A projectile of mass m is fired from the surface of earth at an angle α=60∘ from the vertical. The initial speed of the projectile is u=GMR where M and R are mass and radius of the earth. Find the height by the projectile in kilometer. Neglect air resistance and the rotation of earth. (Given R = 6400 km)

Solution:

Let v be the speed of projectile at the highest point. By conservation of angular momentum and conservation of mechanical energy, we have

$m u R \sin α = m v r_{max}$

$\begin{array}{l} and \frac{1}{2} m u^{2} - \frac{G M m}{R} = \frac{1}{2} m v^{2} - \frac{G M m}{r_{max}} \\ Solving these two equations, we get r_{max} = \frac{3 R}{2} \\ ∴ h_{max} = r_{max} - R = \frac{R}{2} = 3200 km \end{array}$

Solution:

Related content