JEE MAIN - Physics (2020 - 3rd September Morning Slot - No. 16)
A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to
that of the earth’s radius Re
. By firing rockets attached to it, its speed is instantaneously increased
in the direction of its motion so that it become $$\sqrt {{3 \over 2}} $$
times larger. Due to this the farthest distance
from the centre of the earth that the satellite reaches is R. Value of R is :
2Re
3Re
4Re
2.5Re
Explanation
_3rd_September_Morning_Slot_en_16_1.png)
V0 = $$\sqrt {{{GM} \over {{R_e}}}} $$
Applying Conservation of Angular Momentum,
mVRe = mV'R
$$ \Rightarrow $$ m$$\sqrt {{3 \over 2}} {V_0}{R_e}$$ = mV'R
$$ \Rightarrow $$ V' = $$\sqrt {{3 \over 2}} {{{V_0}{R_e}} \over R}$$
Applying Conservation of Energy,
$${ - {{GMm} \over {{R_e}}}}$$ + $${1 \over 2}m{V^2}$$ = $${ - {{GMm} \over R_{max}}}$$ + $${1 \over 2}mV{'^2}$$
$$ \Rightarrow $$ $${ - {{GMm} \over {{R_e}}}}$$ + $${1 \over 2}m\left( {{3 \over 2}V_0^2} \right)$$ = $${ - {{GMm} \over R_{max}}}$$ + $${1 \over 2}m \times $$$${\left( {V'} \right)^2}$$
$$ \Rightarrow $$ $${ - {{GMm} \over {{R_e}}} + }$$ $${1 \over 2}m \times {3 \over 2}{{GM} \over {{R_e}}}$$ = $$ - {{GMm} \over R_{max}} + {1 \over 2}m \times {3 \over 2}{V_0}{{R_e^2} \over {{R^2}}}$$
$$ \Rightarrow $$ $${ - {{GMm} \over {{R_e}}} + }$$ $${1 \over 2}m \times {3 \over 2}{{GM} \over {{R_e}}}$$ = $$ - {{GMm} \over R_{max}} + {1 \over 2}m \times {3 \over 2}{{GM} \over {{R_e}}}{{R_e^2} \over {{R^2}}}$$
$$ \Rightarrow $$ $$ - {1 \over {{R_e}}} + {3 \over {4{R_e}}}$$ = $$ - {1 \over R} + {{3{R_e}} \over {4{R^2}}}$$
$$ \Rightarrow $$ $${1 \over {4{R_e}}}$$ = $$ - {1 \over R} + {{3{R_e}} \over {4{R^2}}}$$
$$ \Rightarrow $$ R = 3Re
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