Question

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

Answer 1

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 $ m $ due to another mass $ M $ separated by distance $ R $ is given by, $ F = \dfrac{{GMm}}{{{R^2}}} $ . Where, $ G $ 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 $ m $ due to another mass $ M $ separated by distance $ R $ is given by, $ F = \dfrac{{GMm}}{{{R^2}}} $ .
Here, we have two masses $ m $ and $ M - m $ .
Now, here the gravitational force that each of the mass give to each other becomes, $ F = \dfrac{{G(M - m)m}}{{{R^2}}} $
Which can be written as, $ 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 $ m $ we get the maxima condition.
Therefore, differentiating $ F $ with respect to $ m $ we get,
 $ \dfrac{{dF}}{{dm}} = \dfrac{d}{{dm}}\left[ {\dfrac{G}{{{R^2}}}(Mm - {m^2})} \right] $
Therefore it becomes,
 $ = \dfrac{G}{{{R^2}}}\left[ {\dfrac{d}{{dm}}(Mm) - \dfrac{d}{{dm}}{m^2}} \right] $ [Since, $ G $ and $ R $ are constants]
Or, $ \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. $ \dfrac{{dF}}{{dm}} = 0 $
Hence, $ \dfrac{G}{{{R^2}}}\left[ {M - 2m} \right] = 0 $
Or, $ M - 2m = 0 $
Or, $ M = 2m $
Or, $ \dfrac{m}{M} = \dfrac{1}{2} $
Now, for maxima to exist $ {\left. {\dfrac{{{d^2}F}}{{d{m^2}}}} \right|_{m = \dfrac{M}{2}}} < 0 $
So, $ \dfrac{{{d^2}F}}{{d{m^2}}} = \dfrac{d}{{dm}}\dfrac{G}{{{R^2}}}\left[ {M - 2m} \right] $
Or, $ \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.
 $ \dfrac{{{d^2}F}}{{d{m^2}}} = \dfrac{{ - 2G}}{{{R^2}}} $
Therefore, $ \dfrac{{{d^2}F}}{{d{m^2}}} $ at $ m = \dfrac{M}{2} $ is $ \dfrac{{ - 2G}}{{{R^2}}} $ which is always less than zero. Hence, it has a maxima at $ m = \dfrac{M}{2} $
Therefore, for the gravitational force to be maximum the ratio of $ \dfrac{m}{M} $ should be $ \dfrac{1}{2} $ or $ 1:2 $
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 $ m $ . Hence, we can say that the gravitational force, $ F = \dfrac{{G(M - m)m}}{{{R^2}}} $ contains no minima.