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Question

A spaceship is sent to investigate a planet of mass $$M$$ and radius $$R$$. While hanging motionless in space at a distance $$5R$$ from the centre of the planet, the spaceship fires an instrument package of mass $$m$$, which is much smaller than the mass of the spaceship. The angle $$\theta $$ for which the package just grazes the surface of the planet is
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A
sin1121+2GM5v20R
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B
sin1151+8GM5v20R
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C
sin1181+3GM5v20R
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D
sin1131+6GM5v20R
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Solution

The correct option is B $$\sin ^{ -1 }{ \left( \dfrac { 1 }{ 5 } \sqrt { 1+\dfrac { 8GM }{ 5{ v }_{ 0 }^{ 2 }R } } \right) } $$
Since no external torque  is present around the centre of the planet. angular momentum about the centre will be conserved.
  $$ \Rightarrow mV_o\sin\theta (5R)=mVR$$
 $$ \Rightarrow V=5V_osin\theta.............(i)$$
Energy conservation (potential+kinetic) of the system
$$ \frac{1}{2}mV_o^{2}- \dfrac{GMm}{5R}= \dfrac{1}{2}mV^{2}- \dfrac{GMm}{R}  $$  
Substituting $$V$$ from eq. $$(i)$$
 $$ \Rightarrow  \dfrac{V_o^{2}}{2}- \dfrac{GM}{5R}= \dfrac{25V_o^{2}\sin^{2}\theta}{2}- \dfrac{GM}{R}    $$
  
$$ \Rightarrow  \dfrac{25\sin^{2}\theta-1}{2}V_o^{2}= \dfrac{4GM}{5R}  $$
 $$ \Rightarrow \sin\theta =  \dfrac{1}{5} \sqrt{1+ \dfrac{8GM}{5V_o^{2}R} }  $$
or $$ \Rightarrow \theta = \sin^{-1}\left(\dfrac{1}{5} \sqrt{1+ \dfrac{8GM}{5V_o^{2}R} }\right)$$
Thus (B) is the correct option.

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Physics

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