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**[SOLVED] Kepler's Laws**

## Homework Statement

Consider an asteroid with a radius of 11 km and a mass of 3.8×1015 kg.

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 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? (answer in hours)

## Homework Equations

gA = G(MA/RA^2)

w (angular velocity) = v esc (escape speed)/RA

*note just some notational stuff: MA = mass of asteroid and RA = its radius

T=2pi/w (angular velocity)

## The Attempt at a Solution

I got part A just fine. Its just a good ol' plug and chug. I got gA= 2.1x10^-3 m/s^2 which is right.

For part B however, I was thinking that if i calculated the angular speed (since its rotational motion) and then just used the formula for a period of rotational motion I could get the seconds and then just easily convert that into hours. Anyway here's my work.

w = sq root (2GM/RA)/RA

since GM/RA is just equal to gA

w=sq root (2gA/RA)

w= sq root(2*2.1X10^-3/11000)= sq root (3.81E-7) = 6.179E-4 rad/s

then i plugged that number into T=2pi/w

T= 2pi/6.179E-4 = 10168.375 s

10168.375 (1/3600) = 2.8 hours

Unfortunately, that answer is wrong. What am i doing wrong?