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Why is kinetic energy due to a planet's rotation treated as negative?

  1. Jan 25, 2015 #1
    1. The problem statement, all variables and given/known data
    I am trying to answer the following question:
    1. Among the animals that appear in the zoo of the Universe there are black holes and neutron stars. The mass of each of these is often the order of the mass of the Sun. The radius of a neutron star is about 10 km and a certain neutron star rotates at about 100π s-1.
      Calculate the ratio of the polar diameter to the equatorial diameter for a typical neutron star. Assume that liquid or solid neutron material is incompressible. How do we know that it is rotating at this frequency?
    2. Relevant equations


    3. The attempt at a solution

    I was finding it ok, but there was just one part that I cannot understand. In the answers, when the surface of the neutron star was considered as an equipotential, the kinetic energy due to rotation of the star was treated as negative, in the same way as the gravitational potential energy. But surely it should be positive, because many space probes are launched from the equator because the kinetic energy due to rotation gives them some energy to start overcoming the gravitational potential energy, so less fuel is needed. So I don't understand why the gravitational potential energy and kinetic energy are of the same sign... My iPad is not allowing me to copy and paste the derivation here due to formatting, but I have attached the solutions document, and this is question 5 b)
    http://www.physics.ox.ac.uk/olympia..._2_Paper_2013_ Solutions_selected_answers.pdf
     
  2. jcsd
  3. Jan 25, 2015 #2

    mfb

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    Staff: Mentor

    If you consider this in the rotating frame, the rotation generates an effective potential with a maximum at the rotation axis - you gain energy as you move outwards (along the direction of centrifugal force).
    If you consider this in the non-rotating frame, moving an object from the pole to the equator should give zero energy difference to "use". You can gain some energy from the rotation of the object (exactly the same you can calculate in the rotating frame), so the object there has to reach a higher gravitational potential, otherwise you would gain energy.

    By the way: the solution incorrectly assumes the gravitational potential stays GM/r for an oblate spheroid. This is not true. It is a good approximation for most objects, however.
     
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