Perhaps this is trivial: how does one prove that the work done by a homogeneous gravitational field on a system of particles is equal to the work done on a point mass with the total mass of the system, located at the center of mass of the system.
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ The Shell Theorem proves it for spheres: en.wikipedia.org/wiki/Shell_theorem I don't think it's true for non-spherical bodies as I'm pretty sure you could get a torque as part of the attraction. $\endgroup$– Brandon EnrightCommented Jan 5, 2015 at 0:22
-
$\begingroup$ Gravity is a central force. A consequence of this is that it will not exert a net external torque on a system by its own self. $\endgroup$– kbhCommented Jan 5, 2015 at 5:39
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The total work done by gravity is $$W = \sum_{i=1}^N \int m_i\mathbf v_i\cdot\mathbf g\text dt.$$ Commuting sum and integral one gets $$W = \int\left(\sum_{i=1}^N m_i\mathbf v_i\right)\cdot\mathbf g\text dt.$$ The quantity inside the parentheses is precisely $M\mathbf v_{\text{CoM}}$, so $$W = M\int\mathbf v_{\text{CoM}}\cdot\mathbf g\text dt,$$ which looks like the work done by a particle of mass $M$ moving as the centre of mass of the particles at $\mathbf r_i$
