Here's a question... I don't know if it's Physics or Astronamy.?

Question

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?

Answer 1

Weird, I just answered a nearly identical question for someone else, so here it is:

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.

(BTW, "astronomy" is how it's spelled.)