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The particles $A$ and $B$ of mass $m$ each are separated by a distance $r$. Another particle $C$ of mass $M$ is placed at the midpoint of $A$ and $B$. Find the work done in taking $C$ to a point equidistant $r$ from $A$ and $B$ without acceleration $(G=$ Gravitational constant and only gravitational interaction between $A, B$ and $C$ is considered)
$\frac{G M m}{r}$
$\frac{2 G M m}{r}$
$\frac{3 G M m}{r}$
$\frac{4 G M m}{r}$
Solution
(b)
Since particle $C$ is moved without any acceleration,
$\Rightarrow \Delta K . E =0$
$\Rightarrow$ Work done by external agent $+W_{\text {gravitation }}=0$
$\Rightarrow$ Work done by external agent $=-W g$
$=-(-\Delta U)$
$=\Delta U$
$=U_f-U_{\text {in }}$
$U_f=-\frac{G M m}{r}-\frac{G M m}{r}=-\frac{2 G M m}{r}$
$U_i=-\frac{G M m}{r / 2}-\frac{G M m}{r / 2}=-\frac{4 G M m}{r}$
$\Rightarrow$ Work done $=\frac{2 G M m}{r}$
Similar Questions
Match the column $-I$ with column $-II$ For a satellite in circular orbit,
| Column $-I$ | Column $-II$ |
| $(A)$ Kinetic energy | $(p)$ $ – \frac{{G{M_E}m}}{{2r}}$ |
| $(B)$ Potential energy | $(q)$ $\sqrt {\frac{{G{M_E}}}{r}} $ |
| $(C)$ Total energy | $(r)$ $ – \frac{{G{M_E}m}}{{r}}$ |
| $(D)$ Orbital energy | $(s)$ $ \frac{{G{M_E}m}}{{2r}}$ |
(where $M_E$ is the mass of the earth, $m$ is mass of the satellite and $r$ is the radius of the orbit)
