Question

A rocket of constant mass starts vertically upwards with speed $V_0$. Choose the correct options:
(R is the radius of the earth).

• A. Speed V at a height h is given by $V_0^2-V^2=\frac{2gh}{1+\frac{h}{R}}$
• B. The maximum height reached by the rocket fired with a speed of 90% of escape velocity is 2.26R.
• C. The maximum height reached by the rocket fired with a speed of 90% of escape velocity is 4.26R.
• D. Speed V at a height h is given by $V_0^2-V^2=\frac{2gh}{1+\frac{h}{2R}}$

Answer 1

Answer: (A), (C)

The correct options are
A Speed V at a height h is given by $V_0^2-V^2=\frac{2gh}{1+\frac{h}{R}}$
C The maximum height reached by the rocket fired with a speed of 90% of escape velocity is 4.26R.

To find the height $h$ reached by the rocket, we conserve energy between the starting point and at the height $h$
$\frac{1}{2}m{V}_o^2-\frac{GMm}{R}=-\frac{GMm}{R+h}+\frac{1}{2}m{V}^2$
$\Rightarrow \frac{1}{2}(V_o^2-V^2)=\frac{GM}{R}-\frac{GM}{R+h}$
$\Rightarrow \frac{1}{2}(V_o^2-V^2)=GM(\frac{1}{R}-\frac{1}{R+h})$
$\Rightarrow \frac{1}{2}(V_o^2-V^2)=\frac{GM}{R}\frac{h}{R+h}$
$\Rightarrow \frac{1}{2}(V_o^2-V^2)=\frac{ghR}{R+h}$
$\Rightarrow V_o^2-V^2=\frac{2gh}{1+\frac{h}{R}}$

Option A is correct.

For the maximum height reached, $V$ in the above equation would be zero.
Hence with an initial velocity $0.9V_e$;
$(0.9\times \sqrt{2gR}{)}^2=\frac{2gh}{1+\frac{h}{R}}$
$\Rightarrow 0.81\times 2gR=\frac{2gR}{\frac{R}{h}+1}$
$\Rightarrow 1+\frac{R}{h}=\frac{1}{0.81}$
$\Rightarrow \frac{R}{h}=1.234-1\Rightarrow \frac{R}{h}=0.234\Rightarrow h=4.26R$;

Option C is correct.