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JEE Main 2020 (6 Sep, Shift 2) · Escape Velocity

Escape Velocity question from JEE Main 2020 (6 Sep, Shift 2)

Q61JEE Main 2020 · 6 SEP · SHIFT 2Difficult

Escape Velocity

Two planets have masses M and 16 M and their radii are aa and 2aa, respectively. The separation between the centres of the planets is 10aa. 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 firing speed needed is:

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Answer: (c)

  1. Neutral point at xx from MM: Mx2=16M(10a−x)2⇒x=2a\dfrac{M}{x^2} = \dfrac{16M}{(10a-x)^2} \Rightarrow x = 2a (i.e. 8a8a from the larger planet).
  2. The body only needs to reach this point; after it, the smaller planet pulls it in.
  3. At surface of larger: V1=−16GM2a−GM8a=−65GM8aV_1 = -\dfrac{16GM}{2a} - \dfrac{GM}{8a} = -\dfrac{65GM}{8a}; at neutral point: VN=−GM2a−16GM8a=−20GM8aV_N = -\dfrac{GM}{2a} - \dfrac{16GM}{8a} = -\dfrac{20GM}{8a}.
  4. 12v2=VN−V1=45GM8a⇒v=325GMa\tfrac12 v^2 = V_N - V_1 = \dfrac{45GM}{8a} \Rightarrow v = \dfrac32\sqrt{\dfrac{5GM}{a}}
  5. So the answer is (c).
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