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
$H=\frac{R}{\frac{2gR}{v_1^2}-1}=\frac{R}{\frac{v_e^2}{v_1^2}-1}$
Where $v_e$ is escape velocity
Here $v=\frac{1}{3}{v}_e$
$\Rightarrow \frac{v_e}{v}=3$
$\therefore H=\frac{R}{9-1}=\frac{R}{8}$