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The weight of the moon

 
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Joined: 11 Sep 2008
Posts: 47
Location: Vancouver, WA

PostPosted: Tue Mar 10, 2009 3:14 am    Post subject: The weight of the moon Reply with quote

Just wondering -

If we are so high that we are standing on a space station platform in orbit, we'd be weightless. If we are so low that we are at the very center of the earth, assuming there's a little bubble there that we could exist in we'd be weightless there, too. This leads me to believe that we are the heaviest right on the earth's surface, and if we go higher or lower by whatever means we become lighter.

What I don't know is if we're heaviest at sea level or on Mount Everest. Or if the difference is detectable.
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jabailo



Joined: 20 Mar 2006
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PostPosted: Sun Apr 19, 2009 2:26 am    Post subject: Reply with quote

Maybe another way to look at it is what is the energy required to achieve escape velocity from the point at which we are standing.
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PostPosted: Sun Apr 19, 2009 12:17 pm    Post subject: escape velocity Reply with quote

Interesting. Suppose we dug a theoretical hole 10 miles deep, would it take less energy to achieve escape velocity from that lower starting point? You'd have less gravity to overcome, but you still have to go past ground zero at which point you'd have maximum gravity, but some momentum to slide by it with...

Hmmm
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jabailo



Joined: 20 Mar 2006
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Location: Kent (East Hill), WA

PostPosted: Tue Apr 21, 2009 11:01 pm    Post subject: Reply with quote

F = Gm1m2 / r^2

So, ordinarily you imagine m(ass) 1 and m(ass) 2 as single points.

Now with you (m1) below the Earth (m2), you might imagine spiting the Earth, with a secant and then m2 = m2' + m2'', the mass "above" and "below" you.

So you have

F = F' - F'' = (Gm1m2' - Gm1m2'')/r^2 = Gm1(m2' - m2'') / r^2.

So the total gravity on you is the difference between the mass above and below you.

For escape velocity, you'd have to overcome getting away from m2'' -- but as you lifted off, m2' would be helping you (less and less).

Once you reached the surface you're then dealing with m2 (excluding the air).

Rotation plays a part...that's why they put launch pads as close to the equator as possible.
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Joined: 11 Sep 2008
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Location: Vancouver, WA

PostPosted: Wed Apr 22, 2009 6:35 am    Post subject: The math of it all Reply with quote

Thank you for the formula behind some of that! On first pass reading your formulas I have to admit I sort of glazed over till you said

Quote:
So the total gravity on you is the difference between the mass above and below you.


Which makes complete sense, so the second time reading it through it just made logical sense.

Is r rotation or radius (or something else)? If it's radius it may deal with the factor that much of the mass of the earth is neither above nor below you; it is more to the side of you. On top of Mt Everest for example more of it would be below you and therefore the energy more focused, but also farther away and therefore slightly weaker.

My guess is you would weigh less on Mt Everest but I could see it going the other way too.
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jabailo



Joined: 20 Mar 2006
Posts: 1273
Location: Kent (East Hill), WA

PostPosted: Wed Apr 22, 2009 7:29 am    Post subject: Reply with quote

Yes, I woke up this morning thinking about the r and wondering if I got that wrong!

But again, I think the idea is to tread the masses as single points. Now, being "inside" the earth messes that up, but if you divide it into two masses, above and below, then you can still say they are two points (like being squeezed between two planents on either side of you, one big, one smaller.

Now since the points are right next to you, if you were to do that, the r divisor would be 0 (or approach zero) and F would approach infinity!
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