What is the magnitude of the gravitational force between the Earth and a 1kg object on its surface ? (Mass of the earth is `6xx10^(24)` kg and radius of the Earth is `6.4xx10^(6) m)`.

Explanation

To find the magnitude of the gravitational force between the Earth and a 1 kg object on its surface, we can use the formula for gravitational force:

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

where:
- \( F \) is the gravitational force,
- \( G \) is the gravitational constant, approximately \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \),
- \( m_1 \) is the mass of the Earth, \( 6 \times 10^{24} \, \text{kg} \),
- \( m_2 \) is the mass of the object, \( 1 \, \text{kg} \),
- \( r \) is the radius of the Earth, \( 6.4 \times 10^{6} \, \text{m} \).

Now, let's calculate the gravitational force step by step.

### Step 1: Identify the values
- Gravitational constant \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- Mass of the Earth \( m_1 = 6 \times 10^{24} \, \text{kg} \)
- Mass of the object \( m_2 = 1 \, \text{kg} \)
- Radius of the Earth \( r = 6.4 \times 10^{6} \, \text{m} \)

### Step 2: Substitute the values into the formula
\[ F = \frac{(6.67 \times 10^{-11}) \cdot (6 \times 10^{24}) \cdot (1)}{(6.4 \times 10^{6})^2} \]

### Step 3: Calculate \( r^2 \)
\[ r^2 = (6.4 \times 10^{6})^2 = 40.96 \times 10^{12} \, \text{m}^2 \]

### Step 4: Substitute \( r^2 \) back into the formula
\[ F = \frac{(6.67 \times 10^{-11}) \cdot (6 \times 10^{24})}{40.96 \times 10^{12}} \]

### Step 5: Calculate the numerator
\[ 6.67 \times 10^{-11} \cdot 6 \times 10^{24} = 40.02 \times 10^{13} \]

### Step 6: Calculate the gravitational force
\[ F = \frac{40.02 \times 10^{13}}{40.96 \times 10^{12}} \approx 9.77 \, \text{N} \]

### Final Answer
The magnitude of the gravitational force between the Earth and a 1 kg object on its surface is approximately **9.77 Newtons**.

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