How high above the Earth's surface would satellite have to be to have a revolutional period of 24 hours?

Question

should I use Kepler's 3rd law and how?

Answer 1

You basically set the centrifugal acceleration of the satellite (v^2/r) to the gravitational acceleration due to Earth's gravity (GM/r^2), where v is the velocity of the satellite, r is the radius of the satellite from the center of the Earth, G is the universal gravitational constant, and M is the mass of the Earth. This gives you

v^2 / r = GM / r^2

But, in this case, the velocity of the satellite is simply

v = 2*pi*r/day

since the satellite at a radius r from the center of the earth will need to travel 2*pi*r in one day (=86,400 seconds) to have a revolutional period of 24hrs.

Substitute this into the previous equation and rearrange to get

r^3 = GM * day^2 / 4*pi^2

In the above, G=6.6742E-11Nm^2/kg, M=5.9736E24kg, and day = 86,400 seconds. Plug in these values and solve to get the distance the satellite must be from the CENTER of the Earth. Then, subtract the radius of the Earth from that value to get how high the satellite needs to be above the Earth. The answer is around 22,240 miles.

Answer 2

This particular orbit has a special name, a geosynchronous orbit, which is used for satellites that need to remain above a certain spot or region of the Earth all the time. Weather satellite, satellite radio, satellite TV are some of the applications that require a geosynchronous orbit, which is about 22,300 miles.

Answer 3

22,600 statute miles. It's called a geosynchronous orbit. The mass of the satellite doesn't matter.

Answer 4

a the same height as the tv satellites

Answer 5

You didn't give us the mass.