Kepler's Laws homework

1. Mar 31, 2008

sillybean

[SOLVED] Kepler's Laws

1. The problem statement, all variables and given/known data

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)

2. Relevant 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)

3. 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?

2. Mar 31, 2008

kamerling

What do you mean with "the formula for a period of rotational motion"

I'd use the fact that gA must be equal to the centripetal acceleration, wich gives something ldifferent from: w=sq root (2gA/RA)

3. Mar 31, 2008

sillybean

I meant the period of rotation formula T=2pi/w (angular acceleration)

4. Mar 31, 2008

sillybean

ah that worked. genius! thanks.

5. Sep 2, 2008

PSEYE

Re: [SOLVED] Kepler's Laws

What worked? I've been trying to figure out how you went wrong for an hour...what did you change?

6. Apr 8, 2011

leliz

Re: [SOLVED] Kepler's Laws

What they meant was that you get the right answer if you set the centripetal acceleration equal to the acceleration due to gravity of the asteroid, solve for velocity, find angular momentum, divide by 3600.
I still don't understand why you couldn't use the equation for escape velocity. Because it's too close to the surface?