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(A) The body is projected with escape velocity $v_e = \sqrt{\frac{2GM}{R_e}}$.By conservation of energy at the surface and at a distance $r$ from the

Vedclass · Accepted Answer

$\frac{1}{3} \sqrt{\frac{2 R_{e}}{g}}\left[\left(1+\frac{h}{R_{e}}\right)^{3 / 2}-1\right]$

Vedclass · Answer

(A) The body is projected with escape velocity $v_e = \sqrt{\frac{2GM}{R_e}}$.By conservation of energy at the surface and at a distance $r$ from the center of the Earth:$\frac{1}{2}mv^2 - \frac{GMm}{r} = 0$ (since total energy is zero for escape velocity).$\frac{1}{2}mv^2 = \frac{GMm}{r} \Rightarrow v = \sqrt{\frac{2GM}{r}} = \frac{dr}{dt}$.Separating variables: $dt = \frac{dr}{\sqrt{2GM}} \cdot \sqrt{r}$.Integrating from $t=0$ at $r=R_e$ to time $t$ at $r=R_e+h$:$t = \frac{1}{\sqrt{2GM}} \int_{R_e}^{R_e+h} r^{1/2} dr = \frac{1}{\sqrt{2GM}} \cdot \frac{2}{3} [r^{3/2}]_{R_e}^{R_e+h}$.$t = \frac{2}{3\sqrt{2GM}} [ (R_e+h)^{3/2} - R_e^{3/2} ] = \frac{2}{3\sqrt{2GM}} R_e^{3/2} [ (1 + \frac{h}{R_e})^{3/2} - 1 ]$.Since $GM = gR_e^2$,we have $\sqrt{GM} = \sqrt{g}R_e$.Substituting this: $t = \frac{2 R_e^{3/2}}{3 \sqrt{2} \sqrt{g} R_e} [ (1 + \frac{h}{R_e})^{3/2} - 1 ] = \frac{1}{3} \sqrt{\frac{2R_e}{g}} [ (1 + \frac{h}{R_e})^{3/2} - 1 ]$.

$A$ body is projected vertically upwards from the surface of the Earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height $h$ is $....\,S.$

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