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If an object the mass of the moon was to hit earth's surface at an angle that would be in the opposite direction of earth's spin, how can you find how much momentum the object needs to cause earth to spin in the opposite direction?

Qmechanic
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user43783
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    a celestial collision like that would sooner rip the earth apart than reverse its direction of rotation – anon01 May 19 '16 at 20:05
  • Can we assume the object 'sticks' to the earth after impact? Or does it bounce off? – lemon May 19 '16 at 20:16
  • @lemon assuming it sticks on earth – user43783 May 19 '16 at 20:35
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    If the earth has a moment of inertia $I$ and rotates at a rate $\omega$ then the energy of rotation is $\frac{1}{2}I\omega^2$. If you assume all of the kinetic energy $\frac{1}{2}mv^2$ of the incoming mass goes into stopping the earth from rotating then you can equate the two, giving $v=\sqrt{I\omega^2/m}$ as a lower bound (and a crude estimate) of the impact velocity needed. – lemon May 19 '16 at 20:39
  • https://www.youtube.com/watch?v=Fwl_JBQtH9o – RedGrittyBrick May 19 '16 at 21:54

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A simplified answer is obtained by ignoring all possible effects and dependencies other that the mass of the earth (M), its angular velocity ($w$), the mass of the impacting object($m$,) and its velocity ($v$).

The rotational energy of the earth is $I w^2/2$, the kinetic energy of the object is $mv^2/2$. To stop the rotation and then make it go in reverse, kinetic energy must be twice the rotational energy, therefore :$$mv^2/2 = Iw^2$$ Since the moment of inertia (I) for a solid sphere is $2Mr^2/5$ and $w^2 =v^2/r^2 $, substituting in the above equation, one gets: $$ mv^2/2 = (2Mr^2/5) (v^2/r^2)$$simplifying gives:$$m=.8M$$ The speed $v$ (at the equator) is obtained from the known earth parameters = 464m/s. So, an object 80% of the mass of the earth, with a speed of 464 m/s, hitting tangentially at the equator (and "sticking" to it), should do the job.

Guill
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