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tandoorichicken
Dec18-03, 12:49 AM
A strange new planet that has no atmosphere has a satellite that orbits very close to the planet's surface with a period of 1.50 hours. What is the approximate density of the planet? (Assume that the radius of orbit equals the radius of the planet.)

SEG9585
Dec18-03, 11:26 AM
Maybe I'm wrong...but it sounds like something is missing here:
It only gives you the time it takes to complete a revolution?
I would think u either need to know the planet's radius, mass, or ship velocity to calculate the density...

Because, say a person drives a car around the Earth's equator, and someone else flies a plane around the same distance. Without knowing velocity of the objects, the period of each revolution would be different, even though its the same planet, hence same density...

jamesrc
Dec18-03, 12:06 PM
How about just equating the gravitational force to the centripetal force?
\frac{mGM}{R^2} = m\omega^2R

and since \rho = \frac{M}{R^3} and v = Rω and T = \frac{2\pi}{\omega} you can solve for density as a function of period. So I think:

\rho = \frac{1}{G}\left(\frac{2\pi}{T}\right)^2