# By how much does Haumea's fast rotation affect its surface gravity?

The equation for surface gravity is $$\frac{GM}{r^2}$$ but I'm not sure how to include the effects from its rotation.

• Can you add some more details? – fasterthanlight Nov 2 '20 at 15:02
• – Mike G Nov 2 '20 at 15:32

The formula is simple enough. For force: $$F= \frac {m v^2} {r}$$ where v is the rotational velocity and r is the same r in the gravitational formula. or $$\frac {v^2} {r}$$ if you want the reverse acceleration.
Some Maths, the lifting force on the equator is $$\frac {468.2^2} {1,050,000}$$ = $$0.208 \frac {m}{s^2}$$, slightly over half it's listed equatorial surface gravity of $$0.401 \frac {m}{s^2}$$.