Question

Two planets of mass M and 16m of radius a and 2 respectively, are at distance 10a. find the minimum speed of a particle of mass M at surface of smaller planet so that it can reached from smaller planet to Larger planet.

Answer 1

CORRECT QUESTION:

Two planets of mass M and 16M of radius a and 2a respectively, are at distance 10a. Find the minimum speed of a particle of mass m at surface of smaller planet so that it can reach from smaller planet to Larger planet.

ANSWER:

• Option 1) v = √{5/9 × (GM/a)}

GIVEN:

• Two planets of mass M and 16M of radius a and 2a respectively.

• They are at distance 10a.

TO FIND:

• The minimum speed of a particle of mass at surface of smaller planet so that it can reach from smaller planet to Larger planet.

EXPLANATION:

Let the distance from the point to the small planet where gravitational force of attraction of both the planets be equal be x.

$\leadsto\sf \dfrac{GMm}{x^2} = \dfrac{16GMm}{(10a-x)^2}$

$\leadsto\sf \dfrac{1}{x^2} = \dfrac{16}{(10a-x)^2}$

Take square root on both sides

$\leadsto\sf \dfrac{1}{x} = \dfrac{4}{10a-x}$

$\leadsto\sf 10a - x = 4x$

$\leadsto\sf 5x = 10a$

$\leadsto\sf x = 2a$

Total energy on the surface of the small planet:

Total energy = Potential energy + kinetic energy

$\leadsto \sf E_i = \dfrac{1}{2}m {v}^{2} - \dfrac{GMm}{a} -\dfrac{16GMm}{9a}$

• a is the distance from the centre of the smaller planet to the surface of the smaller planet.

• 9a is the distance from the centre of the larger planet to the surface of the smaller planet.

$\leadsto \sf E_i = \dfrac{1}{2}m {v}^{2} + \dfrac{ - 9GMm - 16GMm}{9a}$

$\leadsto \sf E_i = \dfrac{1}{2}m {v}^{2} - \dfrac{ 25GMm}{9a}$

Total energy at the null point:

Total energy = Potential energy + kinetic energy

$\leadsto\sf E_f=0 - \dfrac{GMm}{2a} -\dfrac{16GMm}{8a}$

• 2a is the distance from the centre of the smaller planet to the null point.

• 8a is the distance from the centre of the larger planet to the null point.

$\leadsto\sf E_f= \dfrac{- 4GMm - 16GMm}{8a}$

$\leadsto\sf E_f= \dfrac{- 20GMm}{8a}$

$\leadsto\sf E_f= \dfrac{- 5GMm}{2a}$

$\boxed{ \large \sf E_i= E_f}$

$\leadsto \sf \dfrac{1}{2}m {v}^{2} - \dfrac{ 25GMm}{9a} = \dfrac{- 5GMm}{2a}$

$\leadsto\sf \dfrac{1}{2}m {v}^{2} = \dfrac{- 5GMm}{2a} + \dfrac{ 25GMm}{9a}$

$\leadsto \sf \dfrac{1}{2} {v}^{2} = \dfrac{- 5GM}{2a} + \dfrac{ 25GM}{9a}$

$\leadsto \sf \dfrac{1}{2} {v}^{2} =\dfrac{- 45GM + 50GM}{18a}$

$\leadsto \sf {v}^{2} =\dfrac{5GM}{9a}$

$\leadsto \sf v = \sqrt{\dfrac{5}{9}\dfrac{GM}{a}}$

$\boxed{ \sf { \large{Hence \ the \ v_{min} = \sqrt{\dfrac{5}{9}\dfrac{GM}{a}}}}}$