Speed of satellite orbiting around the Earth

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Homework Statement


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" ?

Homework 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

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"[/B]
⇒(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!
 

Answers and Replies

  • #2
Quantum Defect
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[snip]

⇒v = √(Gme)/R

[snip]
Set it up as a ratio. v1 => R, v2 => 4R. What is v2/v1 ?
 
  • #3
TSny
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Quantum Defect's comment should really help. If you want to see where you made a mistake:

New speed of satellite "v*" when radius is "4R"
⇒(Gmems)/R^2 = (ms*(v*^2)/4R
What about the the R on the left side?
 
  • #4
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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?
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
TSny
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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
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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
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
Quantum Defect
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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?
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
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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.
Oh, the "radius of the ORBIT." Thank you so much Quantum Defect and TSny. I get it now.
 

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