Science/Math Question!!!!!?

Question

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 the earth, and assuming a earth sidereal day is exactly 24 hours, how far is that geostationary orbit above the surface of earth?

Answer 1

Find the planet's radius:

d = 1600 km --> r = 800 km * 1000m/1km = 8×10^5 m

Find the planet's volume:

V = (4/3)*π*r³ = ...

Find the planet's mass:

ρ = m/V --> m = ρ*V = ...

[ρ is the planet's density]

Find the distance from the planet's center:

R = (Gm/ω²)^(1/3)

That's the cube root of (Gm/ω²).

G is the Universal Gravitational Constant; G = 6.67×10−11 m³/(kg*s²).

ω is the angular speed; for a 24h sidereal day, ω=(2π)/86400s=...

Now, plug & chug, but remember to subtract the planet's radius (8×10^5 m) from the final answer to get a value measured from the planet's surface.

Your answer should be on the order of 10^6m.

Answer 2

The actual text of this week's Neopets "Lenny Conundrum" contest puzzle (Round 271) is as follows:

"Assume the world of Neopia is a spherical planet with a diameter of 1600 km with a uniform density of 5200 kilograms per cubic metre. If the moon of Kreludor is in a geostationary orbit around Neopia, and assuming a Neopian sidereal day is exactly 24 hours, how far is that geostationary orbit above the surface of Neopia?"

To obtain the prize-winning answer value for this latest Neopets "Lenny Conundrum" contest puzzle, try following the instructions here:

http://lennyconundrumsolutions.blogspot....