Work done on a satellite ( Physics Problem )

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Homework Help Overview

The problem involves calculating the work done to place a satellite of mass 600 kg into orbit around the Earth at a specified radius. The context is centered on gravitational forces and energy considerations in orbital mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts various calculations using gravitational force and energy equations but finds discrepancies in their answers. Some participants suggest using conservation of energy principles and question the assumptions made regarding kinetic energy at the Earth's surface.

Discussion Status

Participants are exploring different methods to approach the problem, including energy conservation and gravitational potential energy. There is an ongoing dialogue about the assumptions related to kinetic energy and the nature of the satellite's motion. No consensus has been reached, but several avenues for further exploration have been proposed.

Contextual Notes

There are discussions about the assumptions regarding the satellite's kinetic energy at the Earth's surface and whether it can be considered zero. Additionally, the lack of information about the satellite's speed or angular velocity is noted as a constraint in the problem-solving process.

mathsTKK
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Homework Statement


What is the work done to place a satellite of mass 600kg which orbits the Earth at a radius of 8.00 x 10^6 m.

The Attempt at a Solution


1st attempt :
W= F x d
= GmM/(r^2) x r
= GmM/r where r is the radius of the satellite
The answer I got is 3 x 10^10 J which is incorrect.

2nd attempt :
W= F x ∆d
= GmM/(r^2) x ( r - r.E ) where r.E is the radius of the Earth

The answer I got is 6.2 x 10^9 J which is also incorrect.

3rd attempt :
W= ∆E total
= -GmM/2r - ( -GmM/2r.E )
= -GmM/2 x ( 1/r.E - 1/r )

The answer I got is 3.8 x 10^9 J which is also incorrect.

The answer for the question is 2.3 x 10^10 J.
From my three attempt, I think I haven't understand the concept of work and energy of a satellite. Can someone help me to identify my mistakes and rectify it?
I appreciate your kind assistance^^

Thank you for spending your time to read this post and help me to solve the problem ^^

Thank you once again ;):smile:
 
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The first two calcs are incorrect because they assume the force of gravity is constant as the satellite is lifted through the change in radius, and they fail to take into account the kinetic energy needed to increase the speed.

The third one uses an energy formula that is the sum of kinetic and potential energy for circular orbits. It is useful to see its derivation here:
http://www.pha.jhu.edu/~broholm/l24/node1.html
This method also has a failure. It is for a change in orbit, but the satellite is not in orbit when sitting at the surface of the earth. In particular, it is going far too slow to be in orbit at radius r.E, so the kinetic energy part of the formula is way off.

I suggest you use E = ½ m⋅v² - GMm/r.
Be aware - the textbook answer may be based on an incorrect assumption that the kinetic energy is zero at the surface of the Earth. That may not make a lot of difference, but keep it in mind.
 
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Try using conservation of energy:
Ui + Ki + W = Uf + Kf

Solve for Work, and find out the amount of kinetic energy the object has while it is in orbit. You can use the fact that while a satellite is in orbit it's position is constantly turning with the Earth, so you can use Newton's law to find its tangential speed as it is in constant speed with the Earth. With that speed you can find the kinetic energy of the object.
 
Delphi51 :
Thanks for your concern Delphi51 :)
From your suggestion,can I substitute 1/2mv^2 with GmM/2r ?
and by substituting the equation, I got an answer of 1.5 x 10^10 J ,still faraway from the given answer 2.3 x10^10 J. I really wonder why the value is 2.3 ? How to get to the answer.

And one more thing to ask is that why I can't assume that the kinetic energy of the satellite on Earth's surface to be Zero since it is not moving? Is it because that the Earth's itself also move?

Thank you once again ;)
 
thepatient :

Thanks for your concern thepatient ;)

Erm,, regarding your comment I would like to ask, do you mean that I will have to assume that the satellite is geostationary?Because the speed or the angular velocity of the satellite isn't stated in the question. And,, if I would like to use your equation,, how does the working look like? ;)

Thank you thepatient for helping me :) I appreciate your kind assistance ^^
 
Fnet = ma

The net force on the object when it is in orbit is equal to it's mass times its acceleration. We want the satellite to have a uniform circular motion though, so it's net force will be just the inward gravitational force, and a = v^2/R. Use that speed to find kinetic energy of the object.

The equation that I posted is the same conservation of energy equation, but with a few terms rearranged. It states that the energy of the system before is equal to the energy in the system after the work is done. Ui stands for initial gravitational potential energy, Ki for initial kinetic energy, W for work, Uf and Kf for final potential and kinetic energies respectively.
 
mathsTKK said:
Erm,, regarding your comment I would like to ask, do you mean that I will have to assume that the satellite is geostationary?Because the speed or the angular velocity of the satellite isn't stated in the question. And,, if I would like to use your equation,, how does the working look like?

You can find velocity and angular vel. by using radius of orbit, mass of earth.

for this, you must use the eqn, Fcentripetal = FGravitational
 
You could use your -GmM/2r formula for the energy in the orbit to save finding the velocity there.
Use ½ m⋅v² - GMm/r for the energy at the surface of the Earth. You may be able to assume v = 0, though in fact the Earth is rotating and has a velocity at the surface. You can find it by using the circumference of the Earth divided by the time for one complete turn.
 

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