# Does angular speed of the Earth around the Sun remain constant?

I presume that angular speed is nothing but angular velocity without direction. So, since the earth always takes nearly 365 days to complete one revolution around the sun, can't we conclude that it's angular speed is constant ($19$ x $10^{-7}$ RPMs) ? Most of the textbooks say that we cannot.

Ignoring the (minor) effects due to the other planets, the angular momentum of the Earth-Sun system must be conserved, and the angular momentum is given by (making the approximation that the Sun is fixed):

$$L = \omega m_e r_e^2$$

where $m_e$ is the mass of the Earth and $r_e^2$ is the Earth-Sun distance. A quick rearrangement to get the formula for the angular velocity gives:

$$\omega = \frac{L}{m_e}\frac{1}{r_e^2}$$

So $\omega \propto 1/r_e^2$, and since the Earth's orbit is an ellipse that means $r_e$ changes throughout the orbit and therefore the angular velocity must change as well.

• Would it mean that for any body rotating about its own axis, any change in its linear velocity must change its angular velocity? In the case of a ball rotating about its own axis, particles that are closer to axis of rotation have greater linear velocity than those farther away, but the angular velocity of every particle is the same. How do we explain this? Sep 17, 2014 at 9:13
• @Swami: in a solid object rotation causes elastic stresses in the object and all the particles (i.e. volume elements) feel these stresses and interact with each other. The result I quote applies only to a single partice moving in a centrally symmetric potential. Sep 17, 2014 at 9:16
• Which means - had the all particles not been in an attached state, every particle of the ball would try to rotate with different angular velocity and thus deform the ball? Sep 17, 2014 at 9:19
• @Swami: if the particles weren't attached to each other they would feel no force and would simply fly off at a tangent. Sep 17, 2014 at 9:21