Universal Gravitation

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Homework Statement



Given: The universal gravitational constant G = 6.67 E-11, the mass of the earth M = 5.98E24, and its radius R = 6.7 E6. How much work must an external force do on the satellite to move it from a circular orbit of radius 2R to 3R, if its mass is 2000kg? Answer in Joules.


Homework Equations



I know that E = -GMm/2r, where M is the mass of the earth and m = satellite.


The Attempt at a Solution


So I calculated the total energy in each radii,
for 2R, -GMm/4R
for 3R, -GMm/6R
then I subtract one from the other to get the net energy change. I don't know what to do after that. Please help. Thank you!
 

Answers and Replies

  • #2
Kurdt
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There is a considerably simpler way of doing this. The work done in moving an object from one orbit to another is:

[tex] W=\int_{r_1}^{r_2} F(r) dr [/tex]

EDIT: Sorry forgot to say that your method is fine for potential energy not total mechanical energy as you have used.
 
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  • #3
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thank you so much! :)
 
  • #4
Dick
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There is a considerably simpler way of doing this. The work done in moving an object from one orbit to another is:

[tex] W=\int_{r_1}^{r_2} F(r) dr [/tex]

EDIT: Sorry forgot to say that your method is fine for potential energy not total mechanical energy as you have used.

Actually, I think he is ok. The kinetic energy at radius R is GMm/(2R). The potential is -GMm/R. So the sum is, as he states, -GMm/(2R).
 
  • #5
Kurdt
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Was just the procedure in the attempt at the solution that made me think it was total mechanical energy.
 
  • #6
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I have the exact same question. This is one of the method that I used:
E intial = Ki + Ui = 0 + (-GMm/2R)
E final = Kf + Uf = (1/2)(GMm/3R) + (-GMm/3R)
then Work = E final - E intial

this make sense right? Then why it doesn't work

I also try the integrate method, but nothing work! Can somebody please tell me why?
and how to fix it?
 
  • #7
Dick
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If it's in orbit at radius 2R, how can Ki=0?
 
  • #8
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ok, so Ki is not = O; Then why is it the integration method didn't work either?
 
  • #9
Dick
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They both work. You just have to do them right.
 

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