Question

Maximum height reached by a rocket fired with a speed equal to $50\%$ of the escape velocity from earth’s surface is:

• A. $R/2$
• B. $16R/9$
• C. $R/3$
• D. $R/8$

Answer 1

Answer: (C)

$V=\frac{50}{100}{V}_e=\frac{1}{2}\sqrt{\frac{2GM}{R}}$

Apply energy conservation

$\Rightarrow -\frac{GMm}{R}+\frac{1}{2}m{V}^2=-\frac{GMm}{\left(R+h\right)}$

$v^2=\frac{2GM}{R}-\frac{2GM}{R+h}$

$\frac{1}{4}\cdot \frac{2GM}{R}=2GM\left(\frac{1}{R}-\frac{1}{R+h}\right)$

$\therefore \frac{1}{4R}=\frac{h}{R\left(R+h\right)}$

$\therefore R+h=4h$

$\Rightarrow h=\frac{R}{3}$