A hypothetical spherical planet consists entirely of iron (p=7860kg/m^3). Calculate the period of a satellite that orbits just above its surface.

Question

I really can't solve this! No other information was provided. Please and thank you

Answer 1

As I told you a few days ago, you can't solve it unless you know the radius of the sphere. if that was all the information you were provided, complain to your teacher. Perhaps you were supposed to assume it has the same raius as earth, but they should have said so. In that case, the orbital period is (as it is for earth, which has similar average density) about 90 minutes.

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