# Spin of planets, bigger means faster?

1. ### valdar

19
Quick question, do bigger planet spin faster?

2. Astronomy news on Phys.org

2,069
Look up the data.

4. ### Nabeshin

2,200
In general, I don't think there's any relation.

5. ### Chronos

9,874
You must look at planet formation to understand this issue. The short answer is yes.

6. ### Oscar Wilde

78
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

8,952
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. ### Oscar Wilde

78

Very interesting. Thank you for providing this

9. ### stevebd1

660
A good way of comparing spin in stars is to compare angular momentum and mass in geometric units where-

$$a=J/mc$$

$$M=Gm/c^2$$

where

$$J=vmr\,k$$

where $a$ 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 $a$ 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 $a$ 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