Escape Velocity of a spherical asteroid

[SoccaCrazy24]

Consider a spherical asteroid with a radius of 5 km and a mass of 8.65x10^15 kg.
(a) What is the acceleration of gravity on the surface of this asteroid?
ANSWER: ___ m/s2
(b) Suppose the asteroid spins about an axis through its center, like the Earth, with an angular speed . What is the greatest value can have before loose rocks on the asteroid's equator begin to fly off the surface?
ANSWER: ___ rad/s

For (a) which I got right... I used the equation g=(G*M)/R^2
g=(6.67e-11*8.65e15)/(5000^2)=.023078 m/s2

For (b) I used the equation Wf^2= Wi^2 + 2*a*d and i substituted the numbers into get... Wf^2 = 0 + 2(.023037*2pi)(31415.9m*2pi) and I got it to be 95.45 rad/s and for some reason this is not right... am I even using the right equation or is there another equation i can use to get the answer such as maybe... F= (G*M*m)/R^2 ...

 

[Päällikkö]

I've got no idea what you've done in (b), but have you taken gravity into account at all?

I'd use centrifugal force.

 

[BerryBoy]

Consider that, in circular motion:

$$a=\frac{v^2}{r}=\omega^2r$$

If a > g, then what would happen?

Does this help you?

Regards,
Sam

 

[SoccaCrazy24]

so you are saying a =w^2 * r... so w = sqroot(a/r) so w= sqroot(.023037/5000) = .0021465 ?

 

[FishieKissie06]

already tried that method

i already tried that method greg and i got the wrong answer...im not liking this topic at all lol