# Thread: physics- circular motion

1. ## physics- circular motion

calculate the orbital height about the earth of a geostationay satellite.
the following data will be required:
earth's mass 6.0x10^24 kg
g.s G=6.67x10^-11
i know the time is going to be 24 hours but i have no idea what equations to use....
thanks.

2. Use the fact that the grativational pull provides the centripetal force.

$F_c = F_g$

Let m be the mass of the satellite,
r the distance from the Earth's centre,
omega the angular velocity of the satellite,
M the mass of the Earth.

$mr\omega ^2 = \dfrac{GMm}{r^2}$

Simplify;

$r^3\omega ^2 = GM$

Now, fill in the values that you know.

$r^3\left(\dfrac{2\pi}{24\times3600}\right)^2 = (6.67\times10^{-11})(6\times10 ^{24})$

You should be able to get the value of r, the distance of the satellite from the centre of the Earth and then, the height above the ground.

Of course, there are assumptions made here to make it easier.

Post the answer that you get!

Answer I got: 3.59 x 10^7 m

3. thank you very much!
i got 3.6x10^7 m.

4. Use your mouse to select my whole post and you'll see the spoiler I hid in there. That's the right answer!