A body is projected vertically upwards from the surface of the earth with a velocity sufficient to carry it…

1. 44.44 min
2. 22.22 min
3. 18.76 min
4. 37.52 min

Solution

If at a distance $r$ from the centre of the earth, the body has velocity $v$, then by conservation of mechanical energy $(\mathrm{TE})_{\text {at weight }}=(\mathrm{TE})_{\text {at surface }}$ $ \frac{1}{2} m v^2+\left(-\frac{G M m}{r}\right)=\frac{1}{2} m v_e^2+\left(-\frac{G M m}{R}\right) $ where, $v_e=$ escape velocity.
$ \begin{aligned} v^2=2 g R+2 g R\left(\frac{R}{r}-1\right) \end{aligned} $
Integrating Eq. (i), we get $ \begin{gathered} \int_0^t d t=\frac{1}{R \sqrt{2 g}} \int_R^{R+h} r^{1 / 2} \cdot d r \\ t=\frac{2}{3} \frac{1}{R \sqrt{2 g}}\left[(R+h)^{3 / 2}-R^{3 / 2}\right] \\ t=\frac{1}{3} \sqrt{\frac{2 R}{g}\left[\left(1+\frac{h}{R}\right)^{3 / 2}-1\right]} \end{gathered} $ According to question, $h=3 R$ $ \begin{aligned} & t=\frac{1}{3} \sqrt{\frac{2 R}{g}\left[\left(1+\frac{3 R}{R}\right)^{3 / 2}-1\right]} \\ & t=\frac{1}{3} \sqrt{\frac{2 R}{g}} \times 7 \end{aligned} $ Putting all values, we get $ \begin{aligned} t= & \frac{1}{3}\left(\frac{2 \times 6400 \times 10^3}{9.8}\right) \times 7 \\ & {\left[\because R=6400 \times 10^3\right] } \\ t & =\frac{80}{3} \times 14.2 \times 7 \mathrm{~s} \\ t & =2666.65 \mathrm{~s} \\ t & =44.44 \mathrm{~min} \end{aligned} $

Asked in: AP EAMCET 2018 (24 Apr Shift 1)

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