Total Energy

  • Thread starter max0005
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  • #1
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Homework Statement


Determine how much energy is needed to move a satellite of mass 500kg to an orbit of height equal to earth's radius.


Homework Equations



Gravitational Potential:

[tex]V = \frac{-GMm}{R}[/tex]

Where M is the mass of Earth (6*10^24kg), m is the mass of the satellite (500kg) and R is the distance from the center of earth.

The Attempt at a Solution



The energy I need is equal to the difference between orbital and surface potential.

Surface Potential:

[tex]V = \frac{-6.67*10^-11*6*1024*500}{6300}[/tex]

I obtain a value of about -3.17*10^12.

Orbital Potential:

[[tex]V = \frac{-6.67*10^(-11)*6*10^(24)*500}{2*6300}[/tex]

I obtain a value of about -1.59*10^12.

I now subtract the two values: (-3.17*10^12)-(-1.59*10^12) to obtain -1.58*10^11 J or
-158,000MJ... The problem is that my textbook gives as an answer -16,000MJ. Disregarding the difference between 1.58 and 1.6, which is surely due to my approximations, why am I getting a result ten times bigger?
 
Last edited:

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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hi max0005! :smile:

(try using the X2 icon just above the Reply box :wink:)

you need to get it up there and to give it a shove so that it stays in orbit! :wink:

(and wouldn't it be easier to use g = 9.8 instead of G ?)
 
  • #3
52
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Hi,

in this case G is the universal constant -6.67*10^-11.

In this part of the problem I'm considering only the energy needed to bring it up there, the second part asks me to calculate the total energy (included kinetic energy to have it orbit), but I'm not there yet. :)

PS: Sorry about that, I'll go and fix the powers!
 
  • #4
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Your radius should be in m, not km.
 
  • #5
tiny-tim
Science Advisor
Homework Helper
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in this case G is the universal constant -6.67*10^-11.

still easier to use g :smile:

then you don't need to look up the mass of the Earth or its radius …

nor risk mistakes in multiplying them! :wink:
 
  • #6
52
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Your radius should be in m, not km.

Even though all my other lengths are expressed in km?

Ignore this, my bad! :S

still easier to use g :smile:

then you don't need to look up the mass of the Earth or its radius …

Ok, I got lost here... :(
 
Last edited:
  • #7
16
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Ok, I got lost here... :(


I think he is asking you to use the easier(approximation) form of gravitational potential energy equation. (the one that involves g)
 
  • #8
52
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Like mgh? I can't use it, apart from the variation in g as r tends to infinity, I'm being graded on it for using the "extended" formula... Kikora I've tried converting 6300km to 6.3*10^6m but I get as a final result -1.6*10^9... Is it a computation mistake (As I hope it is, after 3 hours of this) or is there some other kind of problem in my reasoning?
 
  • #9
16
0
I think you made a computation mistake. Because, i got the answer just fine after substituting all the values. -1.6X10^10 J
 
  • #10
52
0
Ok, I get the same result as well, thanks!!! :D :D
 

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