A satellite of mass 4900 kg orbits the Earth (mass = 6.0 1024 kg) and has a period of 6200 s. PLEASE HELP!!?

Question

A satellite of mass 4900 kg orbits the Earth (mass = 6.0 1024 kg) and has a period of 6200 s

(a) Find the magnitude of the Earth's gravitational force on the satellite.
=......N
(b) Find the altitude of the satellite.
=.......m

Answer 1

The force of gravity on m, the satellite, is W = GmM/R^2; where M is Earth's mass, R is the distance between masses, and G is a constant.

As the satellite is in an orbit at R distance, it's centrifugal force F = mv^2/R offsets its weight W. So F = mv^2/R = GmM/R^2 = W; where weight is that gravitational force you are looking for.

Thus we need to find v, the tangential velocity. C = 2pi R = vT; where the circumference (C) of the orbit at R radius is covered in T = 6200 seconds. Thus, v = 2pi R/T = wR is the tangential velocity, w is the angular velocity = 2pi/T.

Then F = m(wR)^2/R = GmM/R^2 and w^2R = GM/R^2; so that R^3 = GM/w^2 = GM/(2pi/T)^2 and you can solve for R presuming you also know G, the gravitational constant.

a) In which case you have F = m(2pi/T)^2 R = the weight or force of gravity on the satellites since m and T are given, and you just solved for R.

b) The altitude h = R - r; where r is the radius of Earth. I presume you know r as well.

Answer 2

T = 2pi SQRT(a^3 / u)
where
T - Orbital period
pi - 3.14
a - orbital radius = 6371 + 410 km = 6781 km
u - Standard gravitational parameter of earth = 398600 km^3/s^2

That gives a = cuberoot(u * (T / (2pi) )^2) = 7297 km
for an altitude of 926 km

For part (a),
F = Gm1*m1/r^2
= (6.6742E-11) * 4900 * (6E24) / 7297^2
= 36850 N