Escape Velocity


by SoccaCrazy24
Tags: escape, velocity
SoccaCrazy24
SoccaCrazy24 is offline
#1
Nov16-05, 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=(6.67e-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 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 ....
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Päällikkö
Päällikkö is offline
#2
Nov16-05, 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.
BerryBoy
BerryBoy is offline
#3
Nov16-05, 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

SoccaCrazy24
SoccaCrazy24 is offline
#4
Nov16-05, 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 ?????
FishieKissie06
FishieKissie06 is offline
#5
Nov16-05, 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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