JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Gravitation

Two bodies, each of mass

M

, are kept fixed with a separation

2 L

. 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)

(A)The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $4 \frac{GM}{L}$ .
(B)The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $2 \frac{GM}{L}$ .✓ Correct
(C)The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $\frac{2 GM}{L}$ .
(D)The energy of the mass $m$ remains constant.✓ Correct

Answer: (B), (D)

Solution

At the midpoint, distance from each mass M is L. Initial potential energy:

U_{i} = - \frac{GM m}{L} - \frac{GM m}{L} = - \frac{2 GM m}{L}

For escape, final energy at infinity is zero. By energy conservation:

\frac{1}{2} m v_{min}^{2} + U_{i} = 0

\frac{1}{2} m v_{min}^{2} = \frac{2 GM m}{L}

v_{min} = 2 \frac{GM}{L}

✓ (Option B)

Since there is a net gravitational force on mass m (the two forces from M don't cancel when m moves perpendicular), work is done and total mechanical energy remains constant (conservative field). ✓ (Option D)

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · One or more correct options

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JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Gravitation

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