Period of revolution for a communications satellite at 3.59×10^7 m altitude

[pluke100]

Homework Statement

It is proposed to place a communications satellite in a circular orbit rounf the equator at a height og 3.59 x 10^7m above the Earth's surface. Find the period of revolution of the satellite in hours and comment on result.

Values given:
Radius of Earth: 6.37 x 10^6m
Mass of Earth: 5.98 x 10^24kg
Gravitational Constant: 6.67x10^-11 m^3 kg^-1 s^-2

Homework Equations

Not sure which one you can you, but i think two different ones have to be used

The Attempt at a Solution

This is the only question which i couldn't answer, and when i tried several ways out, it got very far away from result. the answer should be :24 hrs

 

[rsk]

You know that the centripetal accelration is provided by the gravitational force.

So,

mr(omega)^2 = GMm/r^2

SO you can calculate omega, the angular velocity.

Then you should have an equation relating the period tn the angular velocity.

 

[pluke100]

thx, i'll try it out. If i get wrong result, i'll post my working to see what's wrong