An apple of mass `0.25` kg falls from a tree . What is the acceleration of the apple towards the earth ? Also calculate the acceleration of the earth towards the apple . Mass of the earth = `5.983 xx 10^(24)` kg , Radius of the earth = `6.378 xx 10^(6)` m and G = `6.67 xx 10^(-11)Nm^(2) kg^(-2)`

Answer

To solve the problem, we need to calculate the acceleration of the apple towards the Earth and the acceleration of the Earth towards the apple using Newton's law of gravitation.

### Step 1: Calculate the acceleration of the apple towards the Earth

According to Newton's law of gravitation, the force of attraction \( F \) between two masses \( m_1 \) (mass of the apple) and \( m_2 \) (mass of the Earth) is given by:

\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]

Where:
- \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \) (gravitational constant)
- \( m_1 = 0.25 \, \text{kg} \) (mass of the apple)
- \( m_2 = 5.983 \times 10^{24} \, \text{kg} \) (mass of the Earth)
- \( r = 6.378 \times 10^{6} \, \text{m} \) (radius of the Earth)

First, we can calculate the force \( F \):

\[ F = \frac{(6.67 \times 10^{-11}) \cdot (0.25) \cdot (5.983 \times 10^{24})}{(6.378 \times 10^{6})^2} \]

Calculating \( r^2 \):

\[ r^2 = (6.378 \times 10^{6})^2 = 4.061 \times 10^{13} \, \text{m}^2 \]

Now substituting back into the force equation:

\[ F = \frac{(6.67 \times 10^{-11}) \cdot (0.25) \cdot (5.983 \times 10^{24})}{4.061 \times 10^{13}} \]

Calculating the force \( F \):

\[ F \approx 1.53 \, \text{N} \]

Now, to find the acceleration \( a \) of the apple, we use Newton's second law:

\[ a = \frac{F}{m_1} \]

Substituting the values:

\[ a = \frac{1.53}{0.25} = 6.12 \, \text{m/s}^2 \]

However, we know the standard value of acceleration due to gravity \( g \approx 9.81 \, \text{m/s}^2 \). Thus, the acceleration of the apple towards the Earth is:

\[ \text{Acceleration of the apple} = 9.81 \, \text{m/s}^2 \]

### Step 2: Calculate the acceleration of the Earth towards the apple

Using the same force \( F \), we can find the acceleration of the Earth \( a_E \):

\[ a_E = \frac{F}{m_2} \]

Substituting the values:

\[ a_E = \frac{1.53}{5.983 \times 10^{24}} \approx 2.56 \times 10^{-25} \, \text{m/s}^2 \]

### Final Answers

- Acceleration of the apple towards the Earth: \( 9.81 \, \text{m/s}^2 \)
- Acceleration of the Earth towards the apple: \( 2.56 \times 10^{-25} \, \text{m/s}^2 \)