Question: Assume the world is a spherical planet with a diameter of 1600 KM. with a uniform density or 5200 KM.

It the moon is in a circular orbit of 630 KM. above the surface of the world,what is it's orbital velocity in metres per second?Round to the nearest WHOLE number.

You have an interesting question. What is important to recognize is the the moon is in a stable orbit. That means that its centripetal acceleration (which comes by way of its orbital velocity) is just the right size to counter the force of gravitation from the planet.

Now there is a formula for the force of gravity between two objects from Newton: F=(GMm)/r².

G is the gravitational constant, r is the radius between the centers of the two bodies, M is the mass (in this case) of the planet and m is the mass (in this case) of the moon. You don't know the mass of the moon, but that doesn't matter, as you'll see.

So that's the force of gravity drawing them together. Let's take a look at the centripetal acceleration of the orbit now.

You can get the centripetal acceleration from the equation a = v²/r, where v is the orbital velocity and r is the radius of the orbit (again, from the center of the planet to the center of the moon). Now Newton also told us that F=ma, so put these together and we have our centripetal force. Again, we don't know "m" - the mass of the moon.

Earlier, I said these two forces have to balance out to make a stable circular orbit, so let's set one equal to the other. Notice that the "m" on both sides cancels out? That's why the mass of the moon doesn't matter. Now you can re-arrange the equation and solve for v, plug in your known values for G, r and M and you've got your answer.

Stephen La Rocque

PS: Your school must be quite advanced. That's the kind of problem I'd not expect to see until secondary school!