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In lugnet.space, Ross Crawford writes:
> In lugnet.space, Steve Bliss writes:
> > In lugnet.space, Kyle D. Jackson writes:
> >
> > > Sweet! That would be *so cool*! Imagine sitting out on the deck
> > > looking up at the full moon, and all of a sudden it starts falling
> > > towards you. If we ignore the gravitational damage to the earth
> > > (tides, crust stresses, etc) and the fact that the earth is still
> > > rotating, how long would you get to watch the moon before it landed
> > > on you? The first person to answer will get a cookie(*)!
> >
> > I get 1 hour, 13 minutes. Or 2 hours, 26 minutes. If I could remember the
> > derivitive of y = x^2, I'd be more precise.
>
> That'd be dy/dx = 2x. (I knew that calculus'd come in handy one day!)
Then it should be 2:26. Assuming that the relative acceleration between the
Earth and the Moon is the sum of their local accelerations due to gravity.
And assuming that acceleration is directly proportional to the force of
gravity. Ie, when it's said that the Moon has 1/6 the gravity of Earth,
that means the acceleration due to gravity on the Moon is 1/6 of the
acceleration on Earth.
Earth 1G = 9.8m/s^2
Moon 1/6G = 1.63m/s^2
Combined = 11.43m/s^2
Ack! I made a mistake. Drat. Well, let's go on, and see what new answer I
find.
a = 11.43 m/s^2
v = 11.43t m/s
d = 5.717t^2 m
And the average Earth-Moon distance is 38440100 meters.
So, at d = 38440100 meters, t = 2593 seconds. Joiks! That's only 43 minutes!
So there's my new answer. Now you can all poke holes in my work.
Steve
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Message has 1 Reply:
 | | Re: Couldn't resist
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| (...) Umm... It's not a poke. honest. Your value for the Earth-moon distance is off by a factor of ten. which means your answer should be 2.25ish days. And... The force of the moon's gravity doesn't really enter into this. We know Force = (...) (23 years ago, 3-Jul-01, to lugnet.space, lugnet.off-topic.geek)
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