Calculate Gravitational Self-Energy of the Sun

[stunner5000pt]

Calculate the gravitational self energy of the sun (without using numbers)

Textbook says that this is the nergy needed to take every particle from the sun's surface to infinity

there are an infinite number of particles on the sun ...
would there be some sort of integral that relates the energy needed to take one particle from the sun's surface ?

something like this - for one particle mass i, energy needed ot take it out from the sun's surface U_{i}(r) = -G \frac{M_{s} m_{i}}{r_{si}}
but since this is a 3D object how would one go about setting up an integral like this??

 

[shyboy]

First, let's assume that the sun density \rho is constant.

then, let's slice the sun from the center. If we add a layer of d R, at a distance R from the center, then the layer mass will be dM_R=4\pi \rho R^2 dR. Now we need to find the gravitational energy between that layer and the outer shell. So let's slice the outer shell as well: dU_R=-GM_R\int{ \frac{4 \pi \rho r^2 dr}{r}}, the integration from R to R_0, the sun's radius. After that we need to find out \int {dU_R}.

 

[learningphysics]

There is gravitational energy between the layer and the inner sphere (at a particular R). The mass of the inner sphere is just (4/3)pi*R^3*density

If we integrate GM_{layer}M_{innersphere}/R over all R from 0 to the radius of the sun... I think we get the answer.

EDIT: Shyboy's method is also exactly right. I got the same answer both ways. Apologies.

 

[stunner5000pt]

what is
\int dU_{R}

what would the integration variable be then??

p.s. The answer is supposed to be \frac{3}{5} G \frac{M^2}{R}

did you guys get that??

 

[shyboy]

looks like so. I missed d before M. It should be dM_R instead of M_R. And I wish I can put limits in the integral but dont
know how (to lasy to find it) :(

 

[quasar987]

int_{lower bound}^{upper bound}

ex:

\int_a^{b-\epsilon_0}

 

[learningphysics]

what is
\int dU_{R}

what would the integration variable be then??

p.s. The answer is supposed to be \frac{3}{5} G \frac{M^2}{R}

did you guys get that??

Yes. Here's what I did:

M_{layer} = 4\pi r^2\rho dr

M_{inner} = \frac{4}{3}\pi r^3\rho

Remove the layers of the sun away to infinity starting with the outermost layer.

Energy required to take a single layer out to infinity=
0 - (- \frac{GM_{layer}M_{inner}}{r})=
\frac{16G\pi^2 \rho^2 r^4 dr}{3}

If we integrate this from r=0 to r=R (giving the energy to take away all the layers), we get
\frac{16\rho^2 \pi^2 R^5}{15}

If we use M=(4/3)\pi R^3\rho, solve for \rho and plug into the above... we get
\frac{3}{5} G \frac{M^2}{R}