Gravitational Potential

  • Thread starter athymy
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  • #1
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Hey guys, I'm having some trouble with the energy topic. So the question is:

"Calculate the gravitational potential of earth and of Mars. Then use that information to calculate the gravitational potential energy required to life yourself completely off each planet"

So I got the gravitational Potential on earth to -62.5 MJ/kg and on Mars it is 12.7 MJ/kg. So this is the first part of the questions, I don't get how you calculate the energy required to lift my enitre mass (which is about 70kg) off each planet.

I would really appreciate it if anyone could help me out.
 

Answers and Replies

  • #2
gneill
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Hey guys, I'm having some trouble with the energy topic. So the question is:

"Calculate the gravitational potential of earth and of Mars. Then use that information to calculate the gravitational potential energy required to life yourself completely off each planet"

So I got the gravitational Potential on earth to -62.5 MJ/kg and on Mars it is 12.7 MJ/kg. So this is the first part of the questions, I don't get how you calculate the energy required to lift my enitre mass (which is about 70kg) off each planet.

I would really appreciate it if anyone could help me out.

You have the potential energy in terms of Joules per kilogram, and you have your mass in kg. What's the problem?
 
  • #3
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You have the potential energy in terms of Joules per kilogram, and you have your mass in kg. What's the problem?

Well, I'm not sure how I should interpret "completely. I know that if I multiply my mass times the GP I will get an answer in terms of energy but how does the "lift completely" affect it?
 
  • #4
gneill
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Well, I'm not sure how I should interpret "completely. I know that if I multiply my mass times the GP I will get an answer in terms of energy but how does the "lift completely" affect it?

Ah. Well, if you lift a body very, very far away indeed ( r → ∞ ), what happens to the gravitational potential energy betwixt them?
 
  • #5
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Ah. Well, if you lift a body very, very far away indeed ( r → ∞ ), what happens to the gravitational potential energy betwixt them?

It approaches 0 right? But I thought that the question was asking for the PE required to lift myself from the surface of the earth/mars. I'm so confused right now... :cry:

EDIT: It's asking for the PE required to " take you comepletely off each planet".
 
  • #6
gneill
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It approaches 0 right? But I thought that the question was asking for the PE required to lift myself from the surface of the earth/mars. I'm so confused right now... :cry:

EDIT: It's asking for the PE required to " take you comepletely off each planet".

So presumably "completely off each planet" implies that you would no longer fall back to it; you would no longer be influenced (gravitationally) by it. That would occur at effectively infinite separation (if you and the planet were the only bodies in the universe).
 
  • #7
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Got ya! So W=∫rFdr, I let r approach ∞, but won't that give me 0J?
 
  • #8
gneill
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Got ya! So W=∫rFdr, I let r approach ∞, but won't that give me 0J?

The reference point for gravitational potential energy is at infinity (which is why you have increasingly negative values as you approach a body). The gravitational potential energy represents the work required to bring a body from infinite distance to some distance r from another body. The negative value means that you actually get energy out of the field in the process. Gravity is a conservative field, so it will take the same amount of energy to remove the body...
 
  • #9
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The reference point for gravitational potential energy is at infinity (which is why you have increasingly negative values as you approach a body). The gravitational potential energy represents the work required to bring a body from infinite distance to some distance r from another body. The negative value means that you actually get energy out of the field in the process. Gravity is a conservative field, so it will take the same amount of energy to remove the body...

Ah, I think I'm getting this now, so no matter what planet I'm on, I'll need (technically) ∞J of energy to "lift completely" because if I'm somewere in the universe, the planets will pull me towards them (even though that pull will be extremely small).
 
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  • #10
gneill
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Ah, I think I'm getting this now, so no matter what planet I'm on, I'll need (technically) ∞J of energy to "life completely" because if I'm somewere in the universe, the planets will pull me towards them (even though that pull will be extremely small).

Nope. The gravitational potential energy represents the energy you'd get dropping from infinite distance to r (presumably it being converted to kinetic energy during the fall). That's all you'd get. It would therefore take exactly the same amount of energy to start at r and be removed to infinite distance.
 
  • #11
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Ughh, so it would take 60MJ for me to lift completely and leave earths gravitational field?
 
  • #12
gneill
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Ughh, so it would take 60MJ for me to lift completely and leave earths gravitational field?

How did you arrive at that number? What's your potential energy at the Earth's surface?
 
  • #13
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I used the gravitational potential formula V=-Gm/r and that gave me -60MJ/kg
 
  • #14
gneill
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I used the gravitational potential formula V=-Gm/r and that gave me -60MJ/kg

In your first post you stated -62.5 MJ/kg. But okay, use 60. But that's 60 MJ PER KILOGRAM. How many kilograms do you need to lift?
 
  • #15
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Oops, sorry about that. I need to lift my mass which is 70kg.
 
  • #16
gneill
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Oops, sorry about that. I need to lift my mass which is 70kg.

So then the result is...
 
  • #17
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60MJ/Kg*70Kg=4200MJ (?)
 
  • #18
gneill
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60MJ/Kg*70Kg=4200MJ (?)

Yup, if 60 MJ/kg is indeed the value you wish to use.
 
  • #19
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Alrighty! Thank you so much for all the help, I really appreciate it :)
 

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