Practicing Success
The minimum speed of a particle projected from earth’s surface so that it will never return is/are |
$\sqrt{\frac{2 G M}{R}}$ 11.2 km/sec $\sqrt{2 g_0 R}$ all of these |
all of these |
Let the minimum speed of projection of the particle of mass m be v ⇒ KE of the particle KR = $\frac{1}{2}~mv^2$ Gravitational potential energy of the particle. $=U_R=-\frac{G M m}{R}$ where M = mass of earth and R = radius of earth. The particle will have to reach the infinity to escape from earth’s surface for ever. At infinity for very large distance of separation (r → ∞) gravitation potential at infinity is zero ⇒ U∞ = 0 Since the particle just reach the infinity, its KE at infinity is zero ⇒ K∞ = 0. Conservation of total mechanical energy of the particle between the infinity and earth’s surface, we obtain, $U_R+K_R=U_{\infty}+K_{\infty}$ $\Rightarrow-\frac{G M m}{R}+\frac{1}{2} m^2=0+0$ $\Rightarrow v=\sqrt{\frac{2 G M}{R}}=\sqrt{\frac{2 G M}{R^2}} R=\sqrt{2 g_0 R}$ $\Rightarrow v=\left(\sqrt{2 \times 9.8 \times 6.410^6}\right)$ m/sec = 11.2 km/sec Since $g_0=\frac{G M}{R^2}=V=\sqrt{\frac{2 G M}{R}}$. Therefore all the choices are correct. |