Two bodies, each of mass M, are kept fixed with a separation 2L . A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G. The correct statement(s) is (are)

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 2\sqrt GM \over L, The energy of the mass m remains constant.

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 4\sqrt GM \over L

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 2\sqrt GM \over L

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is \sqrt 2GM \over L

The energy of the mass m remains constant.

Gravitation PYQ — JEE Advanced 2013 · Shift 2 · Q42 with Solution

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Correct answer: B, D

Set up the configuration

Two fixed masses, each of mass $M$ , are separated by distance $2L$ . The particle of mass $m$ starts from the midpoint $O$ and is projected perpendicular to the line joining the two masses.

At the midpoint, the distance of the particle from each mass is $L$ .

So the initial gravitational potential energy of the particle is

U_0=-\frac{GMm}{L}-\frac{GMm}{L}=-\frac{2GMm}{L}.

Condition for escape

For escape, the total mechanical energy must be at least zero, since at infinity the gravitational potential energy tends to $0$ and the minimum escape case corresponds to final speed $0$ .

Initial kinetic energy is

K_0=\frac12 mv^2.

Thus initial total energy is

E=\frac12 mv^2-\frac{2GMm}{L}.

For minimum escape speed,

E=0.

Hence,

\frac12 mv^2=\frac{2GMm}{L}.

Cancelling $m$ ,

\frac12 v^2=\frac{2GM}{L}

v^2=\frac{4GM}{L}

v=2\sqrt{\frac{GM}{L}}.

Therefore, the minimum initial speed required for escape is

2\sqrt{\frac{GM}{L}}.

So Option B is correct.

Check the other speed options

Option A: $4\sqrt{\frac{GM}{L}}$ is larger than necessary, so it is not the minimum escape speed. False.
Option C: $\sqrt{\frac{2GM}{L}}$ is smaller than the required value. False.

Is energy conserved?

The two masses are fixed, so they create a static gravitational field. Gravitational force is conservative, and no non-conservative force is mentioned.

Hence the mechanical energy of the particle remains constant during motion:

E = K + U = \text{constant}.

So Option D is correct.

Final evaluation of options

A: False
B: True
C: False
D: True

Thus the correct statements are B and D.

More from Gravitation

View written solutionFree

Correct answer: B, D

Set up the configuration

Two fixed masses, each of mass $M$ , are separated by distance $2L$ . The particle of mass $m$ starts from the midpoint $O$ and is projected perpendicular to the line joining the two masses.

At the midpoint, the distance of the particle from each mass is $L$ .

So the initial gravitational potential energy of the particle is

U_0=-\frac{GMm}{L}-\frac{GMm}{L}=-\frac{2GMm}{L}.

Condition for escape

For escape, the total mechanical energy must be at least zero, since at infinity the gravitational potential energy tends to $0$ and the minimum escape case corresponds to final speed $0$ .

Initial kinetic energy is

K_0=\frac12 mv^2.

Thus initial total energy is

E=\frac12 mv^2-\frac{2GMm}{L}.

For minimum escape speed,

E=0.

Hence,

\frac12 mv^2=\frac{2GMm}{L}.

Cancelling $m$ ,

\frac12 v^2=\frac{2GM}{L}

v^2=\frac{4GM}{L}

v=2\sqrt{\frac{GM}{L}}.

Therefore, the minimum initial speed required for escape is

2\sqrt{\frac{GM}{L}}.

So Option B is correct.

Check the other speed options

Option A: $4\sqrt{\frac{GM}{L}}$ is larger than necessary, so it is not the minimum escape speed. False.
Option C: $\sqrt{\frac{2GM}{L}}$ is smaller than the required value. False.

Is energy conserved?

The two masses are fixed, so they create a static gravitational field. Gravitational force is conservative, and no non-conservative force is mentioned.

Hence the mechanical energy of the particle remains constant during motion:

E = K + U = \text{constant}.

So Option D is correct.

Final evaluation of options

A: False
B: True
C: False
D: True

Thus the correct statements are B and D.

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