Spin of planets, bigger means faster?

  1. Quick question, do bigger planet spin faster?
     
  2. jcsd
  3. Dadface

    Dadface 2,080
    Gold Member

    Look up the data.
     
  4. Nabeshin

    Nabeshin 2,200
    Science Advisor

    In general, I don't think there's any relation.
     
  5. Chronos

    Chronos 9,975
    Science Advisor
    Gold Member

    You must look at planet formation to understand this issue. The short answer is yes.
     
  6. I always thought that the smaller planets would rotate more rapidly than larger ones, under identical conditions of course. Much like a figure skater draws in his or her figure and seemingly rotates faster. However, I may be (and more than likely) am wrong. If someone could address this I would be interested to learn
     
  7. mgb_phys

    mgb_phys 8,952
    Science Advisor
    Homework Helper

    Planet Speed at equator (km/h)
    Mercury 10.9
    Venus 6.5
    Earth 1670
    Mars 867
    Jupiter 45600
    Saturn 37000
    Uranus 10900
    Neptune 8460
    Pluto 47

    You would expect small planets to spin faster - from conservation of angular momentum

    But larger planets were formed from larger clouds of stuff.
    As stuff contracted, then because of conservation of angular momentum it speeded up, the bigger/faster the original cloud the faster it ended up. Of course if you took the existing planets and made each of them smaller - they would speed up even more.

    Then there are effects that have happened since. Mercury's rotation is slowed by tidal friction with the sun so has a very slow speed (long day), the Earth's is slowed a little by friction with the moon.
    Uranus probably got hit by something in the past - which is why it has a weird axis tilt.
    Mars might also have been affected by whatever caused the asteroid belt.
     

  8. Very interesting. Thank you for providing this
     
  9. A good way of comparing spin in stars is to compare angular momentum and mass in geometric units where-

    [tex]a=J/mc[/tex]

    [tex]M=Gm/c^2[/tex]

    where

    [tex]J=vmr\,k[/tex]

    where [itex]a[/itex] is the spin parameter in metres, v is the equatorial rotation velocity, m is mass, r is the equatorial radius and k is the moment of inertia coefficient (0.4 for an idealized sphere of uniform density).

    a/M produces a unitless figure between 0 and 1, the higher the number, the higher the spin. For the Sun (k=0.06), a/M=~0.188, for a 2.2 sol neutron star with a frequency of 1500 Hz (k=0.35), a/M=~0.488.

    This doesn't appear so straightforward with planets as M works out considerably smaller than [itex]a[/itex] but there should still be a way of comparing spin geometrically.


    EDIT:
    In the case of planets, you could probably get away with just considering the results of [itex]a[/itex] which is considered to be the amount of angular momentum per unit of mass (sometimes expressed as J/M). In this case, Jupiter is the clear winner and Mercury has the least 'spin' per unit of mass.
     
    Last edited: May 2, 2009
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