geosynchronous orbit?

Question

Situation: A 4000 kg satellite orbits around mars

a) what is it's height?
b) what is the force of gravity?
c) what is the speed of the satellite?

the satellite is in geosynchronous orbit with mars btw

Answer 1

To be 'geosynchronous' I think it needs to be orbiting earth, but that's irrelevant pedantry on my part.
In an orbit, the outward 'centripetal acceleration' must be balanced by the inward force of gravity.
The gravitational force acting on a mass 'm' at a distance 'R' from a point source (the centre of the planet in this case), where the mass of the point source (the planet) is 'M', is given by:
ma = GMm/R^2.
G is the Universal Gravitational Constant. (It is not the same as 'g', which is just acceleration due to gravity on the surface of the earth, and is not relevant here).
M: mass of planet mars in kg
m: mass of satellite: 4000 kg
R: Is the radius of the orbit, from the centre of mars.

For an orbit, you need to balance this force against centripetal acceleration, which generates force by the following equation:

ma = mv^2/R.

m: mass of satellite again
v: speed of the staellite in its orbit, which is the asnwer to part c) of your question.
R: radius of the orbit again.

So, for these forces to balance, the two equations must be equal, ie:

GMm/R^2 = mv^2/R.
The mass of the satellite 'm cancels from this equation, as does one 'R', and it simplifies thus:
GM/R = v^2
Rearrange: (GM/R)^1/2 = v.

So, we see that 'v' depends on 'R'.
What else do we know about 'v'?
We know that the velocity 'v' is distance divided by time.
What distance does the satellite cover in its orbit? It tracks a big circle. The radius of a circle is 2(pi)R, where R is that radius of the orbit again.
So, v is distance/time = 2(pi)R/t, (where t is 1 martian day, measured in seconds).
So, substituting this equation into the previous one:
(GM/R)^1/2 = v = 2(pi)R/t
(GM/R)^1/2 = 2(pi)R/t
Square both sides:
GM/R = 4(pi)^2R^2/t^2
Rearrange:
GMt^2 = 4(pi)^2R^3
Rearrange:
(GMt^2/(4(pi)^2))^1/3 = R

Now, G is the Universal gravitational constant- look it up.
M is the mass of mars in kg- look it up.
t is the length of a martian day in seconds- look it up.
Substitute these numbers into the above equation, to calculate 'R' in metres.
R is then the radius of the orbit.
a) is then R - x, where x is the radius of Mars in metres (look it up).
b) Is just GMm/R^2. You should now have all the numbers- substitute them in.
c) We established earlier, that (GM/R)^1/2 = v. Substitute in the numbers again.

Answer 2

Hm... If only I could remember how long a Martian day is.......