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QJEEM20Ph4ff7: JEE Main 2020 Physics - Class 11 JEEM20Ph4ff7 Solution: A satellite is moving in a low nearly circular orbit around the …

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 32\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 :

Multiple Choice Options

  • A 2Re
  • B 3Re
  • C 4Re
  • D 2.5Re

Step-by-Step Verified Solution

JEE Main 2020 (Online) 3rd September Morning Slot Physics - Gravitation Question 138 English Explanation

V0 = GMRe\sqrt {{{GM} \over {{R_e}}}}

Applying Conservation of Angular Momentum,

mVRe = mV'R

⇒\Rightarrow m 32V0Re\sqrt {{3 \over 2}} {V_0}{R_e} = mV'R

⇒\Rightarrow V' = 32V0ReR\sqrt {{3 \over 2}} {{{V_0}{R_e}} \over R}

Applying Conservation of Energy,

−GMmRe{ - {{GMm} \over {{R_e}}}} + 12mV2{1 \over 2}m{V^2} = −GMmRmax{ - {{GMm} \over R_{max}}} + 12mV′2{1 \over 2}mV{'^2}

⇒\Rightarrow −GMmRe{ - {{GMm} \over {{R_e}}}} + 12m(32V02){1 \over 2}m\left( {{3 \over 2}V_0^2} \right) = −GMmRmax{ - {{GMm} \over R_{max}}} + 12m×{1 \over 2}m \times (V′)2{\left( {V'} \right)^2}

⇒\Rightarrow −GMmRe+{ - {{GMm} \over {{R_e}}} + } 12m×32GMRe{1 \over 2}m \times {3 \over 2}{{GM} \over {{R_e}}} = −GMmRmax+12m×32V0Re2R2- {{GMm} \over R_{max}} + {1 \over 2}m \times {3 \over 2}{V_0}{{R_e^2} \over {{R^2}}}

⇒\Rightarrow −GMmRe+{ - {{GMm} \over {{R_e}}} + } 12m×32GMRe{1 \over 2}m \times {3 \over 2}{{GM} \over {{R_e}}} = −GMmRmax+12m×32GMReRe2R2- {{GMm} \over R_{max}} + {1 \over 2}m \times {3 \over 2}{{GM} \over {{R_e}}}{{R_e^2} \over {{R^2}}}

⇒\Rightarrow −1Re+34Re- {1 \over {{R_e}}} + {3 \over {4{R_e}}} = −1R+3Re4R2- {1 \over R} + {{3{R_e}} \over {4{R^2}}}

⇒\Rightarrow 14Re{1 \over {4{R_e}}} = −1R+3Re4R2- {1 \over R} + {{3{R_e}} \over {4{R^2}}}

⇒\Rightarrow R = 3Re

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