# Satellite orbiting mars

NASA wants to fire a satellite into a circular orbit around Mars that will maintain 'station' above the Martian equator.
what is the distance from the surface of mars?

ok first of all does the term 'station' refer to when g = 0?

and the equation ive used is:
Fc = Fg
mv²/r = GMm/r²

the two 'm' values cancel out and then im left with:

v²/r = GM/r²

is it then possible to transform the equation into:

r = (GM/r²)*v²

is this procedure correct?

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By stationary it means that it will remain at that altitude, not falling or going up at all. Also I think given the context it means that it will be in a geostationary orbit, so it's period is the same as the planets rotational period. (I.e for the earth, it's period would be 24 hours, you'll have to look up the value for mars).

The first step you've used is correct;

$$\frac{v^2}{r} = \frac{GM}{r^2}$$

but check the next step...remember you're multiplying by r...not v.

Next can you think of a way of expressing the period of the shuttle in terms of the velocity? (Hint: think of the orbit as a circle, with radius 2pi r, if it's going at speed v, what is the time period T?) You'll have to think of a way of getting that T into the equation above to solve for r.

period of rotation of mars = 8.86 x 10^4 s
mass of Mars = 6.42 x 10^23 kg
Radius of Mars = 3.40 x 10^6 m

if i use v=2πr/T
v = 2*π*3.4e+6 / 8.86e+4
=241.1 m/s

wait, so would my new equation be:
r³ = GM/v² ?

or would i have to go back a few steps and turn it into:
4π²r/T² = GM/r²

if i took that step my equation could then become:
T²/R³ = 4π²/GM

or would i have to go back a few steps and turn it into:
4π²r/T² = GM/r²

if i took that step my equation could then become:
T²/R³ = 4π²/GM
This is spot on, so now you know the time period, you can find out the radius of its orbit.

ok thank you so much for your help =)