Question

A projectile is fired from the surface of the Earth of radius $R$ with a velocity $\eta v_e$ , where $v_e$ is the escape velocity and $\eta <1$ . Neglecting the air resistance, the orbital velocity of the projectile is

• A. $v_e\sqrt{1-{\eta }^2}$
• B. $v_e\sqrt{\frac{{\eta }^2}{5}}$
• C. $\frac{2}{5}{v}_e\sqrt{\eta }$
• D. $\frac{2\eta }{5}{v}_e$

Answer 1

Answer: (A)

Velocity at the surface $=v_s=\eta v_e$
Velocity in orbit $=v_0$
By conversation of mechanical energy,
$\frac{1}{2}m{v}_s^2+\left(-\frac{GMm}{R}\right)=\frac{1}{2}m{v}_0^2-\frac{GMm}{r}\dots \left(1\right)$
By Newton's law, $\frac{GMm}{r^2}=\frac{m{v}_0^2}{r}\dots \left(2\right)$
Solving equations (1) and (2)
$v_0=v_e\sqrt{1-{\eta }^2}$