Posted by **ALan** on Wednesday, March 3, 2010 at 12:49am.

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

- Physics -
**ALan**, Wednesday, March 3, 2010 at 12:49am
Also, teacher gave me a hint that "r(radius)" was suppose to cancel during the formula manipulation process..

- Physics -
**Marianne**, Tuesday, April 12, 2011 at 6:02pm
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

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