What is the Velocity of a Satellite Orbiting Earth at a Distance R?

[dvvv]

Homework Statement

Find the velocity of a satellite which circles the Earth at a distance R from the Earth's centre.

Homework Equations

Fc = (m*v^2)/r
W = mg

The Attempt at a Solution

I put Fc = W (I think I can do that?)
so (v^2)/R = g
v = sqrt(10R)

I'm not sure if my answer is right or if I'm supposed to actually work out with R is. Can anyone help?

 

[gneill]

g is not constant over large radial distances. The assumption of constant g also assumes that the physics is taking place 'near the Earth's surface'.

Instead of using W = mg, use the full expression for the Newtonian force due to gravity.

 

[Ksitov]

Hi.

Use this relation:

F = G*m*M/d²

Here: - m is the mass of satellite.
- M is the mass of the Earth.
- d is the distance between the center of the Earth and the Satellite.
- G is the gravitationnal constant.

Warning for the distance ;) !

Good Luck.

 

[dvvv]

Fg = Gm1m2/d^2
I think I can just use one m since the mass of the Earth is very large realtive to the satellite.
so Fg = Gm/R^2
put Fg = Fc
and I get v= sqrt(G/R)
where G is the gravitational constant

I presume I'm supposed to have the variable R in my answer since they didn't give it in the question?

 

[gneill]

Fg = Gm1m2/d^2
I think I can just use one m since the mass of the Earth is very large realtive to the satellite.

No, m1 and m2 are multiplied, not added. A small number multiplied by a much larger number is an even larger number.

Also consider the units of the equation. While F = G*m1*M2/r² is a force, G*m1/r² is not -- it's an acceleration (because F/m2 is an acceleration by Newton's second law). But this is a big clue! You can calculate the acceleration due to gravity of the satellite. What other acceleration is operating here to balance it?

I presume I'm supposed to have the variable R in my answer since they didn't give it in the question?

That's right.

 

[dvvv]

g = G*M/R^2
centripetal acceleration a = (v^2)/R
g = a
I get v = sqrt(G*M/R)

 

[gneill]

Don't be confused, that's a perfectly fine result!

 

[dvvv]

Ok, thanks for your help. :)