A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the Sun and the Earth. The Sun is $3 \times 10^5$ times heavier than the Earth and is at a distance $2.5 \times 10^4$ times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is $v_e = 11.2\,\text{km s}^{-1}$. The minimum initial velocity ($v_s$) required for the rocket to be able to leave the Sun-Earth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet)

Step 1: Define the given parameters.

The mass of the Sun $M_S$ and the distance $d$ between the Sun and the Earth are expressed in terms of Earth's mass $M_E$ and Earth's radius $R_E$.

$$M_S = 3 \times 10^5 M_E$$ $$d = 2.5 \times 10^4 R_E$$

The escape velocity from Earth's gravitational field is given.

$$v_e = 11.2 \text{ km s}^{-1}$$

Step 2: Apply conservation of energy for escape.

For the rocket to escape the Sun-Earth system, its total mechanical energy must be greater than or equal to zero.

$$\frac{1}{2} v_s^2 - \frac{G M_E}{R_E} - \frac{G M_S}{d + R_E} \ge 0$$

The minimum initial velocity $v_s$ required to escape satisfies the boundary condition.

$$\frac{1}{2} v_s^2 = \frac{G M_E}{R_E} + \frac{G M_S}{d + R_E}$$

Step 3: Relate the gravitational potential terms to Earth's escape velocity.

The escape velocity from Earth's surface is defined by the gravitational potential of the Earth.

$$\frac{G M_E}{R_E} = \frac{v_e^2}{2}$$

Since $d \gg R_E$, the distance from the Sun to the rocket is approximated as $d + R_E \approx d$.

$$\frac{G M_S}{d + R_E} \approx \frac{G M_S}{d}$$

Substitute the given ratios of mass and distance.

$$\frac{G M_S}{d} = \frac{G (3 \times 10^5 M_E)}{2.5 \times 10^4 R_E}$$ $$\frac{G M_S}{d} = 12 \frac{G M_E}{R_E}$$ $$\frac{G M_S}{d} = 6 v_e^2$$

Step 4: Solve for the minimum initial velocity $v_s$.

Substitute the simplified gravitational terms back into the energy equation.

$$\frac{1}{2} v_s^2 = \frac{v_e^2}{2} + 6 v_e^2$$ $$\frac{1}{2} v_s^2 = 6.5 v_e^2$$ $$v_s^2 = 13 v_e^2$$ $$v_s = \sqrt{13} v_e$$

Step 5: Calculate the numerical value.

Substitute the value of $v_e$.

$$v_s \approx 3.606 \times 11.2 \text{ km s}^{-1}$$ $$v_s \approx 40.4 \text{ km s}^{-1}$$

The closest value to $40.4 \text{ km s}^{-1}$ is $42 \text{ km s}^{-1}$.

Answer: $42 \text{ km s}^{-1}$