A man of mass 50 kg starts moving on the earth and acquires speed of 1.8 m/s. With what speed does the earth recoil? Mass of earth =`6xx10^24kg.`

Explanation

To solve the problem, we will use the principle of conservation of momentum. According to this principle, the total momentum of an isolated system remains constant if no external forces act on it.

### Step-by-Step Solution:

1. **Identify the masses and speeds:**
- Mass of the man, \( m_1 = 50 \, \text{kg} \)
- Speed of the man, \( v_1 = 1.8 \, \text{m/s} \)
- Mass of the Earth, \( m_2 = 6 \times 10^{24} \, \text{kg} \)
- Speed of the Earth, \( v_2 \) (to be determined)

2. **Apply the conservation of momentum:**
The total momentum before the man starts moving is zero (since both the man and the Earth are initially at rest). Therefore, the total momentum after the man starts moving must also be zero:
\[ m_1 v_1 + m_2 v_2 = 0 \]

3. **Substitute the known values into the equation:**
\[ 50 \times 1.8 + (6 \times 10^{24}) v_2 = 0 \]

4. **Calculate the momentum of the man:**
\[ 50 \times 1.8 = 90 \, \text{kg m/s} \]
So, the equation becomes:
\[ 90 + (6 \times 10^{24}) v_2 = 0 \]

5. **Solve for \( v_2 \):**
Rearranging the equation gives:
\[ (6 \times 10^{24}) v_2 = -90 \]
\[ v_2 = \frac{-90}{6 \times 10^{24}} \]

6. **Calculate \( v_2 \):**
\[ v_2 = -1.5 \times 10^{-23} \, \text{m/s} \]

7. **Interpret the result:**
The negative sign indicates that the Earth recoils in the opposite direction to the man's movement.

### Final Answer:
The speed at which the Earth recoils is:
\[ v_2 = 1.5 \times 10^{-23} \, \text{m/s} \]