Question

What is the maximum height attained by a body projected with a velocity equal to one-third of the escape velocity from the surface of the earth? (Radius of the earth $=R$ )

• A. R / 2
• B. R / 3
• C. R / 5
• D. R / 8

Answer 1

Answer: (D)

From conservation of energy, $\frac{1}{2} m v_{1}^{2}-\frac{G M_{e} m}{R}$
$=\frac{1}{2} m v_{2}^{2}-\frac{G M_{e} m}{R+H}$
Here, $v_{2}=0$ (At maximum height)
$\therefore \frac{1}{2} m v_{1}-\frac{G M_{e} m}{R}$
$=-\frac{G M_{e} m}{R+H}$
Solving, we get
$H=\frac{R}{\frac{2 g R}{v_{1}^{2}}-1}=\frac{R}{\frac{v_{e}^{2}}{v_{1}^{2}}-1}$
Where $v_{e}$ is escape velocity
Here $v=\frac{1}{3} v_{e}$
$\Rightarrow \frac{v_{e}}{v}=3$
$\therefore H=\frac{R}{9-1}=\frac{R}{8}$