Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Orbital speed

  1. Apr 5, 2007 #1
    I have a question about orbital speed.
    Imagine a shuttle moving in an orbit near the earth surface,
    its total energy is "-GMm/2r", so its total energy is directly proportional to -1/r

    In order to spiral into another orbit of smaller radius which mean it would have a larger angular speed, the shuttle should increase or decrease its orginal speeed so that it can get into a lower orbital ?

    1) increase
    As orbital speed = (GM/r)^1/2, speed should increase in order to get a smaller r.

    2) decrease
    As total energy directly proportional to -1/r,
    and total energy = PE + KE
    As r decrease, total energy is more negative and KE should decrease
    So the speed should decrease.

    Are there anything wrong in these two contradicting concept ?
    thank you!
  2. jcsd
  3. Apr 5, 2007 #2
    Imagine you want put the shuttle in a higher orbit. You must increase its energy. For this, you accelerate the shuttle, which gets more speed (it is still at the same height). Having a bigger speed, the shuttle does not follow the same orbit, but takes an elliptical trajectory with a bigger apogee. In this trajectory the earth gravitational force brakes the shuttle (do a drawing) and its speed diminishes. At the apogee the energy is always the same but the speed is lower. Now, if you want the shuttle to take a circular orbit with this radius as apogee, you must increment its speed. Otherwise it will follow the elliptical trajectory with the previous perigee.
    You cannot just say "to increase the radius you increase speed". To change radius, you must pass through an elliptical trajectory, and then the speed changes.
  4. Apr 5, 2007 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The total kinetic energy is more properly expressed as

    [tex]E_t = \frac{GMm}{2a}[/tex]

    Where a is the semi-major axis of the orbit( or average radius of the orbit). This allows one to consider elliptical orbits as well.
    Total energy can also be expressed as
    [tex]E_t = \frac{mv^2}{2}- \frac{GMm}{r}[/tex]
    Where r is the radius of your orbit at any given moment.

    If you decrease v, then KE drops, The total energy drops, and 'a' decreases. (You drop into an elliptical orbit with a smaller average altitude.)

    As you start to "fall" you lose PE. To compensate, your velocity increases to increase your KE. When you reach perigee, your velocity is so large, that you start to climb back out, losing velocity as you do so, unitl you return to the point where you decreased v, and you satrt the cycle all over again.
    Last edited: Apr 5, 2007
  5. Apr 5, 2007 #4


    User Avatar

    The total energy is negative (-GMm/2r)as you have written. The potential energy also is negative (-GMm/r). The kinetic energy is positive and is equal to +GMm/2r. Therefore, the kinetic energy is greater in an orbit of smaller radius.
    Once you note the significance of the negativeness of the potential energy and the total energy, your doubt will be cleared.
    Note that the total energy and the kinetic energy are equal in value, but the total energy is negative where as the kinetic energy is positive.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?