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In lugnet.space, Steve Bliss 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(*)!
> 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.
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 = -Gravitationalconstant x Mass(Earth) x Mass(moon) / (seperation)^2
GravitationalConstant = 6.67259x10^-11 Nm^2/kg^2
Mass(Earth) = 5.98x10^24 kg
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.
From the working above
a=0.002
d=0.001t^2
for d=384,400km = 384,400,000m
t=620,000s = 7.17 days.
You'd notice this after about half a day, when the moon would be around
twice as big. After a day, it'd appear four times as big.
At 3hrs before impact, people below would be weightless (i.e between two
bodies exerting equal gravitational forces).
James (who's up past his bedtime figuring numbers.)
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Message has 2 Replies:
 | | Re: Couldn't resist
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| (...) 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 (...) (23 years ago, 3-Jul-01, to lugnet.space, lugnet.off-topic.geek)
| | | Re: Couldn't resist
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| (...) As regard the math, I wouldn't know where to start, but surely even if the moon we're sitting on the earths surface the earth would still exert more force than the moon so you wouldn't be weightless. I don't know if you mean't, at the point (...) (23 years ago, 4-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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| (...) 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 (...) (23 years ago, 3-Jul-01, to lugnet.space, lugnet.off-topic.geek)
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