- #1

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much difference?

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- Thread starter stany
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- #1

- 24

- 0

much difference?

- #2

- 103

- 1

(Wikipedia)[...] this causes the Moon to slowly recede from Earth at the rate of approximately 38 millimetres per year.

- #3

- 24

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so what would dat be?

- #4

russ_watters

Mentor

- 20,413

- 7,015

You can't calculate it yourself? Make an effort. This isn't a place to have things spoon-fed to you.

- #5

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you think that would make the moon appear much closer? i mean if suddenly the moon appeared 5 miles closer would you notice?

- #6

- 1,033

- 1

384,403 kilometers is the number i got for an average. 4.7 miles is about 7 kilometers.

would you notice a change of 7 kilometers if you were almost 400 000 kilometers away?

- #7

- 17

- 0

- #8

- 24

- 0

384,403 kilometers is the number i got for an average. 4.7 miles is about 7 kilometers.

would you notice a change of 7 kilometers if you were almost 400 000 kilometers away?

yes.

- #9

- 1,033

- 1

you have fantastic depth perception then.

- #10

russ_watters

Mentor

- 20,413

- 7,015

The effect of the impacts is miniscule becaues they make up a tiny fraction of the mass of the moon. 200,000 years is a pretty short time, so it is unlikely that it has changed much in that time.

- #11

- 103

- 1

Interesting question, even though, I think, it's based on a misconception. The recession is not an inertial but a friction effect, namely the result of a constant conversion of lunar orbital momentum into tidal effects on Earth. So, counteracting this effect by bombarding the moon with impactors would be not like changing the direction of a rolling car by throwing balls at it, which is tedious but easily doable given enough balls, but like trying to keep the car from eventually stopping due to energy loss, which is much harder in the long run.

Quantitatively, the figure means that the Moon's distance from Earth increases by one part in 10^10 per year, which roughly corresponds to a change in orbital speed by the same fraction, so the annual change in momentum is

delta p ~ lunar mass * change in lunar orbital speed ~ 10^23 kg * 10^3 m/s / 10^10 ~ 10^16 kg m/s

The typical speed of an impactor is on the order of tens of kilometres per second, so counteracting this would require an asymmetric bombardment on the order of 10^12 kg per year. I wasn't able to find a cumulative figure anywhere, but several publications gave total impact masses for certain impactor size ranges, and all of those were in the range of 10^3 to 10^6 kg per year per mass decade (i.e. impactor mass between 10^n and 10^(n+1) kg), so I'm guessing the grand total won't exceed 10^9 kg. And most of that is symmetric, so the momentum contributions will cancel each other out.

So, the answer seems to be that the effect of impacts is negligible in comparison with that of tidal losses, unless I made a major mistake somewhere.

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