Math, Geo, planetary orbit?

Question

Assume the world is a spherical planet with a diameter of 1600 km with a uniform density of 5200 kilograms per cubic metre. If the moon is in a geostationary orbit around the world, and assuming a worldly sidereal day is exactly 24 hours, how far is that geostationary orbit above the surface of the world?

I am so confused. Can you please put your answer in kilometers too. Thanks.

Answer 1

F = - GmM/r^2 = - (mv^2)/r

The m's cancel, leaving

GM/r^2 = (v^2)/r

v = rω

GM/r^2 = ((rω)^2)/r

GM/r^2 = rω^2

ω = 2π/P

GM/r^2 = r(2π/P)^2

GM/r^2 = r(4π^2)/P^2

r^3 = GM(P^2)/(4π^2)

r^3 = Gρ(4/3)π(R^3)(P^2)/(4π^2)

r^3 = Gρ(R^3)(P^2)/(3π)

G ≈ 6.67428e-11 (m^3)/(kg∙s^2)

r^3 ≈ (6.67428e-11 (m^3)/(kg∙s^2))(5200 kg/m^3)((800 km)^3)((24 h)^2)/(3π)

r^3 ≈ (6.67428e-11)(5200)((800)^3)(km^3)/s^2)(... h)(3600 s/hr))^2)/(3π)

r^3 ≈ (6.67428e-11)(5200)((800)^3)((24)(3600))...

r^3 ≈ 1.407453594137813062528693e+11 km^3

r ≈ 5,201.693 km ≈ 5,202 km

d = r - R

d ≈ 5202 - 800 ≈ 4,402 km

Answer 2

This can be solved by using the following tools:

- Newtonian gravitation.  The gravitational force on the satellite is given by F = G m1 m2/r^2, where G is 6.6742E-11 m^3/(kg sec^2).  The mass of the planet can be inferred from its size and density; the mass of the satellite doesn't matter (it cancels out).

- Centrifugal force, which will equal the gravitational force.  This is given by the satellite's mass, multiplied by its radius times the square of the angular velocity in radians per second.

Answer 3

six other ppl have asked precisely this problem in the last day or two. please collaborate with them.