Question

A hypothetical spherical planet consists entirely of iron (p=7860 kg/m^3). Calclate the period of a satellite that orbits just above its surface.
I started out with the density formula P=M/V, but i kept subbing in other formulas, but i get stuck..can someone help me? thanks.
I posted this before, but i made a mistake in the question and I am now fearing that whoever helped me will not go 3 pages into the history just to find this question and help me again. Thanks =P

Answer 1

Also, teacher gave me a hint that "r(radius)" was suppose to cancel during the formula manipulation process..

Answer 2

Time required for 1 revolution,
T= (2πr^(3/2))/√(G∙M) (p. 146 of 7th ed Cutnell & Johnson)

Volume of a sphere,
V=4πr^3 (book inside cover)

Formula for mass density,
M=ρ∙V (p.321)

Where
T: the time period
G: universal gravitational constant, 6.673 E-11 (N∙m^2)/kg^2
note: N=(kg∙m)/s
r: distance from center of planet to satellite, aka planet’s radius
ρ: mass density of planet.
Iron density = 7860 kg/m^3
V: Volume of planet

Substitute formula for volume of a sphere into equation for mass:
M = ρ∙ 4πr^3
Then substitute this into the equation for time period
T= (2πr^(3/2))/√(G∙ρ∙ 4πr^3)
The r^(3/2) in the numerator cancels the √(r^3) in the denominator,
So the equation simplifies to
T= 2π/√(G∙ρ∙4π)
Plug in the known value of G and given value of ρ.
For a planet made of iron, satellite period
T ≈ 2447.39 seconds

~Marianne