# Homework Help: Momentum of a moon after a meteoroid strike

1. Aug 1, 2017

### Ron Burgundypants

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

Phobos is a small moon of Mars. For the purpose of the following problem, assume that Phobos has a mass of 5.8x10^15 kg and that it has a shape of a uniform sphere of radius 7.5x10^3 m. Suppose that a meteoroid strikes Phobos 5.0x10^3 m off center and remains stuck. If the momentum of the meteoroid was 3x10^13 kg m/s before impact and the mass of the meteoroid is negligible compared with the mass of Phobos, what is the change in the rotational angular velocity of Phobos?

Below is a list of all the variables extracted from the text and labelled

Phobos mass: (Mp) = 5.8x10^15 kg
Phobos radius: (Rp) = 7.5x10^3 m
Phobos initial momentum: (Lpi) = Unknown
Phobos final momentum: (Lpf) = Unknown
Moment of inertia of phobos: (Ip) = 2/5 MR2
Meteoroid initial momentum: (Lmi) = 3x10^13 kg m/s
Meteoroid final momentum: (Lmf) = 0
Moment of inertia of meteoroid: (Im) = 2/5 MR2 ( although this could also be that of a point particle (MR2)

Meteoroid Strike zone: (D) = 5.0x10^3 m

2. Relevant equations

Conservation of angular momentum = Li = Lf. (i = initial and f = final)
Angular momentum: L = Iω
Moment of inertia of a sphere (phobos): (Ip) = 2/5 MR2
Parallel axis theorem: I + MpD2

3. The (first) attempt at a solution

So the first question or assumption I had to make was that the the angular velocity that the question is asking for is not that found from phobos' orbit of the moon, but the rotation of it on its own axis. Is this fair to assume? If it were the orbit we would need figures for the moons gravitation pull right? Onward!

Step 1.
So at first I thought the conservation laws were a good place to start and declared that the initial momentum of phobos + the initial momentum of the meteoroid should be equal to the final momentum of phobos. As the meteoroids mass is negligible it disappears from the right hand side of the equation

Lpi + Lmi = Lpf + Lmf
Lpi + Lmi = Lpf

Step 2.
From here it was pretty straightforward to just substitute L for Iω, subtract it from the left and then collect the like terms (assuming that the moment of inertia stays the same). By then dividing through you would get the change in angular velocity.

Lmi = Ipfi)
Lmi / Ip = Δω

However, This clearly can't be correct as I haven't used the impact distance of the meteor (meteor strike zone: D), the question is where does it come in? Can it be used in the parallel axis theorem?

If it were to be subbed into the moment of inertia for phobos then it gets messy... So. Thoughts, hints, tips?

Tak

2. Aug 1, 2017

### CWatters

The meteoroid strikes Phobos off center so it has angular momentum about the centre of Phobos.

3. Aug 1, 2017

### Ron Burgundypants

Ok, well that confirms the first point. Might need a bit more help with this one though...

4. Aug 1, 2017

### CWatters

Where are you stuck?
You can calculate the moment of inertia of Phobos.
You can also calculate the angular momentum that the meteor adds to Phobos.

5. Aug 1, 2017

### Ron Burgundypants

Something isn't right in my second step. Can you tell me where the distance that the meteor impact on Phobos comes in to the equation? Otherwise I don't see how to get a solution without the initial angular velocity of Phobos.

6. Aug 1, 2017

### haruspex

That's a bit sloppy. Taken literally, that would mean there is no change worth mentioning. Better to say it is small compared with the mass of Phobos.
I assume you are using the centre of Phobos as your axis for angular momentum.
How are you calculating the initial angular momentum of the asteroid about that point?

7. Aug 2, 2017

### Ron Burgundypants

I didn't write the question... Sloppy it may be but that's what it says. The angular momentum of the meteoroid isn't calculated it's just given.

8. Aug 2, 2017

### CWatters

The problem gives the linear momentum not the angular. Check the units.

9. Aug 2, 2017

### CWatters

The offset is used to calculate the angular momentum of the meteor.

PS In case it's not obvious... An body travelling past a point has angular momentum about that point even though it's travelling in a straight line and not rotating.

10. Aug 2, 2017

### Ron Burgundypants

Ah ok, I didn't spot that... doh...

So I was nearly there... is this correct below? I've converted the linear momentum of the meteoroid to angular momentum by multiplying by the offset. So the momentum of the meteoroid is now Pm and the offset (where it impacts phobos) is still D.

pi + PmD = Iωpf
PmD = I (ωpf - ωpi)
PmD / Ip = Δωp