Consider an asteroid with a radius of 17km and a mass of 3.1*10^15 kg. Assume the asteroid is roughly spherical.
Suppose the asteroid spins about an axis through its center, like the earth, with a rotational period T. What is the smallest value T can have before loose rocks on the asteroid's equator begin to fly off the surface?
Suppose the asteroid spins about an axis through its center, like the earth, with a rotational period T. What is the smallest value T can have before loose rocks on the asteroid's equator begin to fly off the surface?
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Centripetal force = m*R*w^2
Grav. force = G*m*M/R^2 where
m = loose rock mass
R = radius of asteroid = 17e3 m
w = radian rotational spin speed of asteroid = 2pi/T
M = mass of asteroid = 3.1e15 kg
G = grav. constant = 6.67e-11
For loose rock to barely fly off surface, centripetal force = grav. force: R*w^2 = G*M/R^2 or
w = sqrt(G*M/R^3) = 2.05e-4 rad/s or T = 2pi/w = 30,618 s = 8.5 hr, a little more than 1/3 Earth day.
Grav. force = G*m*M/R^2 where
m = loose rock mass
R = radius of asteroid = 17e3 m
w = radian rotational spin speed of asteroid = 2pi/T
M = mass of asteroid = 3.1e15 kg
G = grav. constant = 6.67e-11
For loose rock to barely fly off surface, centripetal force = grav. force: R*w^2 = G*M/R^2 or
w = sqrt(G*M/R^3) = 2.05e-4 rad/s or T = 2pi/w = 30,618 s = 8.5 hr, a little more than 1/3 Earth day.