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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.
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.
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 $V = -\frac{GMm}{R}$.
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' ?
How can I handle that then?Notice \(\displaystyle \sqrt{(r'-R)^2}\not =(r'-R) \) unless r'>R
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' ?
YesOk,what do you mean by a ? Is it the radius of M ?
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 ?
That doesn't look quite correct.If we let $R\to\infty$, then $r'\to R$.
So I took r'=R
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.
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*}
Since $0 < r' < a < R$, we can write $\sqrt{(r' - R)^2} = \sqrt{(R - r')^2} = R - r'$??