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In lugnet.space, James Howse writes:
> From Newton's first law,
>
> Force = mass x acceleration
>
> so acceleration = Force(moon)/mass(moon)
>
> Acceleration = -G x Me / r^2
>
> hence, the acceleration at the start of the problem is more like 0.002m/s^2
> rather than 11.43!
> Although the acceleration is dependent on the distance (making the problem
> very hideous) we can assume that it is constant for some 95% of the trip. at
> the value given above.
Actually, when calculating the acceleration and taking the distance into
account, it's not so bad. For my first attempt at solving this, I tried
converting gravitational potential energy into kinetic energy. The resulting
integral was definitely hideous.
I used the gravitational acceleration that you mentioned: a = G x M / r^2
We also know that acceleration, a = dr/dt^2
If you integrate (and ignore constants, since we know we are starting with
zero velocity), you get:
t = [ 2 x R^3 / (3 x G x M) ]^0.5
which we can use to find out how long it would take any distant body to fall
from rest into another, much more massive body.
If we plug in 3.8e08m for R, and 6e24kg for the mass of the earth, we get a
time ~ 3.5 days (ignoring the radii of the earth and moon, and the motion
of the earth toward the moon).
Jeff J
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Message has 1 Reply:
 | | Re: Couldn't resist
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| I should probably try to solve the problem myself, it'll help as prep for my college entrance exam. ;-) (...) Hmm, not bad at all - that gives people time enough to *realize* it's happening, broadcast it all over the world, and let everyone go (...) (23 years ago, 3-Jul-01, to lugnet.space, lugnet.off-topic.geek)
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Message is in Reply To:
| | 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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