The gravitational force of the earth on a freely falling ball of mass one kilogram is 9.8 N. The acceleration of the earth towards the ball is

Options

A. ` 9.8 ms^(-2)`
B. negligible.
C. Slightly less than ` 9.8 ms^(-2)`
D. more than `9.8 ms ^(-2)`

Answer

To find the acceleration of the Earth towards a freely falling ball of mass 1 kg, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information**:
- The gravitational force (F) exerted by the Earth on the ball is 9.8 N.
- The mass of the ball (m1) is 1 kg.
- We need to find the acceleration of the Earth (a) towards the ball.

2. **Apply Newton's Third Law of Motion**:
- According to Newton's Third Law, the force that the ball exerts on the Earth is equal in magnitude and opposite in direction to the force that the Earth exerts on the ball. Therefore, the gravitational force exerted by the ball on the Earth is also 9.8 N.

3. **Use Newton's Second Law of Motion**:
- Newton's Second Law states that \( F = m \cdot a \), where F is the force, m is the mass, and a is the acceleration.
- In this case, we can rearrange the formula to find the acceleration of the Earth:
\[ a = \frac{F}{m_e} \]
- Here, \( F = 9.8 \, \text{N} \) and \( m_e \) is the mass of the Earth.

4. **Substitute the Mass of the Earth**:
- The mass of the Earth (\( m_e \)) is approximately \( 6 \times 10^{24} \, \text{kg} \).
- Now substitute the values into the equation:
\[ a = \frac{9.8 \, \text{N}}{6 \times 10^{24} \, \text{kg}} \]

5. **Calculate the Acceleration**:
- Performing the calculation:
\[ a \approx \frac{9.8}{6 \times 10^{24}} \approx 1.63 \times 10^{-24} \, \text{m/s}^2 \]

6. **Interpret the Result**:
- The calculated acceleration of the Earth towards the ball is approximately \( 1.63 \times 10^{-24} \, \text{m/s}^2 \). This value is extremely small, indicating that the Earth's movement towards the ball is negligible compared to the ball's acceleration towards the Earth.

### Final Answer:
The acceleration of the Earth towards the ball is approximately \( 1.63 \times 10^{-24} \, \text{m/s}^2 \), which is negligible.