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Gravitational Potential Energy of a satellite

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data

    i) Show that when a satellite (or planet) is in a circular orbit it's kinetic energy (positive) is one-half of it's potential energy (negative).

    ii) Show that in order to escape from the earth you need a speed v=sqrt(2gR) where g=9.8 m/s^2. Neglect Friction and effects of earth's rotation.


    2. Relevant equations

    PE= -GMm/R
    KE= (1/2)mv^2


    3. The attempt at a solution

    I feel like I'm doing some kind of basic algebra wrong here.

    I was trying to do part 1 by solving (1/2)KE=PE

    So, (1/2)(1/2)mv^2 = GMm/R
    (1/4)v^2 = GM/R
    v^2 = 4GM/R
    and v=sqrt(4GM/R)

    So in the process of trying to find the answer of part 1, I got close to the answer for part 2 (If I had made KE=PE), but now I'm just confused. Is the "1/2" in the equation for KE already accounting for this? Should I be looking at it like mv^2=PE?
     
  2. jcsd
  3. Apr 25, 2010 #2

    ideasrule

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    Homework Helper

    Kinetic energy is one-half of potential energy. That means KE=1/2PE, not 1/2KE=PE. However, you shouldn't start off by assuming what you're trying to prove; that just leads to confusing logic. Instead, assume the satellite orbits at a radius R. Find its potential energy, find its kinetic energy, and you'll see that KE=1/2PE.
     
  4. Apr 25, 2010 #3
    So just plug in a random mass and a random radius? I don't have any values to use... I'd need a mass, a radius... Which I could pick anything... But what about velocity? I can't just pick something there. And I can't find velocity without a change in distance and time.
     
    Last edited: Apr 25, 2010
  5. Apr 25, 2010 #4
    Just keep your equations with the variables in them such as r and M. For the velocity, do you know what orbital velocity is equal to? Remember it is in circular orbit. What is the force on it?
     
  6. Apr 25, 2010 #5
    I just had to look that up, because I can't find it in the textbook chapter at all.

    But, I found V=sqrt(GM/r)



    So KE=(1/2)*m*(sqrtGM/r)^2 or (1/2)*m*(GM/r)

    So basically that means
    PE= -GMm/R
    KE=(1/2) GMm/R


    I think thats satisfactory for part 1, if I did that right. Because PE is negative, KE is positive, and KE is 1/2 PE.



    Im not sure how to approach Part II though.

    I think that KE=PE to escape earth... so...
    (1/2)mv^2 = GMm/R
    v^2 = 2GM/R
    and v=sqrt(2GM/R)

    But I'm supposed to find v=sqrt(2gR)


    ...Where am I going wrong here? I think Im confused about what needs to happen to escape earth.
     
  7. Apr 25, 2010 #6
    Yep looks right to me for the first part. To get the orbital velocity all you do is set the force of gravity (GMm/r^2) equal to ma. But the acceleration is centripetal (a = v^2/r) since it is a circular orbit. Solve for v and voila.

    For escape velocity it looks right as well. What does g equal in v=sqrt(2gR)?
     
  8. Apr 25, 2010 #7
    g is 9.8 ms/^2....... But I'm not seeing how that makes it equal to v=sqrt(2GM/R)?
     
  9. Apr 25, 2010 #8
    This is only true on Earth's surface. But we also use it because it is a good approximation when near Earth's surface as well. How do you find g?
     
  10. Apr 25, 2010 #9
    Oooooooook! g = GM/R^2

    So v=sqrt(2gR)..... v=sqrt(2(GM/R^2)R)...v=sqrt(2GM/R)

    Success!

    Much thanks. This is a study guide for an upcoming physics exam, so I really appreciate you helping me to understand where everything was coming from.
     
  11. Apr 25, 2010 #10
    Yup no problem ;)
     
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