Motion of planets and satellites

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The radius of the earth is reduced by 4%. The mass of the earth remains unchanged. The escape velocity increases by K%. Find the value of K.

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Solution

$Escape velocity = v = \sqrt{\frac{2 G M}{R}} \Rightarrow v^{2} = \frac{2 G M}{R} . . . . . . (i) v^{2} = (2 G M) R^{- 1} Differentiating both sides, we get, 2 v \frac{d v}{d R} = - \frac{2 G M}{R^{2}} \Rightarrow v \frac{d v}{d R} = \frac{G M}{R^{2}} . . . . . . . (i i) Dividing (ii) by (i), \frac{1}{v} \frac{d v}{d R} = - \frac{1}{2 R} \Rightarrow |\frac{d v}{v}| \times 100 = \frac{1}{2} \times 4 % = 2 % ∴ If the radius decreases by 4 %, escape velocity will increase by 2%.$

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Motion of planets and satellites

Remember concepts with our Masterclasses.

The radius of the earth is reduced by 4%. The mass of the earth remains unchanged. The escape velocity increases by K%. Find the value of K.

Escape velocity =v=2GMR ⇒ v2=2GMR......(i) v2=(2GM)R−1 Differentiating both sides, we get, 2vdvdR=−2GMR2⇒vdvdR=GMR2.......(ii) Dividing (ii) by (i), 1vdvdR=−12R ⇒dvv×100=12×4%=2% ∴ If the radius decreases by 4% , escape velocity will increase by 2%.

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