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QJEEM20Phb0f6: JEE Main 2020 Physics - Class 11 JEEM20Phb0f6 Solution: Two planets have masses M and 16 M and their radii are and 2 ,…

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

Multiple Choice Options

  • A 2GMa2\sqrt {{{GM} \over a}}
  • B GM2ma\sqrt {{{G{M^2}} \over {ma}}}
  • C 325GMa{3 \over 2}\sqrt {{{5GM} \over a}}
  • D 4GMa4\sqrt {{{GM} \over a}}

Step-by-Step Verified Solution

JEE Main 2020 (Online) 6th September Evening Slot Physics - Gravitation Question 131 English Explanation

Let at point P, net gravitational force = 0

∴\therefore G(M)(m)(10a−x)2=G(16M)(m)x2{{G\left( M \right)\left( m \right)} \over {{{\left( {10a - x} \right)}^2}}} = {{G\left( {16M} \right)\left( m \right)} \over {{x^2}}}

⇒\Rightarrow x = 8a

By Conservation of Mechanical Energy,

Ui + Ki = Uf + Kf

⇒\Rightarrow −GMm8a−16GMm2a- {{GMm} \over {8a}} - {{16GMm} \over {2a}} + 12mv2{1 \over 2}m{v^2}

= −16GMm8a−GMm2a- {{16GMm} \over {8a}} - {{GMm} \over {2a}} + 0

⇒\Rightarrow 12mv2{1 \over 2}m{v^2} = GMm[18a+162a−12a−168a]GMm\left[ {{1 \over {8a}} + {{16} \over {2a}} - {1 \over {2a}} - {{16} \over {8a}}} \right]

⇒\Rightarrow 12mv2{1 \over 2}m{v^2} = GMm[1+64−4−168a]GMm\left[ {{{1 + 64 - 4 - 16} \over {8a}}} \right]

⇒\Rightarrow 12mv2=GMm[458a]{1 \over 2}m{v^2} = GMm\left[ {{{45} \over {8a}}} \right]

⇒\Rightarrow v = 90GM8a\sqrt {{{90GM} \over {8a}}} = 325GMa{3 \over 2}\sqrt {{{5GM} \over a}}

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