An object starts vertically with speed Vo. Prove that its speed V at height h is given by V 2 o -V 2 = (2gh/1+h/R) where R is the radius of the earth. The earth is a solid sphere of radius 6.4 x 10 6 m and g = 9.8 ms -2 at its surface. Find the vertical speed with which an object should be thrown so as to reach up to height from the surface of the earth. Briefly explain the concept of escape velocity.

Assume potential energy is zero at the starting point.

Initial energy: kinetic energy + potential energy = $\frac{1}{2} m v_0^2 + 0$

Final energy at height $h$: $\frac{1}{2} m v^2 + m g h$

$$\frac{1}{2} m v_0^2 = \frac{1}{2} m v^2 + m g h$$

$$v^2 = v_0^2 - 2 g h$$

v^2 = v_0^2 - 2 g h

$\frac{1}{2} m v_0^2 - \frac{G M m}{R}$

Final energy at $r = 2R$: $0 - \frac{G M m}{2 R}$

$$\frac{1}{2} m v_0^2 - \frac{G M m}{R} = -\frac{G M m}{2 R}$$

$$v_0^2 = \frac{g R^2}{R} = g R$$ $$v_0 = \sqrt{g R}$$

$g R = 9.8 \times 6.4 \times 10^6$

$9.8 \times 6.4 = 62.72$

$g R = 6.272 \times 10^7 \, \mathrm{m^2/s^2}$

$\sqrt{6.272 \times 10^7} = \sqrt{6.272} \times \sqrt{10^7} \approx 2.504 \times 3162.3 \approx 7920 \, \mathrm{m/s}$

7920 m/s