# A Problem Involving Kepler's Third Law

1. Sep 17, 2007

### ColdFusion85

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

A spacecraft on a mission to the outer planets passed through the asteroid belt and
imaged an asteroid with a small moon. You are on a proposal team to define a
mission to the asteroid’s moon to determine its composition and history and have
been requested to estimate the mass of the moon, which will be used to plan either
a landing or an attachment rendezvous.
You know that the moon’s period of revolution (orbit) around the asteroid has
been observed to be 1.06 days, that the asteroid itself is oblong with the
dimensions of 54x22x20 km and has a density somewhere between 2.1 and 2.9
grams/cc and the farthest separation of the moon from the asteroid in the images
is 100 km in what appears to be a circular orbit. Calculate the mass of the moon.

2. Relevant equations

Kepler's Third Law: P^2 = [(2pi)^2 * a^3]/[k^2 * mu], where P is the orbit period of the body, k^2 is the gravitational constant, mu is the dimensionless reduced mass, and a is the semimajor axis of the orbit.

$$\mu$$ = (1/m2)(m2+m1), where m1 is body in orbit around m2

3. The attempt at a solution

OK. First I got m2 so that I could rearrange Kepler's Third Law equation and solve for m1 (mass of the moon). I did this by multiplying the density given by the volume (dimensons given above). I converted g/cm^3 to kg/m^3 and got a density of 2500 kg/m^3. I converted the volume of the asteroid from km^3 to m^3 and got 23760 km^3 = 2.376 x 10^13 m^3. This gives a mass of 5.94 x 10^16 kg for the asteroid.

I rearranged Kepler's eqn, and got the (m2 + m1) = m2[(2pi)^2 * a^3]/[P^2 * k^2]

I assumed P is measured in seconds (1.06 days = 91584 s), k^2 was given in class notes to be 1.327 x 10^20 m^3/s^2 and and a^3 = (100km)^3 = 1E15 m^3

Carrying this out I get 2106 on the right side, and subtracting m2 I get a negative value for the mass of m1 = -5.94E16. This is obviously wrong and I expect the mass of the moon to be much less than the mass of the asteroid. What am I doing wrong here? Am I using incorrect units/numbers somewhere in my assumptions or conversions above? Thanks for any help you may be able to provide.

2. Sep 17, 2007

### ColdFusion85

Oh, I forgot to mention that we were told to use the average density above, so that is why I used 2.5 g/cm^3

3. Sep 17, 2007

### ColdFusion85

Anyone?

4. Sep 17, 2007

### D H

Staff Emeritus
You are using the Sun's gravitational parameter, not the asteroid's. Some planetary gravitational parameters in m3/s2:
\begin{aligned} \text{Sun}\quad &133\times10^{18} \\ \text{Earth}\quad &399\times10^{12} \\ \text{Moon}\quad &4.9\times10^{12} \\ \text{Ceres}\quad &63\times10^{9\phantom{0}} \end{aligned}

Last edited: Sep 17, 2007
5. Sep 18, 2007

### ColdFusion85

Oh, right. Stupid me. I think I got it now - 1.12 E16 kg. Thanks