The minimum speed of a particle projected from earth’s surface so tha…

Let the minimum speed of projection of the particle of mass m be v

⇒ KE of the particle K_R = $\frac{1}{2}~mv^2$ 

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.