Question

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

• A. $\sqrt{\frac{R_e}{2g}}\lfloor {(1+\frac{h}{R_e})}^{3/2}-1\rfloor$
  
• B. $\sqrt{\frac{2R_e}{g}}\lfloor {(1+\frac{h}{R_e})}^{3/2}-1\rfloor$
• C. $\frac{1}{3}\sqrt{\frac{2R_e}{g}}\lfloor {(1+\frac{h}{R_e})}^{3/2}-1\rfloor$
• D. $\frac{1}{3}\sqrt{\frac{R_e}{2g}}\lfloor {(1+\frac{h}{R_e})}^{3/2}-1\rfloor$

Answer 1

The correct option is C $\frac{1}{3}\sqrt{\frac{2R_e}{g}}\lfloor {(1+\frac{h}{R_e})}^{3/2}-1\rfloor$


Applying energy conservation from (1) to (2),

$\frac{1}{2}m\left(\frac{2GM}{R_e}\right)-\frac{GMm}{R_e}=\frac{1}{2}m{v}^2-\frac{GMm}{R+r}$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{GMm}{R+r}$

$\Rightarrow v=\sqrt{\frac{2GM}{R+r}}=\frac{dr}{dt}$

Let, the time taken by it to reach height $h$ is $t$,

$\Rightarrow \sqrt{2GM}{\int }_0^{t}dt={\int }_{R_e}^{R_e+h}\left(\sqrt{R+r}\right)dr$

$\Rightarrow \sqrt{2GM}\times t=\frac{2}{3}{\left[{(R+r)}^{3/2}\right]}_{R_e}^{R_e+h}$

$\Rightarrow t=\frac{2}{3}\sqrt{\frac{R_e^3}{2GM}}[{(1+\frac{h}{R_e})}^{3/2}-1]$

Since, $\frac{GM}{R_e^2}=g$

$\Rightarrow t=\frac{1}{3}\sqrt{\frac{2R_e}{g}}[{(1+\frac{h}{R_e})}^{3/2}-1]$

Hence, $\left(D\right)$ is the correct answer.