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}mv_{s}^{2}+\left(- \frac{G M m}{R}\right)=\frac{1}{2}mv_{0}^{2}-\frac{G M m}{r}\ldots \left(1\right)$
By Newton's law, $\frac{G M m}{r^{2}}=\frac{m v_{0}^{2}}{r}\ldots \left(2\right)$
Solving equations (1) and (2)
$v_{0}=v_{e}\sqrt{1 - \eta^{2}}$