All of electrostatics follows from the $\mathbf{1}\mathbf{/}{\mathit{r}}^{\mathbf{2}}$character of Coulomb's law, together with the principle of superposition. An analogous theory can therefore be constructed for Newton's law of universal gravitation. What is the gravitational energy of a sphere, of mass M and radius R, assuming the density is uniform? Use your result to estimate the gravitational energy of the sun (look up the relevant numbers). Note that the energy is negative-masses attract, whereas (like) electric charges repel. As the matter "falls in," to create the sun, its energy is converted into other forms (typically thermal), and it is subsequently released in the form of radiation. The sun radiates at a rate of$\mathbf{3}\mathbf{.}\mathbf{86}\mathbf{\times }{\mathbf{10}}^{\mathbf{26}}\mathit{W}$; if all this came from gravitational energy, how long would the sun last? [The sun is in fact much older than that, so evidently this is not the source of its power.]

Write the expression for the Coulomb’s law.

$\mathit{F}\mathbf{=}\frac{\mathbf{k}\mathbf{q}\mathbf{Q}}{{\mathbf{r}}^{\mathbf{2}}}\mathit{r}$ …… (1)

Here, F is the force of attraction or repulsion, kis the Coulombs law, qis the magnitude of the first charge, Qis the magnitude of second charge, r is the distance between two charges.

Now, write the expression for the Newton’s law of universal gravitation.

$\overline{\mathbf{g}}\mathbf{=}\frac{\mathbf{GMm}}{{\mathbf{R}}^{\mathbf{2}}}$ …… (2)

Here, g is the gravitational force, Gis the gravitational constant, Mis t. he mass of the first body, mis the mass of the second body, R is the distance separation between two bodies.

Also radiant power of the sun is given, $\mathit{P}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{86}\mathbf{\times }{\mathbf{10}}^{\mathbf{26}}\mathbf{}\mathit{W}$.

Write the different point for the gravitational filed for the sphere.

$\overline{\mathbf{g}}\mathbf{=}\{\begin{array}{l}\frac{GMr}{R^3}r<R\\ \frac{GM}{r^2}r>R\end{array}$

Write the expression for gravitational energy of the sphere.

$$\begin{gathered} \mathit{w}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi G}}\underset{\mathbf{sphere}}{\mathbf{\int }}{\mathbf{g}}^{\mathbf{2}}\mathbf{d\pi }\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}\mathbf{\pi G}}\underset{\mathbf{0}}{\overset{\mathbf{\infty }}{\mathbf{\int }}}{\mathit{g}}^{\mathbf{2}}\mathit{d}\mathit{\zeta } \end{gathered}$$ …… (3)

Here, wis the gravitational energy.

Substitute the value of $\mathbf{\{}\begin{array}{l}\frac{\mathbf{GMr}}{{\mathbf{R}}^{\mathbf{3}}}\mathbf{r}\mathbf{<}\mathbf{R}\\ \frac{\mathbf{GM}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{r}\mathbf{>}\mathbf{R}\end{array}$for $\overline{\mathbf{g}}$in equation (3)

$$\begin{gathered} \mathit{w}\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}\mathbf{\pi }\mathbf{G}}\left\{\underset{0}{\overset{R}{\int }}{\left(\frac{GMr}{R}\right)}^{2}d\zeta +\underset{R}{\overset{\infty }{\int }}{\left(\frac{GMr}{r^2}\right)}^{2}d\zeta \right\} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}\mathbf{\pi }\mathbf{G}}\left\{\underset{0}{\overset{R}{\int }}\frac{G^2{M}^2{r}^2}{R^6}4\pi r^{2}dr+\underset{R}{\overset{\infty }{\int }}\frac{G^2{M}^2}{r^4}4\pi r^{2}dr\right\} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}\mathbf{\pi }\mathbf{G}}\left\{\underset{0}{\overset{R}{\int }}\frac{4\pi G^2{M}^2{r}^2}{R^6}\left(\frac{R^5}{5}\right)+\left(G^2{M}^2\right)\left(4\pi \right)\left(\frac{1}{R}\right)\right\} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}\mathbf{G}{\mathbf{M}}^{\mathbf{2}}}{\mathbf{5}\mathbf{R}} \end{gathered}$$

Thus, the gravitational energy is $\mathit{w}\mathbf{=}\frac{\mathbf{3}{\mathbf{GM}}^{\mathbf{2}}}{\mathbf{5}\mathbf{R}}$.

Substitute $\mathbf{6}\mathbf{.}\mathbf{67}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{}\mathit{N}\mathbf{.}{\mathit{m}}^{\mathbf{2}}\mathbf{/}\mathit{k}{\mathit{g}}^{\mathbf{2}}$ for $\mathit{G}\mathbf{,}\mathbf{}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{99}\mathbf{\times }{\mathbf{10}}^{\mathbf{30}}\mathbf{}\mathit{k}\mathit{g}$ for M and $\mathbf{6}\mathbf{.}\mathbf{96}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}$ for R.

$$\begin{gathered} \mathit{w}\mathbf{=}\frac{\mathbf{3}\left(6.67\times {10}^{-11}N.m^2/k{g}^2\right){\left(1.99\times {10}^{30}kg\right)}^{\mathbf{2}}}{\mathbf{5}\left(6.96\times {10}^{8}m\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{277}\mathbf{\times }{\mathbf{10}}^{\mathbf{41}}\mathbf{}\mathit{J} \end{gathered}$$

Therefore, the value of gravitational energy is $\mathbf{2}\mathbf{.}\mathbf{277}\mathbf{\times }{\mathbf{10}}^{\mathbf{41}}\mathbf{}\mathit{J}$.

Write the expression for the time taken by the sun to last long.

$\mathit{t}\mathbf{=}{\left(\frac{P}{w}\right)}^{\mathbf{-}\mathbf{1}}$

Here, t is the time taken by the sun, P is the radiant power.

Substitute $\mathbf{3}\mathbf{.}\mathbf{86}\mathbf{\times }{\mathbf{10}}^{\mathbf{26}}\mathbf{}\mathit{W}$for P and $\mathbf{2}\mathbf{.}\mathbf{277}\mathbf{\times }{\mathbf{10}}^{\mathbf{41}}\mathbf{}\mathit{J}$ for w.

$$\begin{gathered} \mathit{t}\mathbf{=}\left(\frac{3.86\times {10}^{26}W}{2.277\times {10}^{41}J}\right) \\ \mathbf{}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{898}\mathbf{\times }{\mathbf{10}}^{\mathbf{14}}\mathbf{}\mathit{s} \\ \mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{87}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}\mathit{y}\mathit{r}\mathit{s} \end{gathered}$$

Hence, the time taken by the sun is $\mathbf{3}\mathbf{.}\mathbf{87}\mathbf{\times }{\mathbf{10}}^7\mathbf{}\mathit{y}\mathit{r}\mathit{s}$.