- #1
Ron Burgundypants
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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 = Ip(ωf-ωi)
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