h .

What is the force acting on this satellite? Answer in units of N.

What is the mass of this planet? Answer in units of kg.

Please show wotk and explanation.

Answer 1

The centripetal acceleration of the satellite acts towards the planet and has a value rw^2, where w is the angular velocity of the satellite

since w = 2pi radians in one day w = 2pi rad / 60 x 60 x 24 s = 7.27 x 10^-5 rad/s

Applying Newton's Second Law force = mass x acceleration
Then F = m x rw^2
We have F = 1320 x 1.93 x 10 ^8 x (7.27 x 10^-5)^2 Newton
So F = 1320 x 1.93 x 10^8 x 52.8529 x 10^-10

gives F = 13.20 x 10"2 x 1.93 x 10^8 x 52.8529 x 10^-10

hence = 13.2 x 1.93 x 52 .8529 = 1346.48 N

YOu can get mass from Kepler's third Law:

T^2 = 4*pi^2*a^3/(GM) where T = period of orbit, a = radius of orbit and M = mass of planet:

M = 4*pi^2*a^3/(GT^2)

T = 24 hrs x 60 mn/hr x 60sec/min = 8.64x10^4 sec

a = 1.93x10^5 km = 1.93 x 10^8 m

M = 4*3.14^2*(1.93*10^8)^3 / 6.67259*10^-11 * (8.64*10^4)^2

M = 5.692x10^26 kg (about 100x more massive than earth)