-3
$\begingroup$

My book says $U_{self}=\dfrac{-GM^2}{2R}$ for the hollow sphere, I tried deriving it as:

Suppose mass constructed is $m$, Work done on bringing mass $dm$ from $\infty$ to $R$ is $$dW=dm(V_{R}-V_{\infty})\\W=\int_{0}^{M}-\dfrac{Gm\cdot dm}{R}\\W=-\dfrac{GM^2}{2R}$$

By definition of Gravitation Self Energy,

The gravitational self-energy is equal to the amount of work done in assembling together its infinitesimal particles initially lying infinite distance apart.

$$\implies U_{self}=-\dfrac{GM^2}{2R}$$

But as $dV=\dfrac{dU}{m}$ by definition, also $dW=-dU$, so why when I used $dW=dm(dV)$, it works, I should have used $dW=dm(-dV)$.

Please help, with what I missed very crucial, Thanks.

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ What is your question? You used + instead of -. Is that the crucial problem? $\endgroup$ – sammy gerbil Jan 30 '18 at 6:25
  • $\begingroup$ @sammygerbil yes, why does not "work done by the internal conservative force is negative of change in potential energy" valid here? $\endgroup$ – mnulb Jan 30 '18 at 8:54
3
$\begingroup$

The work done by an external force is equal to the change in the gravitational potential energy or the work done by the gravitational field is equal to minus the change in gravitational potential energy are two equivalent definitions of gravitational potential energy.

In your first equation $W=-\dfrac{GM^2}{2R}$ on the left hand side $W$ is the work done by an external force and the right hand side is the change in gravitational potential energy of the system $\Delta U$ which in the case is the gravitational potential energy as the zero of potential energy has been taken to be the masses infinitely far apart.

Note that $W=+\Delta U$ and there is no negative sign in this equation if $W$ is the work done by the external force.

There would have been a negative sign if you had evaluated the work done by the gravitational field.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.