Momentum of a moon after a meteoroid strike

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Homework Help Overview

The discussion revolves around a problem involving the rotational dynamics of Phobos, a moon of Mars, after being struck by a meteoroid. The problem requires understanding the change in rotational angular velocity due to the impact, considering the conservation of angular momentum and the moment of inertia of Phobos.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the meteoroid striking Phobos off-center and question how this affects angular momentum calculations. There are discussions about the relevance of the impact distance and how it factors into the moment of inertia and angular momentum equations.

Discussion Status

Participants have provided insights into calculating the moment of inertia and angular momentum, while some express uncertainty about the role of the meteoroid's impact distance. There is an ongoing exploration of how to relate linear momentum to angular momentum in this context.

Contextual Notes

Some participants note that the mass of the meteoroid is negligible compared to Phobos, and there is a discussion about the initial angular velocity of Phobos, which is not provided in the problem statement.

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Homework Statement



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

Homework 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
 
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The meteoroid strikes Phobos off center so it has angular momentum about the centre of Phobos.
 
Ok, well that confirms the first point. Might need a bit more help with this one though...
 
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.
 
Something isn't right in my second step. Can you tell me where the distance that the meteor impact on Phobos comes into the equation? Otherwise I don't see how to get a solution without the initial angular velocity of Phobos.
 
Ron Burgundypants said:
the mass of the meteoroid is negligible compared with the mass of Phobos, what is the change in the rotational angular velocity of Phobos?
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.
Ron Burgundypants said:
Can you tell me where the distance that the meteor impact on Phobos comes into the equation?
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?
 
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.
 
The problem gives the linear momentum not the angular. Check the units.
 
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Ron Burgundypants said:
Something isn't right in my second step. Can you tell me where the distance that the meteor impact on Phobos comes into the equation? Otherwise I don't see how to get a solution without the initial angular velocity of Phobos.
The offset is used to calculate the angular momentum of the meteor.

PS In case it's not obvious... An body traveling past a point has angular momentum about that point even though it's traveling in a straight line and not rotating.
 
  • #10
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

Thanks for your help.
 
  • #11
Yes that looks correct.
 
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