JEE Advanced 2013 · Paper 2 · Physics Q5

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PhysicsMultiple correct+3 / -1Question 5 of 60

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

4\sqrt{\dfrac{GM}L}

The minimum initial velocity of the mass

m

to escape the gravitational field of the two bodies is

2\sqrt{\dfrac{GM}L}

The minimum initial velocity of the mass

m

to escape the gravitational field of the two bodies is

\sqrt{\dfrac{2GM}L}

The energy of the mass

m

remains constant.

Answer and worked solution

Official answer: B, D

Potential energy at the midpoint. $U=-\dfrac{2GMm}L$ .

Escape. $\tfrac12mv^2=\dfrac{2GMm}L$ gives $v=2\sqrt{\dfrac{GM}L}$ .

Energy. Gravity is conservative and the bodies are fixed, so the total mechanical energy of $m$ stays constant.

(B), (D)

Topics:Escape velocity Gravitational field and potential

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JEE Advanced 2013 · Paper 2, Physics question 5