Physics Circular Motion Question!

[Sessumaru]

Hello! Gotta question here about Circular Motion...

 

we all know that

 

a = (v ^ 2) / r

 

f = 1 / T

 

and F = ma...

 

Then how do I know which formula to use? For example,

 

Calculate the force of gravity on a spacecraft 12,800km (2 earth radii) above the Earth's Surface if the mass is 1400kg?

 

[big314mp]

Usually you are given a velocity also with these problems.

 

Otherwise, you could try g=GMm/r^2, although I don't think that will give you a good answer.

[insane_alien]

big314mp, that is the right equation to use. why would you think it would give you a wrong answer? you have M,m and r with G being a constant so you can find F(which you have mysteriously labelled g in this case)

[swansont]

F = ma always applies in these problems

 

a = v^2/r always applies to circular motion

 

However, the "centripetal force" is not it's own kind of force, such as gravity, or tension or normal force, that we identify in a free-body diagram. It's an additional condition that holds true when something moves in a circle.

 

GMm/r^2 (as big314mp has given) is the gravitational force (rather than the acceleration)

 

If you were given the period of orbit, you could find the speed, and use the centripetal acceleration equation.

[big314mp]

big314mp, that is the right equation to use. why would you think it would give you a wrong answer? you have M,m and r with G being a constant so you can find F(which you have mysteriously labelled g in this case)

 

Oops. It is supposed to be F.

 

Anyways, I just did this problem the other day for my physics homework (it was the Space shuttle, and it was orbiting lower) and we were expected to use the orbital velocity given to calculate the force of gravity. I figure this is the same question, just with a bit of info missing.

[Sessumaru]

So I have to use F=ma, as a base, right?

[Flashman]

Classically, the force is gonna be the same however fast it's moving, however the resultant acceleration actually affecting it would change according to it's velocity component at right angles to the gravitational force.

[swansont]

Classically, the force is gonna be the same however fast it's moving, however the resultant acceleration actually affecting it would change according to it's velocity component at right angles to the gravitational force.

 

I can't reconcile those two statements. If the acceleration is different, the force cannot be the same.

[Flashman]

Whoops, you're right, not thinking rigorously, had a glitch in my mental free body diagram. Maybe should have said, the resultant displacement towards the center of mass is dependent on the velocity. At classical velocities the force remains the same, independent of the velocity.