A body of mass $$m$$ is projected with velocity $$\lambda v_e$$ in vertically upward direction from the surface of the earth into space. It is given that $$v_e$$ is escape velocity and $$\lambda < 1$$. If air resistance is considered to be negligible, then the maximum height from the centre of earth, to which the body can go, will be ($$R$$ : radius of earth)

A body of mass $$m$$ is projected vertically upward with velocity $$\lambda v_e$$ (where $$v_e$$ is escape velocity and $$\lambda < 1$$). We need to determine its maximum height measured from the centre of the earth.

Since energy is conserved, at the surface (distance $$R$$ from the centre) the sum of kinetic and gravitational potential energy equals the total energy at the maximum height $$h$$ (where the velocity becomes zero):

$$\frac{1}{2}m(\lambda v_e)^2 - \frac{GMm}{R} = 0 - \frac{GMm}{h}$$

We know that $$v_e = \sqrt{\frac{2GM}{R}}$$, so $$v_e^2 = \frac{2GM}{R}$$, which implies $$GM = \frac{v_e^2 R}{2}$$.

Substituting this expression for $$GM$$ into the energy equation gives

$$\frac{1}{2}m\lambda^2 v_e^2 - \frac{GMm}{R} = -\frac{GMm}{h}$$

Next, dividing through by $$m$$ yields

$$\frac{1}{2}\lambda^2 v_e^2 - \frac{GM}{R} = -\frac{GM}{h}$$

Substituting $$GM = \frac{v_e^2 R}{2}$$ once more into this result gives

$$\frac{1}{2}\lambda^2 v_e^2 - \frac{v_e^2}{2} = -\frac{v_e^2 R}{2h}$$

Dividing both sides by $$\frac{v_e^2}{2}$$ leads to

$$\lambda^2 - 1 = -\frac{R}{h}$$

Rewriting this expression gives

$$1 - \lambda^2 = \frac{R}{h}$$

Therefore, solving for $$h$$ yields

$$h = \frac{R}{1 - \lambda^2}$$

This represents the maximum distance from the centre of the earth.

Answer: Option B: $$\dfrac{R}{1-\lambda^2}$$