# Celestial body integration

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• Nov 28th 2006, 03:27 PM
TriKri
Celestial body integration
I would like to prove that a celestial body's gravitational force is the same as if it would have been a point.

Suppose that the density is the same all over at a certain depth into the body. (It may be less at the surface and more halfway down to the center for example). The density is given by $\displaystyle D(r) \geq 0$, where r is the distance from the center of the body. An object with mass $\displaystyle m$ is laying at the distance d from the center of the celestial body. Suppose it's outside the celestial body, in other words, $\displaystyle r \geq d\ \rightarrow\ D(r) = 0$.

Suppose $\displaystyle F(r)$ is the total force the body is affecting the body with, considering only the mass in the body at a distance from the center within the interval $\displaystyle [0,\ r]$, a mass which still fulfills the density criteria*. Note that $\displaystyle F(0) = 0$, since there is no mass within the interval $\displaystyle [0,\ 0]$, and that $\displaystyle \frac{\partial F(r)}{\partial r} \geq 0$, though I don't want to prove it.

Now, I state that
$\displaystyle \frac{\partial F(r)}{\partial r}\ =\ \frac{1}{\partial r}\cdot\int_{\alpha\ =\ 0}^{2\pi} \left( G\cdot m\cdot\overbrace{ D(r)\cdot\underbrace{ sin(\alpha)\cdot 2\pi r }_\text{ length of latitudinal line }\cdot\overbrace{ r\cdot\partial\alpha }^\text{ its "width" }\cdot\underbrace{ \partial r }_\text{ its "height" } }^\text{mass} \right.$ $\displaystyle \left. /\overbrace{\sqrt{(d+cos(\alpha)r)^2 + sin^2(\alpha)r^2}}^ \text{distance from object to line}\ ^2 \right)$

which can be simplified as

$\displaystyle G\cdot m\cdot D(r)\cdot 2\pi r^2\cdot \int_0^{2\pi}\left( \frac{sin(\alpha)}{(d+cos(\alpha)r)^2 + sin^2(\alpha)r^2} \right)\partial \alpha\ =$

$\displaystyle G\cdot m\cdot D(r)\cdot 2\pi r^2\cdot \int_0^{2\pi}\left( \frac{sin(\alpha)}{d^2 + r^2 + 2cos(\alpha)dr} \right)\partial \alpha$

But according to a clause (I have only heard of it) that states that the celestial body can be treated as if it where only a point but with the same mass, we would get

$\displaystyle F(r) = \frac{\displaystyle{G\cdot m\cdot\overbrace{\int_0^r\left( D(r_1)\cdot 4\pi r_1^2 \right)\partial r_1}^\text{mass}}}{d^2}$

so that

$\displaystyle \frac{\partial F(r)}{\partial r} = \frac{G\cdot m\cdot D(r)\cdot 4\pi r^2}{d^2}$

We now has two different formulas for $\displaystyle \frac{\partial F(r)}{\partial r}$:

$\displaystyle G\cdot m\cdot D(r)\cdot 2\pi r^2\cdot \int_0^{2\pi}\left( \frac{sin(\alpha)}{d^2 + r^2 + 2cos(\alpha)dr} \right)\partial \alpha\ \equiv$
$\displaystyle \equiv\ \frac{G\cdot m\cdot D(r)\cdot 4\pi r^2}{d^2}\ \longrightarrow$

$\displaystyle \longrightarrow\ \int_0^{2\pi}\left( \frac{sin(\alpha)}{d^2 + r^2 + 2cos(\alpha)dr} \right)\partial \alpha\ \equiv\ \frac{2}{d^2}$

Is this true? Correct me if I'm wrong. Or if I have made any error (probably). Or if I am using the mathematical signs wrong. ;)

* The density critera says that the body shall have the same density all over any given distance from the center. And I just made it up.
• Nov 28th 2006, 03:43 PM
TriKri
Comment: I checked it in MS Excel and it seems to be false. :mad:
• Nov 28th 2006, 03:49 PM
topsquark
Quote:

Originally Posted by TriKri
Comment: I checked it in MS Excel and it seems to be false. :mad:

Hmph. I can reply to this post, but I couldn't reply to the other. Odd.

Anyway, your theorem is certainly correct. I would recommend looking up Gauss' Law for gravitation. The main idea is that the flux of gravitational field lines through a closed surface is proportional to only the mass inside the surface. It has nothing to do with the position of the mass, nor the density function. This is a direct consequence of the inverse square nature of the Newtonian gravity law.

-Dan
• Nov 29th 2006, 11:48 AM
TriKri
Oops, I think I forgot something very important in my calculations:

$\displaystyle \frac{\partial F(r)}{\partial r}\ \not\equiv\ G\cdot m\cdot D(r)\cdot 2\pi r^2\cdot \int_0^{2\pi}\left( \frac{sin(\alpha)}{(d+cos(\alpha)r)^2 + sin^2(\alpha)r^2} \right)\partial \alpha$

I forgot to consider the direction of the forces. Every force is now directed directly forward. But that is not the case. It should really look like this:

$\displaystyle G\cdot m\cdot D(r)\cdot 2\pi r^2\cdot$$\displaystyle \int_0^{2\pi}\left( \frac{sin(\alpha)}{(d+cos(\alpha)r)^2 + sin^2(\alpha)r^2}\cdot \overbrace{ \frac{d + cos(\alpha)r} {\sqrt{(d+cos(\alpha)r)^2 + sin^2(\alpha)r^2}} }^{\begin{array}{cc}_\text{new factor because of}\\ ^\text{the direction of the force}\end{array}} \right)\partial \alpha\ = \displaystyle G\cdot m\cdot D(r)\cdot 2\pi r^2\cdot$$\displaystyle \int_0^{2\pi}\left( \frac{sin(\alpha)\cdot(d + cos(\alpha)r)}{((d + cos(\alpha)r)^2 + sin^2(\alpha)r^2)^{1.5}} \right)\partial \alpha\ =$

$\displaystyle G\cdot m\cdot D(r)\cdot 2\pi r^2\cdot$$\displaystyle \int_0^{2\pi}\left( \frac{sin(\alpha)\cdot(d + cos(\alpha)r)} {(d^2 + r^2 + 2cos(\alpha)dr)^{1.5}} \right)\partial \alpha$

So we would get this relation instead:

$\displaystyle \int_0^{2\pi}\left( \frac{sin(\alpha)\cdot(d + cos(\alpha)r)} {(d^2 + r^2 + 2cos(\alpha)dr)^{1.5}} \right)\partial \alpha\ \equiv\ \frac{2}{d^2}$

And that actually does seem to be true. Hm, I wonder how to show it, I'm not that good at integrals. （&#180;-）.｡oO( ... )
• Dec 3rd 2006, 05:01 AM
TriKri
Quote:

Originally Posted by TriKri
$\displaystyle \int_0^{2\pi}\left( \frac{sin(\alpha)\cdot(d + cos(\alpha)r)} {(d^2 + r^2 + 2cos(\alpha)dr)^{1.5}} \right)\partial \alpha\ \equiv\ \frac{2}{d^2}$

And that actually does seem to be true. Hm, I wonder how to show it, I'm not that good at integrals. （´-）.｡oO( ... )

I don't know how to solve integrals, but do you think this is solvable? How can I solve this? Is there any method I can use?
• Dec 3rd 2006, 06:03 AM
ThePerfectHacker
Quote:

Originally Posted by TriKri
I don't know how to solve integrals, but do you think this is solvable? How can I solve this? Is there any method I can use?

First, I have absolutely no idea what $\displaystyle da^{1.5}$ means! Some physics alchemistry.

But when dealing with sines and cosine in rational form it is useful to use "Weierstrass Substitution".

Also, you can use a powerful algorithm The Integrator.

And finally you are dealing with countinous definite integral, why not approximate by Trapezoidal rule.
• Dec 4th 2006, 03:59 PM
TriKri
> First, I have absolutely no idea what $\displaystyle da^{1.5}$ means! Some physics alchemistry.

If you read my post again I'm sure you will understand. It's because I multiply the expression by the square root of the same expression.

> But when dealing with sines and cosine in rational form it is useful to use "Weierstrass Substitution".

Yes, I looked that up, very clever way to go, I tried it but I didn't succeed though. Maybe I didn't try hard enough. :o

> Also, you can use a powerful algorithm The Integrator.

It didn't finish my expression in the given amounth of time, is that a bad sign?

> And finally you are dealing with countinous definite integral, why not approximate by Trapezoidal rule.

Thank you, but I want to prove the equality, so I prefer an exact expression to a value.
• Dec 4th 2006, 04:36 PM
ThePerfectHacker
Quote:

Originally Posted by TriKri
If you read my post again I'm sure you will understand. It's because I multiply the expression by the square root of the same expression.

Not true! That symbol does not make mathematical sense (which I why I never use differencials). But physicsits and engineers use them because they have horrible abstraction though so they rely on this trick which sometimes works and sometimes does not.

Quote:

Thank you, but I want to prove the equality, so I prefer an exact expression to a value.
There is absolutely no purpose of having an exact complicated model in physics. Because all we need an approximation. (In fact there is no such thing as an ideal physics model). So approximating now, or getting it exact and then approximating makes no difference. (As long as the answer works).
• Dec 4th 2006, 05:16 PM
TriKri
Quote:

Originally Posted by ThePerfectHacker
Not true! That symbol does not make mathematical sense (which I why I never use differencials).

Why, is it because it's not an integer exponent? Does it have to be in fraction form or something like that? Does it matter? Not that I'm an expert on integrals, I'm just being curious.

Quote:

Originally Posted by ThePerfectHacker
There is absolutely no purpose of having an exact complicated model in physics. Because all we need an approximation. (In fact there is no such thing as an ideal physics model). So approximating now, or getting it exact and then approximating makes no difference. (As long as the answer works).

Well I would like to prove this else I can't fall asleep when I'm going to bed! For me it matters. If I only can prove this equality, the expression suddenly turns very simple, doesn't it? A much nicer expression. And much less calculating expensive and easier to program if you are going to put it into a computer program. I can't believe you're even questioning it! And I know this is true, that's what makes it even worse.
• Dec 4th 2006, 06:25 PM
ThePerfectHacker
Quote:

Originally Posted by TriKri
Why, is it because it's not an integer exponent? Does it have to be in fraction form or something like that? Does it matter? Not that I'm an expert on integrals, I'm just being curious.

Historically $\displaystyle dy$ (called differencial) was introduced by Leibniz. They are not number and they are not functions. What are they? I have absolutely no idea. They are some kind of magic trick that produces the right answer (sometimes). Mathemations, like myself (in the future), do not favor them. Again because we never defined what differencials mean.
• Dec 5th 2006, 10:18 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
First, I have absolutely no idea what $\displaystyle da^{1.5}$ means! Some physics alchemistry.

Just for the record I would like to state as a Physicist that I have NEVER seen the expression $\displaystyle da^{1.5}$ or anything like it used in Physics, and I can't imagine any reputable work using it. (Physics is not quite the Mathematical Dis that you would like to make it out to be, TPH.)

-Dan
• Dec 5th 2006, 11:08 AM
TriKri
What should I write then, $\displaystyle da\cdot\sqrt{da}$? I don't understand what it is you say I am doing wrong. What do you mean by $\displaystyle da$?
• Dec 5th 2006, 12:30 PM
ThePerfectHacker
Quote:

Originally Posted by TriKri
What do you mean by $\displaystyle da$?

Honestly I have no idea what $\displaystyle da$ mean :confused: .
I use it in the end of my integral because that is the standard way everybody writes them. I do not not want to confuse the readers. They are useful in Multi-variable calculus when you are dealing with several functions they can be used to keep track of which function is first, second ,third ,.... As I said I see them used all over the place. And I have a philosophy if I do not understand something (the proof or definition) I usually do not use it. Which is why I am the only person on this site that integrates using the inverse chain rule instead of a u-substitution.

Historically Leibnize defined them as infinitemal quantities (but then you need to define what infintesmal means!). But mathematicians were not satified with that because it was not rigorous. I have no idea if you were dealt with mathematicans before but now you just have. Be prepared with the pain they put you through. Not because they (the mathemations) want to bother or show off but because it is necessary if you want to prove something completely.
• Dec 5th 2006, 01:23 PM
TriKri
May there has been a small missunderstanding here? :)
I have never written $\displaystyle da$, but I did write $\displaystyle \partial\alpha$ after the equation. I don't know if it is this you meant, and maybe I should call it something else. Then the $\displaystyle \square^{1.5}$ in the denominator is applied on an area turning into a volume. However, I do have a variable $\displaystyle d$ in it, representing a distance. If I still haven't got it right, it's probably not your fault! ;)
• Dec 5th 2006, 01:42 PM
topsquark
Quote:

Originally Posted by TriKri
I don't know how to solve integrals, but do you think this is solvable? How can I solve this? Is there any method I can use?

Quote:

Originally Posted by TriKri
May there has been a small missunderstanding here? :)
I have never written $\displaystyle da$, but I did write $\displaystyle \partial\alpha$ after the equation. I don't know if it is this you meant, and maybe I should call it something else. Then the $\displaystyle \square^{1.5}$ in the denominator is applied on an area turning into a volume. However, I do have a variable $\displaystyle d$ in it, representing a distance. If I still haven't got it right, it's probably not your fault! ;)

No, TriKri, you are correct. I went back and looked at the derivation again and you simply (as you said) had a variable "d" and a variable "r" that you raised to the 3/2 power. You didn't do anything wrong. Though I would like to point out for future reference in you work to try to not use a variable "d" when you are integrating! (See the confusion it can lead to? :) )

TPH and I were guilty of not looking at your previous work carefully enough.

-Dan
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