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Conservation of Energy in Satellite Motion

  1. Sep 5, 2012 #1
    1. The problem statement, all variables and given/known data
    A satellite moves around the Earth in a circular orbit of radius r.
    (a) What is the speed vi of the satellite?
    (b) Suddenly, an explosion breaks the satellite into two pieces, with masses m and 4m. Immediately after the explosion, the smaller piece of mass m is stationary with respect to the Earth and falls directly toward the Earth. What is the speed v of the larger piece immediately after the explosion?
    (c) Because of the increase in its speed, this larger piece now moves in a new elliptical orbit. Find its distance away from the center of the Earth when it reaches the the other end of the ellipse

    2. Relevant equations
    ΔE = ΔK + ΔU = 0
    ΔƩmv = 0

    3. The attempt at a solution
    The apparently correct solution involves using the conservation of momentum. I tried to use the conservation of kinetic energy. My reasoning is that if the smaller piece of mass m is basically at rest after the explosion, it no longer has kinetic energy, thus all of the final kinetic energy is in the moving piece of mass 4m. I assume that the original mass is 5m. Equating the original kinetic energy of the object before the explosion to the kinetic energy of the smaller object, I got the following result:

    [tex] K_i=\dfrac{1}{2}m_iv_i^2=\dfrac{5}{2}m\left(\dfrac{GM_E}{R_E+h}\right) [/tex]
    [tex] K_f=2mv_f^2 [/tex]
    [tex] \dfrac{5}{2}m(\dfrac{GM_E}{R_E+h})=2mv_f^2 [/tex]
    [tex] v_f=\sqrt{\dfrac{5GM_E}{4(R_E+h)}} [/tex]

    This is apparently wrong, but I'm not sure why. One thing that troubles me is that I do not account for potential gravitational energy. Keep in mind the correct answer, according to the solution's manual, is:

    [tex] v_f=\dfrac{5}{4}\sqrt{\dfrac{GM_E}{R_E+h}} [/tex]

    Also I do not know how to begin part (c). Intuitively I'm thinking that if the satellite is on the "other side" of an elliptical orbit, it's distance from the surface of the Earth would be the same as it's distance at it's starting point...suggestions?
  2. jcsd
  3. Sep 6, 2012 #2
    That would be correct - because you do not know whether the explosion transferred any of its energy to the remnants. It very well could.

    The good thing is that after you get the momentum of the larger remnant, you can compute its kinetic energy as well.
  4. Sep 6, 2012 #3
    If the explosion was from fuel on board, then the system would be closed, and energy should be conserved. If the explosion was from a missile, for instance, then the system wouldn't be closed and I wouldn't be able to calculate the momentum of the larger piece of the satellite without first knowing the momentum of the missile at the instant of collision. Yikes!

    But I still don't know how to approach part (c)
  5. Sep 6, 2012 #4
    Total energy would be conserved. But we are talking about the mechanical energy here. Explosions usually convert huge amounts of chemical energy into mechanical.

    I think you will need conservation of energy and angular momentum. Or, if you have already studied motion in a central force field or the two-body problem, you could their results.
  6. Sep 6, 2012 #5


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    Staff: Mentor

    For part (c), you should be able to calculate the orbital total mechanical energy ##\xi## for the object in question. ##\xi## is related to the major axis of an elliptical orbit (how?). Since the pre-explosion condition was a circular orbit, given the conditions of the explosion event it seems that at that instant the object in question will be at the perigee point of its new orbit (why?).

    Note that elliptical orbits do NOT center the ellipse on the orbited body -- the primary is located at one focus of the ellipse. Thus the perigee and apogee are not at the same distance from the Earth!
  7. Sep 6, 2012 #6
    Ah, okay I see what you are saying at the perigee. I wasn't making the connection between how the increase in velocity affected the point on the ellipse (the new orbit) that the planet would now be at. I'll take a closer look at the relationship between the total mechanical energy and the semi-major axis a.

    [tex] E = -\dfrac{GMm}{2a} [/tex]

    I'll give this a shot and see if it comes together.
  8. Sep 6, 2012 #7


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    Staff: Mentor

    ##\mu = G(M + m) \approx GM## since ##M >> m##.
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