Gravitational potential energy: derive expression for energy

  • #1
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Homework Statement


Part A:
Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.

Part B:
Ignoring Earth's rotation, how much energy is needed to get the same object into orbit at height h?

Express your answer in terms of some or all of the variables h, mass of the object m, mass of Earth mE, its radius RE, and gravitational constant G.


Homework Equations


U=-GmmE/RE+h

The Attempt at a Solution


I figured that the energy you would need to bring the object to height h above the surface would equal the difference between the energy at the surface and the energy at h. I assume the object is at rest and the only energy it has is gravitational potential.

E=-GmmE/RE-(-GmmE/RE+h)....E=GmmE/h

This is not correct. Feedback I got was that the answer depends on RE. I've no idea how to go about part B.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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The Attempt at a Solution


I figured that the energy you would need to bring the object to height h above the surface would equal the difference between the energy at the surface and the energy at h. I assume the object is at rest and the only energy it has is gravitational potential.

E=-GmmE/RE-(-GmmE/RE+h)....E=GmmE/h

From this line, how did you reach the right side?

[tex] U = -\frac{Gmm_E}{R_E} - \left( - \frac{Gmm_E}{R_E +h} \right)[/tex]

Your denominators don't cancel out.
 
  • #3
Dick
Science Advisor
Homework Helper
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Homework Statement


Part A:
Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.

Part B:
Ignoring Earth's rotation, how much energy is needed to get the same object into orbit at height h?

Express your answer in terms of some or all of the variables h, mass of the object m, mass of Earth mE, its radius RE, and gravitational constant G.


Homework Equations


U=-GmmE/RE+h

The Attempt at a Solution


I figured that the energy you would need to bring the object to height h above the surface would equal the difference between the energy at the surface and the energy at h. I assume the object is at rest and the only energy it has is gravitational potential.

E=-GmmE/RE-(-GmmE/RE+h)....E=GmmE/h

This is not correct. Feedback I got was that the answer depends on RE. I've no idea how to go about part B.

U=-G*m*mE/(RE+h). Review your algebra about subtracting fractions! You are doing it wrong.
 
  • #4
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0
From this line, how did you reach the right side?

[tex] U = -\frac{Gmm_E}{R_E} - \left( - \frac{Gmm_E}{R_E +h} \right)[/tex]

Your denominators don't cancel out.

E=(-GmmE/Re)+(GmmE/Re)+(GmmE/h) First two terms cancel? I may have forgotten basic algebra.
 
  • #5
Dick
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E=(-GmmE/Re)+(GmmE/Re)+(GmmE/h) First two terms cancel? I may have forgotten basic algebra.

Yes, forgotten basic algebra. a/(b+c) is not the same as a/b+a/c. Find a common denominator etc.
 
  • #6
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0
E=(-GmmE+h/Re+h)+(GmmE/Re+h)

Can't believe I did that. Can I add +h to the numerator and denominator of the first term? Is that finding a common denominator? Then I get E=h/(Re+h).
 
  • #7
Dick
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E=(-GmmE+h/Re+h)+(GmmE/Re+h)

Can't believe I did that. Can I add +h to the numerator and denominator of the first term? Is that finding a common denominator?

No, that's equally bad. a/b-c/d. Find common denominator, bd. Multiply first term by d/d, second by b/b getting ad/bd-bc/bd. Now you have a common denominator bd so you get (ad-bc)/bd. I hope this sounds familiar.
 
  • #8
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No, that's equally bad. a/b-c/d. Find common denominator, bd. Multiply first term by d/d, second by b/b getting ad/bd-bc/bd. Now you have a common denominator bd so you get (ad-bc)/bd. I hope this sounds familiar.

Reworked and got E= [-GmmE(Re+h)+Re(GmmE)]/Re(Re+h)=GmmEh/Re(Re+h)

Doing the algebra correctly, is the answer right? Is my reasoning sound? Sorry for forgetting elementary school, and thanks for your help.
 
Last edited:
  • #9
Dick
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Reworked and got E= [-GmmE(Re+h)+Re(GmmE)]/Re(Re+h)=GmmEh/Re(Re+h)

Doing the algebra correctly, is the answer right? Is my reasoning sound? Sorry for forgetting elementary school, and thanks for your help.

Yes, that looks much better. For part B you need enough energy to reach the radius of the orbit and then more energy to get to the correct velocity to maintain that orbit.
 
Last edited:
  • #10
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0
Yes, that looks much better. For part B you need enough energy to reach the radius of the orbit and then more energy to get to the correct velocity to maintain that orbit.

Very helpful. I got to the right answer. Thank you!
 
  • #11
1
0
For part B, the energy that is extra and used to get the correct velocity, would that be kinetic or gravitational potential?
 

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