The masses and radii of the earth and moon are M₁ , R₁ and M₂ , R₂ respectively. Their centres are distance d apart. The minimum velocity with which a particle of mass m should be projected from a point midway between their centres so that it escape to infinity is

The correct option is 2√(G(M₁+M₂)/d).

To find the minimum velocity required for the particle to escape to infinity, we can apply the principle of conservation of mechanical energy.

Let:
- The mass of the Earth be _M₁
- The mass of the Moon be _M₂
- The distance between their centres be d
- The mass of the particle projected be m

The particle is projected from a point midway between the centres of the Earth and the Moon. Therefore, the distance of the particle from the centre of the Earth is:
$r_1=\frac{d}{2}$
Similarly, the distance of the particle from the centre of the Moon is:
$r_2=\frac{d}{2}$

The total gravitational potential energy (U) of the particle of mass m at this midway point is the sum of the gravitational potential energies due to the Earth and the Moon:
$U=-\frac{G{M}_{1}m}{r_1}-\frac{G{M}_{2}m}{r_2}$
Substituting _r₁ = _r₂ = d/2:
$U=-\frac{G{M}_{1}m}{d/2}-\frac{G{M}_{2}m}{d/2}$
Factoring out the common terms:
$U=-\frac{2Gm}{d}(M_1+M_2)$

If the particle is projected with a minimum escape velocity v, its kinetic energy (K) at the point of projection is:
$K=\frac{1}{2}m{v}^2$

For the particle to escape to infinity, its total mechanical energy at the projection point must be at least zero (since the potential energy and minimum kinetic energy at infinity are both zero):
$E_{\text{total}}=K+U=0$
Substituting the expressions for K and U:
$\frac{1}{2}m{v}^2-\frac{2Gm}{d}(M_1+M_2)=0$

Solving for v by dividing both sides by m and rearranging terms:
$\frac{1}{2}{v}^2=\frac{2G(M_1+M_2)}{d}$
Multiplying by 2:
$v^2=\frac{4G(M_1+M_2)}{d}$
Taking the square root on both sides:
$v=\sqrt{\frac{4G(M_1+M_2)}{d}}=2\sqrt{\frac{G(M_1+M_2)}{d}}$