Satellite orbits and energy

  1. Can't really figure out where to start. Any help would be appreciated.

    1. The problem statement, all variables and given/known data
    A 575 kg satellite is in a circular orbit at an altitude of 550 km above the Earth's surface. Because of air friction, the satellite eventually falls to the Earth's surface, where it hits the ground with a speed of 2.10 km/s. How much energy was transformed to internal energy by means of friction?

    2. Relevant equations
    E = -1/2 G Ms Me / r
    v = 2 pi r / T
    E = - G Me Ms / Re + d

    M = mass
    R = radius

    3. The attempt at a solution
    ok so 575kg, 550000 m above earths surface, falls at 2100 m/s

    5.97 x 10^24 kg is earths mass (Me)
    6378100 for earth radius (Re)
    Ms is mass of satelite
    G is constant

    so for Ek = 1/2mv^2
    = 0.5*575*2100^2
    = 1267875 kJ

    E = - G Me Ms / Re + d
    = (6.67 x 10^-11)(5.97 x 10^24)(575) / (6378100 + 550000)
    = -3.305 x 10^10

    so do I substract E in orbit by Ek at crash (E - Ek)??
     
    Last edited: Feb 11, 2007
  2. jcsd
  3. Well we know energy in these situations is conserved, except that lost to drag and heating of the satellite.

    So the initial energy in orbit equals the final energy at impact plus that lost due to drag. So see if you can set up some equations for the situation based on that. Since it was in orbit, there may be way to simplify the kinetic energy of the satellite.
     
  4. ok, yeah i agree, but im not sure if im using the right equations :uhh:
     
  5. well at least take a stab, post what is the total energy in orbit, then what you compute for the potential energy plus kinetic at impact. From there we can help.
     
  6. i did take a stab and i did calculate it
     
  7. Janus

    Janus 2,424
    Staff Emeritus
    Science Advisor
    Gold Member

    All right let's see if some pointers help

    This will give you the total energy of the satellite in orbit as measured with respect to the center of the Earth. (providing that r is measured from the center of the Earth.)
    This will give you the orbital velocity of the satellite. (not really needed to solve the problem)
    This give you the potential energy of the Satellite while in orbit as measured with respect to the center of the Eartrh
    okay for the kinetic energy when the Satellite strikes the Earth
    Again, this is just the Potential energy part of the total energy of the satellite in orbit.
    What you want to do is find the difference between the Total energy of the Satellite in orbit (the combination of both its kinetic and potential energies), and its total energy when it strikes the Earth (again the combination of both its kinetic and potential energies).

    You've got the formula for finding the total energy in orbit and you've found the kinetic energy at impact. You just need to find the potential energy at impact to complete the picture.
     
    Last edited: Feb 11, 2007
  8. wow thank you,

    so if im understanding correctly

    E = -1/2 G Ms Me / r (total energy in orbit)
    Ek = 1/2mv^2 (kinetic on ground)
    Ep = mgh (potential on ground)

    and therefore E = Ek + Ep + Elost

    Elost = - 0.5(6.67x10^-11)(575)(5.97x10^24) / (6378100 + 550000) - 0.5(575)(2100)^2 - (575)(9.8)(6378100)

    = -5.37 x 10^10

    Im pretty sure thats not right so i probably screwed up sumwhere in my understanding.
     
  9. Janus

    Janus 2,424
    Staff Emeritus
    Science Advisor
    Gold Member

    The problem with this equation, in this situation, is that it assumes that acceleration due to gravity(g) doesn't change with height. This is fine when you are trying to find the potential energy difference between an object sitting on the ground and one 100m above the ground, because g doesn't measurably change btween these two point. This is not the case when it comes to the center of the Earth and its surface.
    Instead, go back and look at the equation that gives the potential energy for the Satellite in orbit. Can you see a way of modifying it so that it gives the potential energy at the surface?
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?