# Astrophysics rotational break up velocity

1. Oct 15, 2008

### nissanztt90

1. The problem statement, all variables and given/known data

Write out expressions for the rotational speed (at the equator) and rotational period of an object mass M and radius R that is spinning at the break up velocity.

2. Relevant equations

$$F= mg$$

$$F = m \omega^2 r$$

$$g = \frac{mG}{r^2}$$

3. The attempt at a solution

So i know that for something to be "flung" off the spining mass, the centrifugal acceleration would need to be greater than the gravitational acceleration.

So i just set them equal to each other, and solve for omega...

which is...
$$\sqrt{\frac{mg}{r^3}} < \omega$$

And for the rotation period...

$$p < \frac{2\pi}{\sqrt{\frac{mG}{r^3}}}$$

Is this correct?

Last edited: Oct 15, 2008
2. Oct 15, 2008

### l'ingenieur

That's correct. You should just write your answers with M, G and R instead to avoid confusion. And the break up velocity is reached with an "=" sign, not ">" or "<".

You're basic force equation for a masse "m" would be:
F = -mg +N = -m(v^2)/r, but since the normal force N equals to 0 you get
g = (v^2)/r
After that you just need to replace with the angular velocity and the gravitational field strenght equations, which you have figured out.

So the earth would need to spin 17 times faster for us to feel no gravity!

3. Oct 15, 2008

### nissanztt90

Thanks!...we had to calculate how fast the sun needed to spin as well...calculated a period of 2.8 hours...i was quite surprised to say the least.