A body of mass (2M) splits into four masses {m, M – m, m, M – m}, which are rearranged to form a square as shown in the figure. The ratio of M m for which, the gravitational potential energy of the system becomes maximum is x : 1. The value of x is

U = − G m ( M − m ) d * 4 + 2 x − ( G m 2 2 d ) For Umax d U d m = 0 ⇒ m = M 2 ⇒ M m = 2

A body of mass (2M) splits into four masses {m, M – m, m, M – m}, which are rearranged to form a square as shown in the figure. The ratio of $\frac{M}{m}$ for which, the gravitational potential energy of the system becomes maximum is x : 1. The value of x is

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Vishal Baghel | Contributor-Level 10

7 months ago

$U = - \frac{G m (M - m)}{d} * 4 + 2 x - (\frac{G m^{2}}{\sqrt[]{2} d})$

For Umax

$\frac{d U}{d m} = 0 \Rightarrow m = \frac{M}{2} \Rightarrow \frac{M}{m} = 2$

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