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- Thread starter dwsmith
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- Mar 5, 2012

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Integrating for $\rho$ (shouldn't that be $\phi$?) is simply multiplying by $2\pi$.Anyone have any smooth ideas for this triple integral?

$$

-\rho Gm\iiint\frac{r'^2\sin\theta}{\sqrt{R^2 + r'^2 - 2Rr'\cos\theta}}d\theta d\rho dr'

$$

where $0<r'<a$, $0<\theta<\pi$, and $0<\rho<2\pi$.

The $\rho$ out front is constant.

Then integrate with respect to $\theta$.

To do so, consider what $\dfrac{\partial}{\partial \theta}\sqrt{R^2 + r'^2 - 2Rr'\cos\theta}$ is.

You'll be left with a pretty simple expression to integrate with respect to r'.

Finally you'll end up with the gravitational energy between a mass $m$ and a sphere (with radius $a$ and density $\rho$) at a distance R.

Last edited:

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Yes, it should be $\phi$.Integrating for $\rho$ (shouldn't that be $\phi$?) is simply multiplying by $2\pi$.

Then integrate with respect to $\theta$.

To do so, consider what $\dfrac{\partial}{\partial \theta}\sqrt{R^2 + r'^2 - 2Rr'\cos\theta}$ is.

You'll be left with a pretty simple expression to integrate with respect to r'.

Finally you'll end up with the gravitational energy between a mass $m$ and a sphere (with radius $a$ and density $\rho$) at a distance R.

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- #4

I keep ending up with

\begin{alignat*}{3}

V & = & -2\pi\rho Gm\int_0^a\left[\sqrt{1 +\left(\frac{r'}{R}\right)^2-2\frac{r'}{R}\cos\theta}\right|_0^{\pi}dr\\

& = & -2\pi\rho Gm\int_0^a\left[\sqrt{\frac{(r' + R)^2}{R^2}} - \sqrt{\frac{(r' - R)^2}{R^2}}\right]dr\\

& = & -\frac{2\pi\rho Gm}{R}\int_0^a[r' + R - r' + R]dr\\

& = & -4\pi\rho Gm\int_0^a dr\\

& = & -4\pi\rho Gma

\end{alignat*}

Then $\rho = \frac{M}{V}$ so $V = -\frac{4\pi GMm a}{V}$ but I need to conclude $V = -\frac{GMm}{R}$.

- Jan 17, 2013

- 1,667

Assuming you did the integral right ,it is impossible to get rid of (a) unless you have a value for it. May I ask what is r' ?I keep ending up with

\begin{alignat*}{3}

V & = & -2\pi\rho Gm\int_0^a\left[\sqrt{1 +\left(\frac{r'}{R}\right)^2-2\frac{r'}{R}\cos\theta}\right|_0^{\pi}dr\\

& = & -2\pi\rho Gm\int_0^a\left[\sqrt{\frac{(r' + R)^2}{R^2}} - \sqrt{\frac{(r' - R)^2}{R^2}}\right]dr\\

& = & -\frac{2\pi\rho Gm}{R}\int_0^a[r' + R - r' + R]dr\\

& = & -4\pi\rho Gm\int_0^a dr\\

& = & -4\pi\rho Gma

\end{alignat*}

Then $\rho = \frac{M}{V}$ so $V = -\frac{4\pi GMm a}{V}$ but I need to conclude $V = -\frac{GMm}{R}$.

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- #6

r' is rAssuming you did the integral right ,it is impossible to get rid of (a) unless you have a value for it. May I ask what is r' ?

- Jan 17, 2013

- 1,667

Notice \(\displaystyle \sqrt{(r'-R)^2}\not =(r'-R) \) unless r'>R

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- #8

Here is what I am working on. It is on page 1. Maybe there is something you will see that will help.

View attachment 230 Notes.pdf

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- #9

How can I handle that then?Notice \(\displaystyle \sqrt{(r'-R)^2}\not =(r'-R) \) unless r'>R

- Jan 17, 2013

- 1,667

Can we deduce from the figure that R>r' ?

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- #11

No because the differential volume can be anywhere. I just drew so that r' > R. As the distance between the planets go to infinity, $r'\to R$ but even doing that only yields:Can we deduce from the figure that R>r' ?

$$

-\frac{Gma\pi(a+2R)\rho}{R}

$$

- Jan 17, 2013

- 1,667

Ok,what do you mean by a ? Is it the radius of M ?

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- #13

YesOk,what do you mean by a ? Is it the radius of M ?

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\begin{alignat*}{3}

V & = & -\frac{2\pi\rho Gm}{R}\int_0^a r'\left[r' + R - \sqrt{(r' - R)^2}\right]dr'\\

& = & \frac{2\pi\rho Gm}{R}\int_0^a 2r'^2dr'\\

& = & -\frac{4Gma^3\rho\pi}{3R}\\

& = & -\frac{GMm}{R}

\end{alignat*}

Since $\rho = \frac{M}{V}$ where $V = \frac{4\pi r^3}{3}$, we have the desired result.

Last edited:

- Jan 17, 2013

- 1,667

Excellent , so you are choosing that R>r' , but what is the meaning of that ? Are you raking some unnecessary restrictions ?

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- #16

If we let $R\to\infty$, then $r'\to R$.Excellent , so you are choosing that R>r' , but what is the meaning of that ? Are you raking some unnecessary restrictions ?

So I took r'=R

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- #17

- Mar 5, 2012

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That doesn't look quite correct.If we let $R\to\infty$, then $r'\to R$.

So I took r'=R

If we let $R\to\infty$, then $r' \le a \ll R$.

But really, there is no need to do this.

Your formula works for any $R$ with $r' \le a \le R$.

Actually, the additionally interesting case is when $R < a$.

You can also calculate what happens then.

It should come out that $V = -\frac 4 3 \pi R^3 \rho \cdot \frac {Gm }{R}$.

That is, the attractive force when inside a solid sphere is completely determined by the subsphere between you and the center.

The forces by the part of the sphere that is on the outside cancel out.

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- #18

How do I go from the first integral to the second then?That doesn't look quite correct.

If we let $R\to\infty$, then $r' \le a \ll R$.

But really, there is no need to do this.

Your formula works for any $R$ with $r' \le a \le R$.

Actually, the additionally interesting case is when $R < a$.

You can also calculate what happens then.

It should come out that $V = -\frac 4 3 \pi R^3 \rho \cdot \frac {Gm }{R}$.

That is, the attractive force when inside a solid sphere is completely determined by the subsphere between you and the center.

The forces by the part of the sphere that is on the outside cancel out.

\begin{alignat*}{3}

V & = & -\frac{2\pi\rho Gm}{R}\int_0^a r'\left[r' + R - \sqrt{(r' - R)^2}\right]dr'\\

& = & \frac{2\pi\rho Gm}{R}\int_0^a 2r'^2dr'

\end{alignat*}

Since $0 < r' < a < R$, we can write $\sqrt{(r' - R)^2} = \sqrt{(R - r')^2} = R - r'$??

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- #19

- Mar 5, 2012

- 9,416

If $R \ge r'$, then $\sqrt{(r' - R)^2} = |r' - R| = (R - r')$.How do I go from the first integral to the second then?

\begin{alignat*}{3}

V & = & -\frac{2\pi\rho Gm}{R}\int_0^a r'\left[r' + R - \sqrt{(r' - R)^2}\right]dr'\\

& = & \frac{2\pi\rho Gm}{R}\int_0^a 2r'^2dr'

\end{alignat*}

Substitute?

Since $0 < r' < a < R$, we can write $\sqrt{(r' - R)^2} = \sqrt{(R - r')^2} = R - r'$??

Yes!