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Consider an asteroid with a radius of 14km and a mass of 3.35 x 10^15kg. Assume the asteroid is roughly spherical.

a. What is the acceleration due to gravity on the surface of the asteroid?

b. Suppose the asteroid spins about an axis through its center, like the Earth, with 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?

Answers and Views:

Answer by Yugiantoro
(a).

g = GM/R² = 6.673 E-11(3.35 E15)/(14 E3)²

g = 1.141 E-3 m/s²

(b).

loose rocks on the asteroid’s equator due to rotational effect of asteroid, it means N = 0

∑F = ma

W – N = m ω² r

m g = m ω² r

g = (2π/T)² r

1.141 E-3 = (2π/T)² (14 E3)

T = 22009 s

T = 6.11 hours

Answer by r9ohcop
Force(gravity) = mMG/r^2

if m <<<< M, then you can ignore m.
G = 6.674 x 10^-11 N(m/kg)^2
M = 3.35 x 10^15kg
r = 14km or 14,000m

so Force(gravity) = (3.35 x 10^15kg)(6.674 x 10^-11 N(m/kg)^2)/(14,000m)^2

Force(gravity) = 0.00114070918 m/s^2

for part b, centrepital acceleration = v^2/r

F = ma
mg = ma
a = g, so the centrepital acceleration is equal to the Force(gravity) calculated.

Solve for v.
v^2= ra
v = sqrt(rg)
v = 3.99623929 m/s

that is the minimum speed the asteroid needs so that the the rocks don't fly off.

next, T = 2pi(r) / v
T = 2pi(14,000m) / (3.99623929m/s)
T = 6.1hrs which equals 6.1(3600)sec