Expert: Jayendra Upadhye - 7/17/2008

Question
Assume the earth is a spherical planet with a diameter of 1600 km with a uniform density of 5200 kilograms per cubic metre. If the moon is in a geostationary orbit around earth and assuming a earth sidereal day is exactly 24 hours, how far is that geostationary orbit above the surface of the earth?  

Answer
Hi Snogale,
Sorry for the delay, the site had been down for a day, so i could not submit my answer which i completed same day.
Posting it now.

Snogale,
I generally hate numericals!
And that too slogging for some one else!

But i am presuming that you are asking in order to know a bit more and not "only to make some one else do the work".

You see what matters is:-
1 - Just the mass of the main gravitating body
2 - The orbital distance where orbital period equals 24 hours.

The mass M (in Kilograms) of the main gravitating body = (4/3)*(pi)*(r cubed), all in mks units.

M = Volume of the sphere * average density           
or
Volume = (4/3)*pi*(1600 kilometer * 1000 meter/kilometer) cubed]
Volume = (4/3)*pi*[(1600 * 1000)^3] cubic meter
Volume = 17157284678805057473 met cubed
density = 5200 Kg/Met cubed

M = 17157284678805057473 met cubed * 5200 Kg/met cubed
= 17157284678805057473*5200
= 89217880329786298859600 Kg.

To those at the planet's surface or outside of it, all this mass will appear to be concentrated at the center of the planet, and gravitational acceleration will follow the inverse square law from there onwards.

eqn 1) -
g = G*M/(r*r)
where G = 6.67*(10^-11)
And g is centripetal accleration due to gravity at an orbital radius r units from center of planet.
But centripetal acceleration also g = (V*V)/r for any stable orbit.

Thus we get
eqn 2) -
g = (V*V)/r = G*M/(r*r) or
(V*V) = G*M/r or
V = sqrt{G*M/r}
But V = 2*pi*r/T where T is orbital period.
In our case, T = 24 hours * 3600 sec/hour = 86400 sec.
thus our V = r*7.2722 * 10^-5 met/sec

Substituting this for V in V*V = G*M/r, we get
(r*r)*5.2885*10^-9 = G*M/r or
r cube = G*M/(5.2885*10^-9)
OR
r cubed = (6.67*10-11)*89217880329786298859600/(5.2885*10^-9) or
r cubed = 6.67*89217880329786298859600/(5.2885*100) or
r = 10401159 met or
r = 10401.159 Kilometer.

Check:-
Orbital speed at geostationary level is
V = 2*pi*r/86400 (because T = 86400 sec).
V = 756.3936 met/sec

Centripetal acceleration corresponding to this speed at r units radius
g = V*V/r = (756.3936*756.3936)/10401159 = 0.055 met/sec squared.

This should also match g calculated from inverse square law at r units distance from center
g = G*M/(r*r) or
g = (6.67*10^-11) * 89217880329786298859600 / (10401159*10401159)
g = 0.055 met/sec squared!!!

Thus geostationary orbit lies at radius 1040.159 kilometer, with a gravitational centripetal acceleration of 0.055 met/sec suqared and an orbital period of exactly 86400 seconds.

Formulae used
M = [(4/3)*pi*Re*Re*Re]*(average density) where Re = earth radius
g = G*M/(r*r) - where G = 6.67*10^-11
g = (V*V)/r
V = 2*pi*r/T where T is known as 86400 secs.
as (V*V)/r = G*M/(r*r) or 4*pi*r/(T*T) = G*M/(r*r), we get
r cube = G*M*(T squared) / (4*pi)

This is the famous keplerian T squared varies as r cubed statement!
meaning the square of the orbital period varies as the cube of the orbital radius!

hope that suffices.

please do rate the answer if you found it helpful.

regards
Jayen