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Two hypothetical planets of masses ${m_1}$ and ${m_2}$ are at rest when they are at infinite distance apart. Because of the gravitational force they move towards each other along the line joining their centers. What is their speed if their separation is‘d’? (Speed of ${m_1}$ is ${v_1}$ and that of ${m_2}$ is ${v_2}$ )
(A)${v_1} = {v_2}$
$\eqalign{ & {v_1} = {m_2}\sqrt {\dfrac{{2G}}{{d({m_1} + {m_2})}}} \cr
  & {v_2} = {m_1}\sqrt {\dfrac{{2G}}{{d({m_1} + {m_2})}}} \cr} $
$\eqalign{ & {v_1} = {m_1}\sqrt {\dfrac{{2G}}{{d({m_1} + {m_2})}}} \cr
  & {v_2} = {m_2}\sqrt {\dfrac{{2G}}{{d({m_1} + {m_2})}}} \cr} $
$\eqalign{ & {v_1} = {m_2}\sqrt {\dfrac{{2G}}{{d{m_1}}}} \cr
  & {v_2} = {m_1}\sqrt {\dfrac{{2G}}{{d{m_2}}}} \cr} $

Last updated date: 17th Jun 2024
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Hint: We solve this problem by considering the gravitational potential energy of the system of planets and also use the conservation of energy principle. We choose the gravitational potential energy at infinity to be 0 and then carry on to calculate the speed at the separation ‘d’ of the two planets.

Complete answer:
The reference point is chosen at infinity where the total energy of the system is 0.
So, when they are moving towards each other, the total energy , that is the sum of the kinetic and the potential energy is given by
$\dfrac{1}{2}{m_1}{v^2}_1 + \dfrac{1}{2}{m_2}v_2^2 - G\dfrac{{{m_1}{m_2}}}{d}$
So, by the conservation of energy we have,
$\dfrac{1}{2}{m_1}{v^2}_1 + \dfrac{1}{2}{m_2}v_2^2 - G\dfrac{{{m_1}{m_2}}}{d} = 0$
$ \Rightarrow \dfrac{1}{2}{m_1}{v_1}^2 + \dfrac{1}{2}{m_2}v_2^2 = \dfrac{{G{m_1}{m_2}}}{d}$ …………………….. (1)
According to the principle of conservation of linear momentum, we have
  & {m_1}{v_1} = {m_2}{v_2} \cr
  & \Rightarrow {v_2} = - \dfrac{{{m_1}}}{{{m_2}}}{v_1} \cr} $ ……………………….. (2)
Then we will put the value of equation (2) in (1) and hence we get:
  & \dfrac{1}{2}{m_1}{v_1}^2 + \dfrac{1}{2}{m_2}v_2^2 = \dfrac{{G{m_1}{m_2}}}{d} \cr
  & \Rightarrow \dfrac{1}{2}{m_1}{v_1}^2 + \dfrac{1}{2}{m_2}{\left( { - \dfrac{{{m_1}}}{{{m_2}}}{v_1}} \right)^2} = \dfrac{{G{m_1}{m_2}}}{d} \cr
  & \Rightarrow \dfrac{1}{2}{m_1}{v_1}^2 + \dfrac{1}{2}{m_2}\left( {\dfrac{{{m_1}^2}}{{{m_2}^2}}{v_1}^2} \right) = \dfrac{{G{m_1}{m_2}}}{d} \cr
  & \Rightarrow \dfrac{1}{2}{m_1}{v_1}^2\left( {1 + \dfrac{{{m_1}}}{{{m_2}}}} \right) = \dfrac{{G{m_1}{m_2}}}{d} \cr} $
Then simplifying this further by rearranging, we get,
  & {v_1}^2\left( {\dfrac{{{m_2} + {m_1}}}{{{m_2}}}} \right) = \dfrac{{2G{m_2}}}{d} \cr
  & \Rightarrow {v_1}^2 = \dfrac{{2G{m_2}^2}}{{d\left( {{m_1} + {m_2}} \right)}} \cr
  & \therefore {v_1} = {m_2}\sqrt {\dfrac{{2G}}{{d({m_1} + {m_2})}}} \cr} $
Similarly, if we substitute the value of ${v_1}$ instead of ${v_2}$, we get:
${v_2} = - {m_1}\sqrt {\dfrac{{2G}}{{d({m_1} + {m_2})}}} $
Thus, we get the same values of velocities as option (B) but with the exception that the magnitude of ${v_2}$ is negative. Since, we are only asked of speed in the question and not velocity, the values of velocities can be considered without their speeds to get the correct answer as (B)
Hence, the correct answer is option (B).

Additional Information:
Gravitational potential energy is the type of energy that a body possesses due to its position in a gravitational field. The general form of gravitational potential energy of two bodies separated by a distance d is given as U=$G\dfrac{{{m_1}{m_2}}}{d}$.

Here, when we say total energy, it is important to note that we consider the sum of the gravitational potential energy and the kinetic energy. The next important thing to remember is that we even consider the conservation of linear momentum to obtain a relationship between the masses and the velocities.