# Plz Help!!!

• Jul 18th 2008, 10:28 AM
charl17728
Plz Help!!!
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 it, hours, how far is that geostationary orbit above the surface of the earth after 24 hours???????
• Aug 6th 2008, 09:08 PM
TwistedOne151
First, use that the volume of a sphere or radius R is $V=\frac{4}{3}\pi{R^3}$, and multiply by the density to get the mass M of your planet.

Then, for a moon of mass m, the force balance for a circular orbit of radius r is $\frac{GMm}{r^2}=\frac{mv^2}{r}$, where v is the velocity of the orbiting moon, and G is the universal gravitational constant [=6.673x10^(-11) m^3 kg^(-1) s^(-2)]. The distance traveled in one orbit is $2\pi{r}$, so the period T (time to make one orbit) is $T=\frac{2\pi{r}}{v}$, so $v=\frac{2\pi{r}}{T}$, and substituting that into the force balance equation,
$\frac{GMm}{r^2}=\frac{4\pi^2mr}{T^2}$
We can cancel m from both sides:
$\frac{GM}{r^2}=\frac{4\pi^2r}{T^2}$
so the mass of the moon does not matter (this is a form of Kepler's Third Law as applied to a circular orbit).
Bringing the r terms to one side,
$GM=\frac{4\pi^2r^3}{T^2}$
$\frac{GMT^2}{4\pi^2}=r^3$
$r=\sqrt[3]{\frac{GMT^2}{4\pi^2}}$
Now, the definition of geostationary orbit is that the orbiting body orbits at the same speed at the body it's orbiting rotates (so it remains over the same point on the "planet"). Thus the period T must be one day=24 hours=86400 seconds. Thus, we have G, M, and T, and so you can plug in those values to get r, which is the distance from the center of the planet to the moon. r-R is thus the distance from the planet's surface to the moon, the answer you want.

--Kevin C.