Orbital motion question

  • Thread starter starbaj12
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  • #1
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Show that the escape speed from a planet is related to the speed of a circulat orbit just above the surface of the planet by Ve = square root(2Vc), where Vc is the speed of an object in the circular orbit.

I'm looking at GMem(object)/r^2 = m(o)v^2/r

But I do not know if this equation will help or how to connect it together with anything else. Any help would be great.
 

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  • #2
pervect
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OK, here's some questions that may help us help you

Do you know the formula for the potential energy of an object a distance R away from a planet - and how this relates to the escape energy?

Do you know how to relate energy to velocity, to translate escape energy into escape velocity?
 
  • #3
arildno
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First of all, let's agree that the equation you've set up gives Vc, since it simply says that the force of gravity produces the necessary centripetal acceleration for circular motion at distance r.

Hence, what you are lacking, is an equation which yields the escape velocity Ve.
Unless you have that, you cannot compare Vc with Ve!

So, question:
How can you find an equation which can be said to yield Ve?
Post a few of your own thoughts on it, particularly what you think is MEANT with the concept "escape velocity"!
 
  • #4
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Are you talking about U(r) = -GMm/r
The bigger the radius the smaller the potential energy

Ve = square root ( 2GMe/Re) = square root(2gRe)

Is this what you are talking about?
 
  • #5
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escape velocity to me means the speed it takes to overcome gravity (air friction negligible) leaving orthogonal from the surface of the planet
 
  • #6
arildno
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starbaj12 said:
Are you talking about U(r) = -GMm/r
The bigger the radius the smaller the potential energy

Ve = square root ( 2GMe/Re) = square root(2gRe)

Is this what you are talking about?
This is correct!
Now compare Ve and Vc!
 
  • #7
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I got it!
Thank you arildno
 

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