JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Gravitation

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Two planets have masses M and 16 M and their radii are a and 2a, respectively. The separation between the centres of the planets is 10a. A body of mass m is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach at the surface of smaller planet, the minimum firings speed needed is:

(A) $2 \frac{GM}{a}$
(B) $4 \frac{GM}{a}$
(C) $\frac{G M ^{2}}{ma}$
(D) $\frac{3}{2} \frac{5 GM}{a}$ ✓ Correct

Answer: (D)

Solution

The body need only be driven as far as the neutral point between the planets, where the two gravitational pulls cancel; beyond that the smaller planet draws it in on its own. Setting

\frac{GM}{x ^{2}} = \frac{G ( 16 M )}{( 10 a - x ) ^{2}}

places that point at

x = 2 a

from the smaller planet. At the launch point on the larger planet's surface the potential is

V_{A} = - \frac{GM}{8 a} - \frac{16 GM}{2 a} = - \frac{65 GM}{8 a}

, and at the neutral point it is

V_{C} = - \frac{GM}{2 a} - \frac{16 GM}{8 a} = - \frac{5 GM}{2 a}

. Energy conservation,

\frac{1}{2} v^{2} = V_{C} - V_{A} = \frac{45 GM}{8 a}

, then fixes the minimum launch speed at

v = \frac{3}{2} \frac{5 GM}{a}

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Gravitation

Solution

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