Imagine that you are in a spacecraft orbiting around the earth in a circle of radius $7000 k m$ (from the centre of the earth). If you decrease the magnitude of mechanical energy of the spacecraft-earth system by $10$ % by firing the rockets, then what is the greatest height you can take your spacecraft above the surface of the earth? [ $R_{e} = 6400 k m$ ]

6400 k m

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540 k m

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1377 k m

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3000 k m

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Solution

The correct option is C $1377 k m$
Energy before firing $P E_{1} = - \frac{G M m}{R_{1}}$
Energy after firing $P E_{2} = 0.9 \times - \frac{G M m}{R_{1}} = - \frac{G M m}{R_{2}}$
$⟹ + \frac{0.9 G M m}{7000} = + \frac{G M m}{R_{2}}$
$⟹ R_{2} = \frac{7000}{0.9} = 7777.8 k m$
$∴$ Height= $7777.8 - 6400$
$= 1377.8 k m$

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The correct option is C 1377kmEnergy before firing PE1=−GMmR1Energy after firing PE2=0.9×−GMmR1=−GMmR2⟹+0.9GMm7000=+GMmR2⟹R2=70000.9=7777.8km∴ Height=7777.8−6400=1377.8km