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Homework Help: General Gravitational Potential Energy (Help)

  1. Feb 16, 2007 #1
    1) 2 stars with the mass and radius of the Sun are separated by distance 10R_s between their centers of mass. (where R_s is the radius of the Sun). A 10000kg space shuttle moves along a line between the stars. Draw a graph of the space shuttle's potential energy (as a function of position) along a line passing through the 2 stars. Let one star be at x=0 and the other at x=10R_s

    I have no clue...Can anyone give me some hints?

    I know the formula for the grav. potential energy between TWO masses is -GMm/r. But in this case, the system has THREE masses, what should I do? (I have never seen something similar to this.) How would the "shape" of the graph look like?

    I hope that someone can help me out. Thanks a lot!
  2. jcsd
  3. Feb 16, 2007 #2


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    Potential energy adds up like scalar quantities. Each of the stars does work on the shuttle as it approaches/recedes from them. Therefore add the potential energy of the two stars point by point as a function of the distance from the midpoint between the two stars.
  4. Feb 16, 2007 #3
    Thanks for your response...

    So would the resulting graph look something like a parabola that opens down?
    Last edited: Feb 16, 2007
  5. Feb 17, 2007 #4
    If I place the shuttle 1m closer to one star than the other, how is it possible to calculate the final impact speed with the closer star? How can I account for the fact that there are 3 objects in this system rather than two? I just don't know how to set up the "potential energy" part in the conservation of mechanical energy equation...
  6. Feb 18, 2007 #5
    Can someone please help me out?
  7. Feb 22, 2007 #6
    Can someone please help me out?
  8. Feb 23, 2007 #7
    First calculate the gravitational pull between the shuttle and the star at x=0 this would be F1= -GM1m/r...then with the star at 10R_s, this would give
    F2=-GM2m/(10R_S-r).M1=M2, u can replace them by M. and since the forces are opposite each other u would have to substract them from each other, and no u don't get a parabola because when u'r close to one star the pull is opposite to x, but when u pass the midpt the force reverses, so u must get a sloped curve.
  9. Feb 23, 2007 #8


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    The shuttle can be thought of like a "test charge" used to map out the potential field. Its presence do not alter the gravitational field significantly. This means that the potential can be obtained by just adding the potentials generated by the two stars from one point to the other.

    The graph will most likely look like you suggests. It drops down steeply from the midpoint towards each star, that is the shuttle is falling deeper and deeper into the potential well created by the stars. The graph will be symmetrical towards each star so you can calculate it towards one star only and mirror the other part of the graph. Once you go inside of a star the contributing mass will be just that amount which is "below" the shuttle, that is just the mass that is located in a circle with a radius up to the point where the shuttle is located (or don't you need to calculate it inside?).
    Last edited: Feb 23, 2007
  10. Feb 23, 2007 #9
    wait a minute here,.isn't Potential energy the mass*gravitational pull*height?
    when i say gravitational pull i mean w.r.t the star its falling towards. shouldn't it be like the energy that this object will have if it started falling towards a certain star considering the other star was suddenly removed. Cause when u calculate ur P.E w.r.t earth u don't take into account other forces that are being applied to you. Example: ur tied up to a plane that is dragging u and stopping u from falling, but ur potential energy will be calculated according to the theory of energy u will have if the string suddenly brakes. so when calculating P.E u can't take into account 2 gravitational pulls, u have to take 2 P.Es one w.r.t each star.
  11. Feb 23, 2007 #10
    Think about what form the gravitational force takes. You should see that the height factor you included in your last post cancels, so you end up with mass*gravitational potential.

    Gravitational potential and gravitational field strength are NOT the same thing.
  12. Feb 23, 2007 #11
    can u explain more where the equation is wrong, and how to correct it?
  13. Feb 23, 2007 #12


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  14. Feb 23, 2007 #13
    this confirms what i said in my second post ! the potential gravitational pull is only related to the pulling mass and the mass in question, no relation with the other mass, which in this case is the second star
  15. Feb 23, 2007 #14
    quote"this confirms what i said in my second post ! the potential gravitational pull is only related to the pulling mass and the mass in question, no relation with the other mass, which in this case is the second star"

    :confused: :confused:
    Remember that both are pulling even if the acceleration is toward one of the stars.

    Also, why can't I choose to insert a quote in my response? Sorry to thread jack.
  16. Feb 23, 2007 #15
    u r considering potential !! don't forget. Lets reconsider u tied up by a rope from a building at 5m height. the rope is the gravity of the second star that's holding u from fallin back. but when i ask u what is ur P.E u say mgh and u take height that u r currently on and forget whether another force is pulling u in another direction....i'm also trying to convince myself with this theory, but i think it makes sense.
  17. Feb 23, 2007 #16
    btw. both r pulling also in my example. the gravity of the earth is pulling and the rope is pulling opposite to it
  18. Feb 23, 2007 #17
    Whether you are trying to find the total potential or net force you need to consider all masses.
  19. Feb 23, 2007 #18
    how do u calculate P.E with respect to two masses?
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