Escape Speed of a Satellite

In summary: Hope that helps!In summary, the minimum amount of energy required to move a satellite of mass 7500 kg from its circular orbit of radius 7.3(10^6) m to a location very far away from the Earth is approximately 4.13(10^11) J. This can be calculated using the Energy Principle, taking into account the kinetic and potential energy of the satellite. However, this value needs to be divided by 2 to account for the satellite's initial kinetic energy.
  • #1
brometheus
5
0
A satellite of mass 7500 kg orbits the Earth in a circular orbit of radius of 7.3(10^6) m (this is above the Earth's atmosphere). The mass of the Earth is 6.0(10^24) kg.

What is the minimum amount of energy required to move the satellite from this orbit to a location very far away from the Earth?


We are supposed to employ the Energy Principle to solve this problem, so we start with:

K_i + U_i = K_f + U_f

We know that K (at low speeds) = (1/2)*m*(v^2) and U = -GmM/r


Using the Energy Principle, we know that

K_f - K_i = U_i - U_f

ΔK = -ΔU = -GmM[(1/r_f) - (1/r_i)]

Since r_f is very large, ΔK = GmM(1/r_i)

Using accepted and aforementioned values,
ΔK = [6.7(10^-11) * 7500 * 6(10^24)]/[7.3(10^6)]

This got me approximately 4.13(10^11)J, which is apparently incorrect. What am I doing wrong?
 
Physics news on Phys.org
  • #2
brometheus said:
A satellite of mass 7500 kg orbits the Earth in a circular orbit of radius of 7.3(10^6) m (this is above the Earth's atmosphere). The mass of the Earth is 6.0(10^24) kg.

What is the minimum amount of energy required to move the satellite from this orbit to a location very far away from the Earth?We are supposed to employ the Energy Principle to solve this problem, so we start with:

K_i + U_i = K_f + U_f

We know that K (at low speeds) = (1/2)*m*(v^2) and U = -GmM/rUsing the Energy Principle, we know that

K_f - K_i = U_i - U_f

ΔK = -ΔU = -GmM[(1/r_f) - (1/r_i)]

Since r_f is very large, ΔK = GmM(1/r_i)

Using accepted and aforementioned values,
ΔK = [6.7(10^-11) * 7500 * 6(10^24)]/[7.3(10^6)]

This got me approximately 4.13(10^11)J, which is apparently incorrect. What am I doing wrong?
You are forgetting that it already has kinetic energy. If it gained GmM/R in kinetic energy it would have more than enough energy to escape.

In order to escape, it has to have just enough kinetic energy to get to an arbitrarily large distance from the Earth ie. where it has 0 kinetic and 0 potential energy.

So the condition for escape is: KE + U = 0

Can you write the equation for KEr+ Ur (hint: if there are no external forces acting, does KE + U change?)

Welcome to PF by the way!

AM
 
  • #3
the binding energy of a satellite moving in an orbit is GMm/2r. so i think you ought to divide your answer by 2. [4.13*(10^11)] will be the answer if it is at rest on the Earth's surface.
PS: i am new here. could you please tell me how to start a new thread?
 
  • #4
Ah, okay, that makes sense now. Thanks for your help, and the warm welcome!

To start a new thread I just went to the specific forum (Introductory Physics in this case) and clicked "New Topic" at the top right. It's in the same location as "New Reply" on this page.
 
  • #5


It seems like you have made a slight error in your calculation. The correct value for the change in kinetic energy should be 4.13(10^7)J, not 4.13(10^11)J. This is because the units for the mass of the satellite should be in kilograms (kg), not grams (g). So the correct calculation would be:

ΔK = [6.7(10^-11) * 7500 * 6(10^24)]/[7.3(10^6)] = 4.13(10^7)J

This means that the minimum amount of energy required to move the satellite from its current orbit to a location very far away from the Earth is 4.13(10^7)J. This energy is known as the escape energy or escape velocity.

To put this into perspective, the escape velocity for the Earth is approximately 11.2 km/s. This means that the satellite would need to be accelerated to a speed of at least 11.2 km/s to escape the Earth's gravitational pull and travel to a location far away from the Earth.

It is important to note that this calculation assumes there is no external force acting on the satellite, such as the thrust from a rocket. In reality, a satellite would need to overcome not only the Earth's gravitational pull, but also the drag and atmospheric resistance from the Earth's atmosphere.
 

1. What is escape speed of a satellite?

The escape speed of a satellite is the minimum speed required for it to break free from the gravitational pull of a celestial body, such as a planet or moon.

2. How is escape speed calculated?

Escape speed is calculated using the formula v = √(2GM/r), where v is the escape speed, G is the gravitational constant, M is the mass of the celestial body, and r is the distance from the center of the body to the satellite.

3. Why is escape speed important for satellite missions?

Escape speed is important for satellite missions because it determines the minimum amount of energy needed to launch a satellite into space and maintain its orbit around a celestial body. It also helps scientists plan and execute missions more efficiently.

4. How does escape speed differ for different celestial bodies?

Escape speed varies depending on the mass and size of the celestial body. For example, the escape speed for Earth is about 11.2 km/s, while the escape speed for the Moon is only 2.4 km/s.

5. Can escape speed be exceeded?

Yes, escape speed can be exceeded by a satellite. When this happens, the satellite will escape the gravitational pull of the celestial body and continue on a trajectory into outer space.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
429
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Mechanics
Replies
13
Views
906
  • Introductory Physics Homework Help
2
Replies
39
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
829
Back
Top