How Much Force Keeps a Satellite in Orbit at Three Earth Radii?

[chiurox]

Homework Statement

What is the force of gravity acting on a 4500kg satellite when it orbits three Earth radii from the Earth's center?

Homework Equations

Gm1m2/r²
mass of the earth=5.98*10^24kg
radius of the earth=6370km

The Attempt at a Solution

This is a multiple choice question, with answers:
15000N
45000N
500N
1500N
5000N

However, what I get from plugging in is:
[(6.67*10^-11)(4500)(5.98*10^24)]/(3*6370)²
which equals roughly: 5.0*10^9 which is completely off any of the answers... can anyone help me?

hahaha, never mind! I forgot to convert the km to m... -_-

 

[Snazzy]

You're not using SI units for distance.

 

[Moetasim]

The force of gravity acting on a 4500kg satellite when it orbits three Earth radii from the Earth's center is approximately 15000N. This can be calculated using the equation F = Gm1m2/r^2, where G is the gravitational constant (6.67*10^-11 Nm^2/kg^2), m1 is the mass of the satellite (4500kg), m2 is the mass of the Earth (5.98*10^24kg), and r is the distance between the satellite and the center of the Earth (3 Earth radii, which is equivalent to 3*6370km = 19110km = 1.911*10^7m). Plugging in these values gives us F = [(6.67*10^-11)(4500)(5.98*10^24)]/(1.911*10^7)^2 = 1.5*10^4 N, which is approximately equal to 15000N. This is the force of gravity that keeps the satellite in orbit around the Earth.