Question 194: Gravitation

Question

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 :

Accepted Answer

V
0
= $$\sqrt {{{GM} \over {{R_e}}}} $$
Applying Conservation of Angular Momentum,
mVR
e
= 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,
$${ - {{GMm} \over {{R_e}}}}$$ + $${1 \over 2}m{V^2}$$ = $${ - {{GMm} \over R_{max}}}$$ + $${1 \over 2}mV{'^2}$$
$$ \Rightarrow $$ $${ - {{GMm} \over {{R_e}}}}$$ + $${1 \over 2}m\left( {{3 \over 2}V_0^2} ight)$$ = $${ - {{GMm} \over R_{max}}}$$ + $${1 \over 2}m 	imes $$$${\left( {V'} ight)^2}$$
$$ \Rightarrow $$ $${ - {{GMm} \over {{R_e}}} + }$$ $${1 \over 2}m 	imes {3 \over 2}{{GM} \over {{R_e}}}$$ = $$ - {{GMm} \over R_{max}} + {1 \over 2}m 	imes {3 \over 2}{V_0}{{R_e^2} \over {{R^2}}}$$
$$ \Rightarrow $$ $${ - {{GMm} \over {{R_e}}} + }$$ $${1 \over 2}m 	imes {3 \over 2}{{GM} \over {{R_e}}}$$ = $$ - {{GMm} \over R_{max}} + {1 \over 2}m 	imes {3 \over 2}{{GM} \over {{R_e}}}{{R_e^2} \over {{R^2}}}$$
$$ \Rightarrow $$ $$ - {1 \over {{R_e}}} + {3 \over {4{R_e}}}$$ = $$ - {1 \over R} + {{3{R_e}} \over {4{R^2}}}$$
$$ \Rightarrow $$ $${1 \over {4{R_e}}}$$ = $$ - {1 \over R} + {{3{R_e}} \over {4{R^2}}}$$
$$ \Rightarrow $$ R = 3R
e

Answer

2Re

Answer

3Re

Answer

4Re

Answer

2.5Re

Gravitation

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