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(A) Let the body be at a distance $r$ from the centre of the earth with velocity $v$. By the law of conservation of mechanical energy:$(TE)_{	ext{at

Vedclass · Accepted Answer

$44.44$

Vedclass · Answer

(A) Let the body be at a distance $r$ from the centre of the earth with velocity $v$. By the law of conservation of mechanical energy:$(TE)_{	ext{at surface}} = (TE)_{	ext{at distance } r}$$\frac{1}{2} m v_e^2 - \frac{GMm}{R} = \frac{1}{2} m v^2 - \frac{GMm}{r}$Since $v_e = \sqrt{\frac{2GM}{R}}$,we have $\frac{1}{2} v_e^2 = \frac{GM}{R} = gR$.Substituting this,$\frac{1}{2} v^2 = gR - gR + \frac{GM}{r} = \frac{GM}{r} = \frac{gR^2}{r}$.Thus,$v = \frac{dr}{dt} = R \sqrt{\frac{2g}{r}}$.Rearranging and integrating from $r = R$ to $r = R + h = 4R$:$\int_0^t dt = \int_R^{4R} \sqrt{\frac{r}{2gR^2}} dr = \frac{1}{R\sqrt{2g}} \int_R^{4R} r^{1/2} dr$$t = \frac{1}{R\sqrt{2g}} \left[ \frac{2}{3} r^{3/2} ight]_R^{4R} = \frac{2}{3R\sqrt{2g}} \left[ (4R)^{3/2} - R^{3/2} ight]$$t = \frac{2}{3R\sqrt{2g}} R^{3/2} [8 - 1] = \frac{2}{3} \sqrt{\frac{R}{2g}} 	imes 7 = \frac{7}{3} \sqrt{\frac{2R}{g}}$Substituting $R = 6.4 	imes 10^6 \ m$ and $g = 9.8 \ m/s^2$:$t = \frac{7}{3} \sqrt{\frac{2 	imes 6.4 	imes 10^6}{9.8}} = \frac{7}{3} \sqrt{1.306 	imes 10^6} \approx \frac{7}{3} 	imes 1142.8 \approx 2666.5 \ s$$t = \frac{2666.5}{60} \approx 44.44 \ min$.

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