A small body of mass is projected with a velocity just sufficient to make it reach from the surface of a planet (of radius 2R and mass 3M) to the surface of another planet (of radius R and mass M). The distance between the centers of the two spherical planets is 6R. The distance of the body from the center of bigger planet is 'x' at any moment. During the journey, find the distance x where the speed of the body is maximum.
Assume motion of body along the line joining centres of planets.

Question

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Answer 1

The force on the body at a distance x would be

$F = \frac{G 3 M m}{x ^{2}} - \frac{G M m}{( 6 R - x ) ^{2}} = G M m (\frac{3}{x ^{2}} - \frac{1}{( 6 R - x ) ^{2}})$

It is obvious that the velocity of the body will first devrease and then increase. Hence the maximum velocity will either be at $x = 2 R$ or $x = 5 R$ .

Conserving energy,

$\frac{1}{2} m v_{1}^{2} - \frac{3 G M m}{2 R} - \frac{G M m}{4 R} = \frac{1}{2} m v_{2}^{2} - \frac{G M m}{R} - \frac{3 G M m}{5 R}$

$\frac{1}{2} m (v_{1}^{2} - v_{2}^{2}) = \frac{G M m}{R} (\frac{3}{2} - \frac{3}{5} + \frac{1}{4} - 1) > 0$

Thus, $v_{1} > v_{2}$

Hence, answer is $2 R$

Answer 2

The force on the body at a distance x would be

F = \frac{G 3 M m}{x ^{2}} - \frac{G M m}{( 6 R - x ) ^{2}} = G M m (\frac{3}{x ^{2}} - \frac{1}{( 6 R - x ) ^{2}})

It is obvious that the velocity of the body will first devrease and then increase. Hence the maximum velocity will either be at

x = 2 R

or

x = 5 R

.

Conserving energy,

\frac{1}{2} m v_{1}^{2} - \frac{3 G M m}{2 R} - \frac{G M m}{4 R} = \frac{1}{2} m v_{2}^{2} - \frac{G M m}{R} - \frac{3 G M m}{5 R}

\frac{1}{2} m (v_{1}^{2} - v_{2}^{2}) = \frac{G M m}{R} (\frac{3}{2} - \frac{3}{5} + \frac{1}{4} - 1) > 0

Thus,

v_{1} > v_{2}

Hence, answer is

2 R