
#1
Nov1605, 11:05 AM

P: 27

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=(6.67e11*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 in to 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 .... 



#2
Nov1605, 11:34 AM

HW Helper
P: 482

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. 



#3
Nov1605, 11:35 AM

P: 183

Consider that, in circular motion:
[tex]a=\frac{v^2}{r}=\omega^2r[/tex] If a > g, then what would happen? Does this help you? Regards, Sam 



#4
Nov1605, 02:16 PM

P: 27

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




#5
Nov1605, 05:15 PM

P: 6

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



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