Solve Equatorial Orbit: 12 Hours, Radius & Speed

[LadiiX]

Homework Statement

A satellite is to be put into an equatorial orbit with an orbital period of 12 hours.
Given: 12 Hours = 12 X 60 X 60 seconds

What is the radius?
What is the orbital speed?
How many times a day will the satellite be over the same point on the equator if the satellite orbits in the same direction of the Earth's rotation? If it orbits in the opposite direction?

Homework Equations

r= (GMT^2)/(4pi^2 )^1/3
G = Gravitational Constant = 6.67 X 10^-11 N m^2/kg^2
M = Mass of Earth 5.98 X 10^24 kg
T = time

V= Radical (GM/r)
r = radius

The Attempt at a Solution

Well I started with the radius equation and plugged everything in

r= ((6.67×10^(-11) N∙m^2∕kg^2 × 5.98×10^24 kg × (12×60×60 s)^2)/(4pi^2 ))^(1/3)

However I had trouble working it out. Then the other problems just went over my head.

 

[AtticusFinch]

Well you already got the first question, just multiply the numbers.

You already stated that the orbital speed is [tex] \sqrt{\frac{GM_{earth}}{r}} [/tex]

You know radius already and the other values are constants so find the orbital speed.

Convert orbital speed into angular velocity and do the same for a point on the equator. Then come up with an equation for the angular displacement of the satellite and the point on the equator (angular velocity * time). Then graph these equations and find their intercepts on the domain 0 < t < 86400 (seconds in a day)(note that if the angular displacement differs by a integer multiple of 360 they are technically in the same place). Do something similar for the other scenario.