$A$ body of mass $m$ is moving in a circular orbit of radius $R$ about a planet of mass $M$. At some instant,it splits into two equal masses. The first mass moves in a circular orbit of radius $\frac{R}{2}$,and the other mass moves in a circular orbit of radius $\frac{3R}{2}$. The difference between the final and initial total energies is

  • A
    $ - \frac{GMm}{2R}$
  • B
    $ + \frac{GMm}{6R}$
  • C
    $ - \frac{GMm}{6R}$
  • D
    $ \frac{GMm}{2R}$

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Similar Questions

State whether the following statements are true or false:
$(a)$ The value of '$g$' is slightly less at the poles than at the equator.
$(b)$ The unit $\text{Newton/kilogram}$ $(N/kg)$ is also a unit of acceleration due to gravity.
$(c)$ The escape velocity $v_e$ for a stationary object on the surface of a planet is directly proportional to the mass and radius of the planet.

Three equal masses of $m \; kg$ each are fixed at the vertices of an equilateral triangle $ABC.$
$(a)$ What is the force acting on a mass $2 \; m$ placed at the centroid $G$ of the triangle?
$(b)$ What is the force if the mass at the vertex $A$ is doubled?
Take $AG = BG = CG = 1 \; m$.

$A$ particle of mass $M$ is situated at the centre of a spherical shell of same mass $M$ and radius $a$. The gravitational potential at a point situated at a distance of $\frac{a}{2}$ from the centre will be:

Two identical particles of mass $1 \, kg$ each go round a circle of radius $R$,under the action of their mutual gravitational attraction. The angular speed of each particle is:

Match the $\text{LIST-I}$ with $\text{LIST-II}$:
$\text{LIST-I}$ $\text{LIST-II}$
$A$. Gravitational constant $I$. $[LT^{-2}]$
$B$. Gravitational potential energy $II$. $[L^2 T^{-2}]$
$C$. Gravitational potential $III$. $[ML^2 T^{-2}]$
$D$. Acceleration due to gravity $IV$. $[M^{-1} L^3 T^{-2}]$

Choose the correct answer from the options given below:

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