# Speed of satellite orbiting around the Earth

• bijou1
In summary, the satellite needs to maintain a speed of 2v in order to maintain a new orbit of radius 4R.
bijou1

## 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!

bijou1 said:
[snip]

⇒v = √(Gme)/R

[snip]

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

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

bijou1 said:
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?

TSny said:
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!

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

TSny said:
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?

bijou1 said:
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.

Quantum Defect said:
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.

## 1. What is the speed of a satellite orbiting around the Earth?

The speed of a satellite orbiting around the Earth depends on its altitude and the mass of the Earth. On average, a satellite in low Earth orbit (LEO) has a speed of about 17,500 miles per hour (28,163 kilometers per hour).

## 2. How is the speed of a satellite calculated?

The speed of a satellite is calculated using the formula v = √(G*M/R), where v is the speed in meters per second, G is the gravitational constant, M is the mass of the Earth in kilograms, and R is the radius of the orbit in meters.

## 3. Can the speed of a satellite change while in orbit?

Yes, the speed of a satellite can change while in orbit due to factors such as atmospheric drag, gravitational pull from other celestial bodies, and adjustments made by thrusters.

## 4. What is the fastest speed a satellite can reach while orbiting the Earth?

The fastest speed a satellite can reach while orbiting the Earth is approximately 27,600 kilometers per hour (17,100 miles per hour). This is known as the orbital speed, and it is achieved by satellites in a circular orbit close to the Earth's surface.

## 5. How does the speed of a satellite affect its orbit?

The speed of a satellite affects its orbit by determining the shape and altitude of the orbit. A faster speed will result in a larger orbit, while a slower speed will result in a smaller orbit. If the speed is too slow, the satellite may not be able to maintain its orbit and could eventually fall back to Earth.

• Introductory Physics Homework Help
Replies
16
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
17
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
894
• Introductory Physics Homework Help
Replies
4
Views
997
• Introductory Physics Homework Help
Replies
39
Views
3K