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Question: A mass \( M \) is split into two parts \( m \) and \( M - m \) . Which are then separated by a certa...

A mass MM is split into two parts mm and M−mM - m . Which are then separated by a certain distance. The ratio mM\dfrac{m}{M} which maximizes the gravitational force between the parts is,
(A) 1:41:4
(B) 1:31:3
(C) 1:21:2
(D) 1:11:1

Explanation

Solution

Hint : Find the gravitational force between the masses and use the condition to find maxima of a given function. The gravitational force acting on a mass mm due to another mass MM separated by distance RR is given by, F=GMmR2F = \dfrac{{GMm}}{{{R^2}}} . Where, GG is the gravitational constant.

Complete Step By Step Answer:
We know the gravitational force between the masses and use the condition to find maxima of a given function. The gravitational force acting on a mass mm due to another mass MM separated by distance RR is given by, F=GMmR2F = \dfrac{{GMm}}{{{R^2}}} .
Here, we have two masses mm and M−mM - m .
Now, here the gravitational force that each of the mass give to each other becomes, F=G(M−m)mR2F = \dfrac{{G(M - m)m}}{{{R^2}}}
Which can be written as, F=GR2(Mm−m2)F = \dfrac{G}{{{R^2}}}(Mm - {m^2}) .
Now, we know for a function to have maxima or minima the first order derivative of the function must be zero. Hence, if we differentiate the force with respect to mm we get the maxima condition.
Therefore, differentiating FF with respect to mm we get,
dFdm=ddm[GR2(Mm−m2)]\dfrac{{dF}}{{dm}} = \dfrac{d}{{dm}}\left[ {\dfrac{G}{{{R^2}}}(Mm - {m^2})} \right]
Therefore it becomes,
=GR2[ddm(Mm)−ddmm2]= \dfrac{G}{{{R^2}}}\left[ {\dfrac{d}{{dm}}(Mm) - \dfrac{d}{{dm}}{m^2}} \right] [Since, GG and RR are constants]
Or, dFdm=GR2[M−2m]\dfrac{{dF}}{{dm}} = \dfrac{G}{{{R^2}}}\left[ {M - 2m} \right]
Now, we know for maxima/minima condition this must be equal to zero i.e. dFdm=0\dfrac{{dF}}{{dm}} = 0
Hence, GR2[M−2m]=0\dfrac{G}{{{R^2}}}\left[ {M - 2m} \right] = 0
Or, M−2m=0M - 2m = 0
Or, M=2mM = 2m
Or, mM=12\dfrac{m}{M} = \dfrac{1}{2}
Now, for maxima to exist d2Fdm2∣m=M2<0{\left. {\dfrac{{{d^2}F}}{{d{m^2}}}} \right|_{m = \dfrac{M}{2}}} < 0
So, d2Fdm2=ddmGR2[M−2m]\dfrac{{{d^2}F}}{{d{m^2}}} = \dfrac{d}{{dm}}\dfrac{G}{{{R^2}}}\left[ {M - 2m} \right]
Or, d2Fdm2=GR2[ddmM−ddm2m]\dfrac{{{d^2}F}}{{d{m^2}}} = \dfrac{G}{{{R^2}}}\left[ {\dfrac{d}{{dm}}M - \dfrac{d}{{dm}}2m} \right]
So, we get the second order differentiation as.
d2Fdm2=−2GR2\dfrac{{{d^2}F}}{{d{m^2}}} = \dfrac{{ - 2G}}{{{R^2}}}
Therefore, d2Fdm2\dfrac{{{d^2}F}}{{d{m^2}}} at m=M2m = \dfrac{M}{2} is −2GR2\dfrac{{ - 2G}}{{{R^2}}} which is always less than zero. Hence, it has a maxima at m=M2m = \dfrac{M}{2}
Therefore, for the gravitational force to be maximum the ratio of mM\dfrac{m}{M} should be 12\dfrac{1}{2} or 1:21:2
Hence, option ( C) is correct.

Note :
The gravitational force between the masses depends only on the masses if the distance is kept constant. From the second order derivative we have seen that it is always a negative quantity or does not depend on the mass mm . Hence, we can say that the gravitational force, F=G(M−m)mR2F = \dfrac{{G(M - m)m}}{{{R^2}}} contains no minima.