# How can you find the impact necessary to change the direction of Earth's spin?

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?

• a celestial collision like that would sooner rip the earth apart than reverse its direction of rotation May 19, 2016 at 20:05
• Can we assume the object 'sticks' to the earth after impact? Or does it bounce off? May 19, 2016 at 20:16
• @lemon assuming it sticks on earth May 19, 2016 at 20:35
• 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. May 19, 2016 at 20:39
• youtube.com/watch?v=Fwl_JBQtH9o May 19, 2016 at 21:54

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.