A mass M at rest is broken into two pieces having masses m and (M-m). The two masses are then separated by a distance r. The gravitational force between them will be the maximum when he ratio of the masses [m:(M-m)] of the two parts is

Question

A mass M at rest is broken into two pieces having masses m and (M-m). The two masses are then separated by a distance r. The gravitational force between them will be the maximum when he ratio of the masses [m:(M-m)] of the two parts is (A) 1: 1 (B) 1: 2 (C) 1: 3 (D) 1: 4

Tardigrade Admin · Accepted Answer

The gravitation force between two masses,

F = \frac{G m _{1} m _{2}}{r}

here,

m_{1} = m

and

m_{2} = (M - m)

∴

F

= \frac{G m ( M - m )}{r ^{2}}

For maximum gravitational force,

\frac{d F}{d m} = 0

∴ \frac{d}{d m} [m (M - m)] = 0

By solving, we get

m = \frac{M}{2}

So,

\frac{m}{( M - m )} = \frac{1}{1}

Q. A mass $M$ at rest is broken into two pieces having masses $m$ and $(M - m)$ . The two masses are then separated by a distance $r$ . The gravitational force between them will be the maximum when he ratio of the masses $[m : (M - m)]$ of the two parts is

1 : 1

1 : 2

1: 3

1 : 4

Q. A mass M at rest is broken into two pieces having masses m and (M−m). The two masses are then separated by a distance r. The gravitational force between them will be the maximum when he ratio of the masses [m:(M−m)] of the two parts is

1 : 1

1 : 2

1: 3

1 : 4

The gravitation force between two masses, F=rGm1​m2​​ here, m1​=m and m2​=(M−m) ∴ F =r2Gm(M−m)​ For maximum gravitational force, dmdF​=0 ∴dmd​[m(M−m)]=0 By solving, we get m=2M​ So, (M−m)m​=11​

Q. A mass $M$ at rest is broken into two pieces having masses $m$ and $(M - m)$ . The two masses are then separated by a distance $r$ . The gravitational force between them will be the maximum when he ratio of the masses $[m : (M - m)]$ of the two parts is