Physics problem... Stumped (1 Viewer)

If all you need to find is the force of gravity on the object, it's

Fg=G(m1m2/r^2)

Remember to add the radius of the Earth to 2.90 × 10 ^6 m to get the value for r.

I did not have m2

But apparently it did not matter..

Answer was 4.63 m/s^2
 
Fg=G(m1m2/r^2)

That looks correct from what I remember, but what does the big G stand for.

Also, if I remember correctly, the satillite is accelerating. The change in direction does count as acceleration, even if there is no change in velocity. Just as the moon is accelerating around the Earth. I always liked physics, just when it got into electro-magnetics, it was over my head.
 
Assuming the satellite is geosynchronous (not moving with respect to the earth's rotation)

f = GmM/r^2

ma = GmM/r^2

divide by m

a = GM/r^2

Look up G and M, r is the distance to the surface + the radius of the earth (offhand I don't know if its significant). Mass of the satellite is irrelevant because it was cancelled by division.

If its not geosynchrnous, you don't have enough information (need the speed of the satellite)

I figured it was a constant.

The acceleration is constant, but only near the surface of the earth is it close to 9.8 m/s^2 (or 32 feet/s^2)
 
Actually it doesn't have to be geosynchronous, but it couldn't have artificial means of thrust, which satellites don't (except for brief periods for adjustment), and I mixed up geosynchronous with geostationary. Been a long time since that college physics course ...

actually i think as the orbit decays it begins to fall to the earth shortening its orbit length and causing it to increase in speed....
just a thought. no facts to back it up...

In real satellites, the orbit does decay, but we're ignoring that affect.
 
Last edited:

Create an account or login to comment

You must be a member in order to leave a comment

Create account

Create an account on our community. It's easy!

Log in

Already have an account? Log in here.

Users who are viewing this thread

    Back
    Top Bottom