... Consider an asteroid with a radius of 19km and a mass of 3.8x10 15kg .Assume
ID: 2243133 • Letter: #
Question
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Consider an asteroid with a radius of 19km and a mass of 3.8x10 15kg .Assume the asteroid is roughly spherical. g A = 7.0x10 -4 m/s2 Significant Figures Feedback: Your answer 7.021-10-4 = 7.021x10-4 m/s2 was either rounded differently or used a different number of significant figures than required for this part Suppose the asteroid spins about an axis through its center, like the Earth, with a rotational period ' What is the smallest value T can have before loose rocks on the asteroid's equator begin to fly off the surface? At what altitude above the Earth's surface is the acceleration due to gravity equal to g/10?Explanation / Answer
the gravitational force on the surface is
Fg = m*g = m*7.0*10^-4 N
the centrifugal force is
Fc = mw^2*R = mw^2*19000 (w = angular speed)
the rocks begin to fly wen Fg = Fc
m*7.0*10^-4 = mw^2*19000
7.0*10^-4 = w^2*19000
w^2 = 7.0*10^-4/19000
w = 0.0001919 rad/s
now use
T = 2pi/w
T = 2pi/0.0001919 s
T = 32734,64601 s (= 545.57 minutes = 9.093 h )
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