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Author Topic: for you physicists or rocket scientists out there  (Read 2707 times)

Normandy

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Re: for you physicists or rocket scientists out there
« Reply #15 on: March 29, 2011, 08:23:14 pm »

More specifically, he proved that since you can treat a single shell of mass as a point mass, you can treat a spherically symmetric object as a point mass.

http://en.wikipedia.org/wiki/Shell_theorem#Outside_the_shell
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Heron TSG

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Re: for you physicists or rocket scientists out there
« Reply #16 on: March 29, 2011, 09:40:59 pm »

As long as you are outside the surface of the Earth, yes, you can. Newton, back when he outlined the laws of gravitation, realised that you can integrate the position vectors of a planet across its entirety and approximate it to the centre of mass. There are cases where this does not apply, e.g. for very irregularly shaped objects (picture a giant cup shape, etc), but for spherical or near spherical objects, yes, it is a valid approximation.
As long as you are not within an object, its gravitational field is generally calculated from its center of mass.
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G-Flex

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Re: for you physicists or rocket scientists out there
« Reply #17 on: March 29, 2011, 09:43:33 pm »

As long as you are outside the surface of the Earth, yes, you can. Newton, back when he outlined the laws of gravitation, realised that you can integrate the position vectors of a planet across its entirety and approximate it to the centre of mass. There are cases where this does not apply, e.g. for very irregularly shaped objects (picture a giant cup shape, etc), but for spherical or near spherical objects, yes, it is a valid approximation.

Keep in mind that there is a world of difference between equivalency and a "valid approximation".
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Osmosis Jones

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Re: for you physicists or rocket scientists out there
« Reply #18 on: March 29, 2011, 09:56:54 pm »

As long as you are not within an object, its gravitational field is generally calculated from its center of mass.

Yes, unless you are close to a large object of an interesting shape (say, something not close to spherical?), in which case you can't. Hence the excessive verbiage of my earlier post. Back when I was a Physics 101 tutor, we gave students a range of examples where you could and could not do so.

Keep in mind that there is a world of difference between equivalency and a "valid approximation".

The gravitational effects of a spherical object are equivalent to those from a point mass at the object's centre.  It's a valid approximation because Earth is very near spherical.
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G-Flex

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Re: for you physicists or rocket scientists out there
« Reply #19 on: March 29, 2011, 10:05:14 pm »

Right, I see what you're saying. Of course, if the subject of discussion involves nitpicking gravitational variation with increasing elevation, the Earth's spin, oblongness, etc. are worth mentioning.
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Osmosis Jones

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Re: for you physicists or rocket scientists out there
« Reply #20 on: March 30, 2011, 01:14:39 am »

That's good, because I'm not sure I follow you. It's oblong-ness (nice word, btw) was taken into account by the use of the equatorial radius as opposed to the mean radius, the Earth's spin was accounted for in the second part of my first post, and the differing heights was the whole point of the problem.

Everything that is
Spoiler: relevant to mention (click to show/hide)
was accounted for.

So what are you asking for?

FAKEEDIT: Actually, I didn't twig to this before, and I should really test it mathematically first, but I'm pretty damn sure it's valid; the maths for a spheroid (e.g. the Earth) allows it's treatment as a point mass anyway, because we are looking at it directly above the equator.
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The Marx generator will produce Engels-waves which should allow the inherently unstable isotope of Leninium to undergo a rapid Stalinisation in mere trockoseconds.

Virex

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Re: for you physicists or rocket scientists out there
« Reply #21 on: March 30, 2011, 03:35:25 am »

More specifically, he proved that since you can treat a single shell of mass as a point mass, you can treat a spherically symmetric object as a point mass.

http://en.wikipedia.org/wiki/Shell_theorem#Outside_the_shell
Oh well, that settles it then. Guess I should've payed more attention during physics...
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Nadaka

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Re: for you physicists or rocket scientists out there
« Reply #22 on: March 30, 2011, 11:59:50 am »

If you are going to ask about such a small difference, then it matters where on earth you are talking about.  There are non-trivial differences in gravity in different parts of the globe (rule of thumb is things are heavier closer to the equator).  I'm assuming that from 100 km in the air you mean 100km above sea level.

I'm going to have to have to ask for a source on that. I know of two factors that will make a difference from pole to equator off the top of my head.
Distance from center of mass and the centrifugal effect of the earths rotation. Both of those will reduce the effective weight of an object at the equator due to higher velocity and the roughly oblate shape of the earth.
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Re: for you physicists or rocket scientists out there
« Reply #23 on: March 30, 2011, 05:22:03 pm »

I really wish I knew General Relativity for this.
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Osmosis Jones

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Re: for you physicists or rocket scientists out there
« Reply #24 on: March 30, 2011, 10:26:09 pm »

I really wish I knew General Relativity for this.

NO need :) This works fine under the regime of Newtonian physics. General relativity is more for dealing with cases with large objects moving fast or at large distances (so there is a significant timeframe for the gravity waves to get from one object to the next).
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The Marx generator will produce Engels-waves which should allow the inherently unstable isotope of Leninium to undergo a rapid Stalinisation in mere trockoseconds.
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