# Speed of satellite orbiting around the Earth

1. Jan 22, 2015

### bijou1

1. The problem statement, all variables and given/known data
A 1,000 kg satellite traveling at speed "v" maintains an orbit of radius, "R" around the earth. What should be its speed if it is to develop a new orbit of radius "4R" ?

2. Relevant equations
Gravitational force equation (F) = (Gmems)/r^2
G (universal gravitational constant) = 6.67 x 10^-11 N*m^2/kg^2
me = mass of earth
ms = mass of satellite
r= distance between the centers of me and ms

Centripetal force equation (Fc) = (ms*v^2)/r
ms = mass of satellite
v^2/r = centripetal acceleration
r = radius of orbit for satellite

3. The attempt at a solution
The satellite and the earth both exert a mutual attraction for each other through the gravitational force equation and the earth exerts a centripetal force to the satellite, so these two forces can be set equal to each other, since the gravitational force provides the centripetal force necessary for the satellite to orbit the earth.
The speed of satellite "v" when radius is "R"

⇒(Gmems)/R^2 = (ms*v^2)/R
⇒(G*me)/R = v^2
⇒v = √(Gme)/R

New speed of satellite "v*" when radius is "4R"
⇒(Gmems)/R^2 = (ms*(v*^2)/4R
⇒(4Gme)/R = v*^2
⇒v* = 2√(Gme)/R
since speed of satellite "v" when radius is "R" is equal to √(Gme)/R, substitute this into v* equation,
∴ v* = 2v
However my solution is wrong, and I don't understand why. I would like to know conceptually as well as mathematically the reason why. Any help would be great, thanks!
I have trouble with this problem. Please correct my approach and if my understanding of the concept is wrong. Thanks!

2. Jan 22, 2015

### Quantum Defect

Set it up as a ratio. v1 => R, v2 => 4R. What is v2/v1 ?

3. Jan 22, 2015

### TSny

Quantum Defect's comment should really help. If you want to see where you made a mistake:

What about the the R on the left side?

4. Jan 22, 2015

### bijou1

Hi, I may be wrong but I thought the "R" on the left side is the distance between the centers of the masses of the satellite and the earth, so the "4R" would only matter in the right side of the equation. Wait, is the distance between the centers of the masses of the satellite and the earth = the total radii of the satellite and the earth, so for the left side of the equation, the radius is (4R)^2
∴(Gmems)/((4R)^2) = ms*(v*^2)/4R
⇒(Gme)/4R = v*^2
⇒v*^2= 1/2√(Gme)/R
⇒v* = 1/2v
Please correct me if I am not understanding this conceptually and mathematically. Thank you!

5. Jan 22, 2015

### TSny

Yes. In the equation (Gmems)/r^2 = ms*(v*^2)/r, the r on the left and the r on the right stand for the same quantity: the radius of the orbit.

When r = R, you have Gmems)/(R^2) = ms*(v*^2)/R

But when r = 4R you get Gmems)/((4R)^2) = ms*(v*^2)/4R

6. Jan 22, 2015

### bijou1

So, is it correct to say the "R" in the left side of equation (Gmems)/R^2 is the radius and the distance between the centers of the satellite and the earth?

7. Jan 22, 2015

### Quantum Defect

R is the distance between the two centers of mass. They talk about the radius of the orbit -- this is wrt to the earth's com.

8. Jan 22, 2015

### bijou1

Oh, the "radius of the ORBIT." Thank you so much Quantum Defect and TSny. I get it now.