Centripetal Acceleration of Satellite

[Physics53]

Homework Statement

Hi, I am stuck on this question:
A satellite of mass 1200 kg is in orbit around the Earth at a distance of 22000 km from the centre of the Earth.
Calculate the magnitude of the centripetal acceleration of the satellite at this distance.

The Attempt at a Solution

I did:
mv^2/r= GMm/r^2
so v= square root of GM/r
v= square root of 6.67 x10^-11 x 6.0 x 10^24/ 22000 x 10^3
this gave me=
4.266x 10^-3 m/s but the answer at the back is 0.83 m/s^2, can someone please explain to where i went wrong.

Thanks[/B]

 

[vela]

Carefully reread the question and make sure you're answering it.

 

[Physics53]

Carefully reread the question and make sure you're answering it.

Hi, I've reread it many time, but i still feel confused

 

[vela]

What's it asking you to find? What did you actually calculate?

 

[Physics53]

What's it asking you to find? What did you actually calculate?

Oh I see what you mean, are you implying that I solved centripetal velocity instead of centripetal acceleration?

 

[vela]

Essentially. You solved for what's sometimes called the tangential velocity, not centripetal velocity.

 

[Physics53]

Essentially. You solved for what's sometimes called the tangential velocity, not centripetal velocity.

So does this mean I consider gravity of Earth as my centripetal velocity and therefore:
g= GM/r^2, but i still have a query, if you could kindly clear this up for me, why can't we use the centripetal formula in this question.

 

[vela]

The work you did was correct, but you didn't actually solve for the quantity you were asked for. (The fact that the units were different is a big clue.)

I was just pointing out the phrase "centripetal velocity" doesn't really make sense. The velocity you found is tangent to the satellite's path; it doesn't point inward. The acceleration, however, does, so that's why it's referred to as centripetal acceleration.