1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gravitational Potential and Total Energy

  1. Apr 18, 2005 #1
    Two stars with the mass and radius of the sun are separated by distance 10*r_sun, measured between their centers. A 10,000 kg space capsule moves along a line between the stars.

    Suppose the space capsule is at rest 1.0 m closer to one star than the other. What will be the speed of the space capsule when it meets its ultimate fate?

    Can anyone help with this problem? I thought that you need to to use the energy equation K1+U1=K2+U2 but can seem to get the correct answer.

    My work
    let mass capsule= m_c, mass stars= m

    .5*m_c*0^2 - 2((G*m*m_c)/(5*r_sun)) = .5*m_c*v^2 - ((G*m_c*m)/(r_sun)) - ((G*m_c*m)/(9*r_sun))

    Let me know what I did wrong. Thanks
     
  2. jcsd
  3. Apr 18, 2005 #2

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You could simple calculate the work done by the gravitational force of each stars in the capsule going from 0 to 4R (i.e. when the capsule meets the surface of the star). Add those (actually they will substract). This is the change in kinetic energy of the capsule.
     
  4. Apr 18, 2005 #3
    quasar987, I am not following you. Can you explain your method in more detail?

    Any other options out there?
     
  5. Apr 18, 2005 #4

    StatusX

    User Avatar
    Homework Helper

    Your equation looks right, make sure you did the math correctly.
     
  6. Apr 18, 2005 #5
    the change in the potential energy due to interaction with the first star

    U1=-GMm(1/R-1/5R)

    the change in the potential energy due to ineraction with the second satar
    U2=-GMm(1/9R-1/5R)

    the total change in the potential energy u=U1+U2 is equal kynetic energy of the capsule

    P.S. BTW, the answer does not depend on the capsule's mass
     
    Last edited: Apr 18, 2005
  7. Apr 18, 2005 #6
    Thanks to everyone. I got the answer.
     
  8. Apr 19, 2005 #7

    SpaceTiger

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It's worth mentioning that this is not the way the three-body problem is typically treated. Why? Well, it's simply because the situation is not stable. That is, two stationary stars in close proximity will move towards one another. In fact, you could be a smartass and tell your teacher that the "correct" answer to this problem is wrong because you'd have to take into account the changing potential as a result of the stars moving towards one another, but I wouldn't recommend it. :tongue2:

    Generally, when we treat the three-body problem in astronomy, we consider two stars in orbit about their center of mass. In order to simplify the problem, we generally move to a corotating frame; that is, we continually rotate the coordinate system so that the stars remain at the same points in the space. Unfortunately, when we do this, we sacrifice many of the simplicities of classical mechanics, the most important one being the conservation of energy. That is, a particle's (or spaceship's) energy in a time-varying potential is not necessarily conserved.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?