Question

Mass M is divided into two parts X and (1-X). For a given seperation the value of X for which the gravitational attraction between the two pieces becomes maximum is

• A. 44563
• B. 44625
• C. 1
• D. 2

Answer 1

Answer: (A)

44563

Answer 2

Gravitational Force between 2 bodies is given by $F =\frac{ Gm _{1} m _{2}}{ R ^{2}}$ where, $m_{1}$ and $m_{2}$ are the masses of the bodies, $G$ is the universal Gravitational constant and $R$ is the distance between them. For a given distance, $F =\frac{ Gm (1- X ) m ( X )}{ R ^{2}}$ is maximum when $X (1- X )$ is maximum. By differentiation, we get, $\frac{ dF }{ dX }=\frac{ Gm ^{2}}{ R ^{2}} \frac{ d }{ d X }( X (1- X ))$ $\frac{ d F }{ dX }=\frac{ Gm ^{2}}{ R ^{2}} \frac{ d }{ d X }\left( X - X ^{2}\right)$ $\frac{ dF }{ dX }=\frac{ Gm ^{2}}{ R ^{2}}(1-2 X )=0$ $1-2 X =0$ $X =\frac{1}{2}$ The gravitational force of attraction has a maximum value at $X =1 / 2$