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 xx 10^(5)` times heavier than the earth and is at a distance `2.5 xx 10^(4)` times larger than the radius of the earth. the escape velocity from earth's gravitational field is `u_(e) = 11.2 kms^(-1)`. The minmum initial velocity `(u_(e)) = 11.2 kms^(-1)`. the minimum initial velocity `(u_(s))` required for the rocket to be able to leave the sun-earth system is closest to (Ignore the rotation of the earth and the presence of any other planet

Options

A. (a)`v_s=62 km s^(-1)`
B. (b)`v_s=42 km s^(-1)`
C. (c)`v_s=72 km s^(-1)`
D. (d)`v_s=22 km s^(-1)`

Explanation

To solve the problem, we need to determine the minimum initial velocity \( u_s \) required for a rocket to escape the gravitational influence of both the Earth and the Sun.

### Step-by-Step Solution:

1. **Understanding the Masses and Distances**:
- Let the mass of the Earth be \( M_e \).
- The mass of the Sun is given as \( M_s = 3 \times 10^5 M_e \).
- The distance from the Earth to the Sun is \( d = 2.5 \times 10^4 R_e \), where \( R_e \) is the radius of the Earth.

2. **Escape Velocity from Earth**:
- The escape velocity from the Earth's gravitational field is given as \( u_e = 11.2 \, \text{km/s} \).
- The escape velocity formula from a planet is given by:
\[ u_e = \sqrt{\frac{2GM_e}{R_e}} \]
- Where \( G \) is the universal gravitational constant.

3. **Escape Velocity from the Sun-Earth System**:
- To escape the Sun-Earth system, the rocket must overcome the gravitational potential from both the Earth and the Sun.
- The total gravitational potential energy when the rocket is at the Earth's surface is:
\[ U = -\frac{GM_e m}{R_e} - \frac{GM_s m}{d + R_e} \]
- For escape, the total mechanical energy should be zero:
\[ \frac{1}{2} m u_s^2 - \left(-\frac{GM_e m}{R_e} - \frac{GM_s m}{d + R_e}\right) = 0 \]
- Simplifying gives:
\[ \frac{1}{2} u_s^2 = \frac{GM_e}{R_e} + \frac{GM_s}{d + R_e} \]

4. **Substituting Values**:
- Substitute \( M_s = 3 \times 10^5 M_e \) and \( d = 2.5 \times 10^4 R_e \):
\[ \frac{GM_s}{d + R_e} = \frac{G(3 \times 10^5 M_e)}{2.5 \times 10^4 R_e + R_e} = \frac{3 \times 10^5 G M_e}{(2.5 \times 10^4 + 1) R_e} \]
- This simplifies to:
\[ \frac{GM_s}{d + R_e} = \frac{3 \times 10^5 G M_e}{2.51 \times 10^4 R_e} \]

5. **Combining Terms**:
- Now we can write:
\[ \frac{1}{2} u_s^2 = \frac{GM_e}{R_e} + \frac{3 \times 10^5 G M_e}{2.51 \times 10^4 R_e} \]
- Factor out \( \frac{GM_e}{R_e} \):
\[ \frac{1}{2} u_s^2 = GM_e \left( \frac{1}{R_e} + \frac{3 \times 10^5}{2.51 \times 10^4 R_e} \right) \]

6. **Calculating Escape Velocity**:
- Now, substituting \( u_e^2 = \frac{2GM_e}{R_e} \):
\[ u_s^2 = 2GM_e \left( \frac{1}{R_e} + \frac{3 \times 10^5}{2.51 \times 10^4 R_e} \right) = 2u_e^2 \left( 1 + \frac{3 \times 10^5}{2.51 \times 10^4} \right) \]
- Calculate the numerical values:
\[ \frac{3 \times 10^5}{2.51 \times 10^4} \approx 11.95 \]
- Thus:
\[ u_s^2 = 2 \times (11.2)^2 \times (1 + 11.95) \approx 2 \times 125.44 \times 12.95 \approx 3243.84 \]
- Taking the square root gives:
\[ u_s \approx \sqrt{3243.84} \approx 57 \, \text{km/s} \]

### Final Answer:
The minimum initial velocity \( u_s \) required for the rocket to escape the Sun-Earth system is approximately \( 57 \, \text{km/s} \).