Question

A projectile of mass 100 kg is shot from the surface of Earth by means of a very powerful cannon. If the projectile reaches a height of 65,000 m above Earth's surface, what was the speed of the projectile when it left the cannon? (Mass of Earth 5.97x10^24 kg, Radius of Earth 6.37x10^6 m)

Answer 1

Answer:

$v = 1.11 \times 10^3 m/s$

Explanation:

By energy conservation law we will have

$U_i + KE_i = U_f + KE_f$

as we know that as the projectile is rising up then due to gravitational attraction of earth it will slow down

At the highest position the speed of the projectile will become zero

So here we will have

$-\frac{GMm}{R} + \frac{1}{2}mv^2 = -\frac{GMm}{R+h} + 0$

$-\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(100)}{6.37 \times 10^6} + \frac{1}{2}(100)v^2 = -\frac{(6.67 \times 10^{-11})(5.97 \times 10^{-11})(100)}{(6.37 \times 10^{6} + 65000)}$

$- 6.25 \times 10^7 + 0.5 v^2 = -6.19 \times 10^7$

$v = 1.11 \times 10^3 m/s$