JEE MAIN - Physics (2020 - 6th September Evening Slot - No. 22)
Two planets have masses M and 16 M and their radii are $$a$$ and 2$$a$$, respectively. The separation
between the centres of the planets is 10$$a$$. 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 :
$$2\sqrt {{{GM} \over a}} $$
$$\sqrt {{{G{M^2}} \over {ma}}} $$
$${3 \over 2}\sqrt {{{5GM} \over a}} $$
$$4\sqrt {{{GM} \over a}} $$
Explanation
_6th_September_Evening_Slot_en_22_1.png)
Let at point P, net gravitational force = 0
$$ \therefore $$ $${{G\left( M \right)\left( m \right)} \over {{{\left( {10a - x} \right)}^2}}} = {{G\left( {16M} \right)\left( m \right)} \over {{x^2}}}$$
$$ \Rightarrow $$ x = 8a
By Conservation of Mechanical Energy,
Ui + Ki = Uf + Kf
$$ \Rightarrow $$ $$ - {{GMm} \over {8a}} - {{16GMm} \over {2a}}$$ + $${1 \over 2}m{v^2}$$
= $$ - {{16GMm} \over {8a}} - {{GMm} \over {2a}}$$ + 0
$$ \Rightarrow $$ $${1 \over 2}m{v^2}$$ = $$GMm\left[ {{1 \over {8a}} + {{16} \over {2a}} - {1 \over {2a}} - {{16} \over {8a}}} \right]$$
$$ \Rightarrow $$ $${1 \over 2}m{v^2}$$ = $$GMm\left[ {{{1 + 64 - 4 - 16} \over {8a}}} \right]$$
$$ \Rightarrow $$ $${1 \over 2}m{v^2} = GMm\left[ {{{45} \over {8a}}} \right]$$
$$ \Rightarrow $$ v = $$\sqrt {{{90GM} \over {8a}}} $$ = $${3 \over 2}\sqrt {{{5GM} \over a}} $$
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