The first part of this question relies on remembering one key principle which is that for a satellite, or any body orbiting in a circle, centripetal force is provided by gravitational force. The rest is simply algebraic manipulation as long as you can remember the equations for circular motion and gravitational force.

So, begin by equating centripetal force and gravitational force:

mv^2/r=GMm/r^2

You can see immediately that m, which is the mass of the satelite, is on both sides of the equaiton and cancels out. As this term is not in the final equation we are aiming for, this is a good start. You can also see that r is on both sides of the equation so we want to gather these together by multiplying both sides by r, leaving:

v^2=GM/r

This cannot be simplified further, however v, which is the speed of the satellite, is not in the final equation. We know that the speed of an object in circular motion is given by the equation v=2pir/T, where T is the period of the satellite's orbit. Substituting this into v^2=GM/r gives:

4pi^2r^2/T^2=GM/r

If we rearrange for T^2 we get:

T^2=(4pi^2r^3)/GM

which is what the question asked us to find. This is an important equation which can be use to find many features of an orbit and it is used regularly so it is important to learn the above derivation. It's not a difficult derivation with only two key steps, centripetal force=gravitational force and substituting v for v=2pir/T.

The second part of the question requires you to use the equation which we just derived. As when using any equation it is worthwhile checking which values you know and which you need to calculate:

T=Orbital period of a geostationary satellite=24 hours=8.64*10^4 seconds

M=Mass of the Earth 6*10^24kg

G=Gravitational constant=6.67*10^-11 Nm^2/kg^2

pi=3.14 or use calculator value

r= radius of the satellite's orbit, what we are trying to find

We now know all the terms in the equation apart from the one which we wish to calculate. So we can rearrange the equation for r and substitute in the above values. Also note that values of M, G and the number of seconds in a day will be provided in the data sheet in the exam and so don't need to be memorised. We then have:

r^3=(T^2GM)/(4pi^2)

Substituting in the values above we find:

r^3=((8.64*10^4)^2*6.67*10^-11*6*10^24)/(4*3.14^2)=7.57*10^22

Then take the cubic root to find r

r=4.23*10^7 m

This is the orbital radius of any geostationary satellite as we can see that it does not depend on the mass of the satellite, only the mass of the Earth. Also remember that geostationary satellites orbit above the equator and have an orbital period of 1 day.