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In my previous question, I asked how much force it would take to destabilize the Moon's orbit enough for the moon to start falling into the Earth and collide.

Assume this has already happened. Now, the moon is falling towards the Earth.

a) How long would it take until the Moon hits?

b) How much force would it take to stop the moon and bring it back to stable orbit?

c) Would having the moon swing so close change Earth's orbit enough to cause long-term problems? On what magnitude would the tidal forces be?

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  • $\begingroup$ By Newton's third law, the amount of force to return it to the orbit would be equal and opposite to the force required to destabilize it from the previous question. $\endgroup$ – johnpaton Jan 29 '16 at 18:10
  • $\begingroup$ Consider that you have much less time in this scenario. The moon is decelerated over the time frame of a year. Seeing as the moon is now accelerating towards the Earth, you will likely have less time. $\endgroup$ – DevilApple227 Jan 29 '16 at 18:12
  • $\begingroup$ Does "stable orbit" mean - "its old orbit", or "an orbit that doesn't graze the Earth's atmosphere too badly". The latter is more-or-less stable, highly eccentric, and would cause monster tides on Earth. But it would take less effort than putting the moon back in its original orbit - you just have to increase its angular momentum... $\endgroup$ – Floris Jan 29 '16 at 18:56
  • $\begingroup$ @Floris I'm defining a stable orbit as an orbit that doesn't crash into Earth $\endgroup$ – DevilApple227 Jan 29 '16 at 19:22
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Let's simplify the scenario slightly - imagine an object at the distance of the Moon, that has had its angular momentum slowed sufficiently that it will approach the Earth to a distance less than the radius of the Earth plus the radius of the Moon.

Distance Moon-Earth ~ 400,000 km
Radius Moon         ~   1,700 km
Radius Earth        ~   6,300 km
Mass Moon           ~  7.3E22 kg
Mass Earth          ~  6.0E24 kg

The orbit we are interested in has a semimajor axis $a$ less than (400,000+1,700+6,300)/2 = 204,000 km.

The specific orbital energy is given by

$$\epsilon = -\frac{GM}{2a}$$

This means that the energy of the "grazing" orbit of the moon is $6.7\cdot 10^{-11}\cdot 6.0\cdot 10^{24}\cdot 7.3\cdot 10^{22}/(4\cdot 10^8) = 7.2\cdot 10^{28}~\rm{J}$

If you assume that you would need to increase the semimajor axis by 20 km to make sure that the moon just clears the Earth (recognizing that it would give you a horrific tidal pull), then the difference in energy is given by

$$\begin{align} \Delta E &= GMm\left(\frac{1}{2a_1}-\frac{1}{2a_2}\right)\\ &=\frac{GMm}{2a_1}\left(1-\frac{a_1}{a_2}\right) \end{align}$$

So for a change in orbit of 20 km / 400,000 km, you need to add about $3.6\cdot 10^{24}~\rm{J}$. To put this in perspective, the entire world population's energy consumption in 2012 was estimated to be about $5.2\cdot 10^{20}~\rm{J}$; so even nudging the moon a little bit would the world's total energy use for about 7000 years.

As for the time taken - the orbit I described would have a period of $$T=2\pi\sqrt{\frac{a^3}{GM}}\approx 10.6~ \rm{days}$$

It would take half that time to drop from apogee to perigee; you have about 5 days to get that kind of energy together and shoot it to the moon. Not gonna happen.

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  • $\begingroup$ I should have mentioned this in the question, but would coming so close change Earth's orbit around the sun a noticable amount? $\endgroup$ – DevilApple227 Jan 29 '16 at 22:07

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