A mass $M$ is broken into two parts one of which has mass $m$. To have maximum gravitational force of attraction between broken masses ; $1$. $m =2M$
$2$. $m=M/2$
$3$. $m=M/4$
$4$. $m=M/6$
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2 Answers
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Firstly the masses of the two bodies are m and M-m respectively. Now there are two ways to solve this problem,one by simply using the equation F=Gm(M-m)/r² and seeing that the force is directly proportional to the product of the masses while the rest are constant.From here you can clearly see by eliminating other options that the maximum value can only be when m=M/2 i.e; option 2.Note that option 1 cannot be true as you can't get a heavier body by breaking the original body. The second way to solve this problem is by using simple calculus,the concept of maxima-minima in this case. Now,F(m)=GM(M-m)/r² (force is a function of only m as the other quantities are constant in the above stated problem). For Maxima put F'(m)=0 and calculate the value of m,which you will get equal to M/2. To ensure that it is the maximum you can always check by differentiating the function for a second time with respect to m and see that it is less than zero.
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After breaking let us say that the two masses were at $r$ distance and values of mass 1 be $m$ then second one will be $M-m$ Now By newton law of gravitation , $$F_{g}=\frac{Gm(M-m)}{r^2}$$ Now for max mass we Find $$\frac{d (F_{g})}{d m}$$ and put it equal to zero. WE, get finally $$M-2m=0$$ $$m=\frac{M}{2}$$