What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10²⁴ kg and radius of the earth is 6.4 × 10⁶ m.)

The concept of gravitational force is crucial in understanding the force of attraction between two objects with mass. 

In this context, we will calculate what is the magnitude of the gravitational force between the earth and a 1 kg object on its surface.

To find the magnitude of gravitational force, we use Newton's Law of Universal Gravitation, which is expressed by the formula:

F = G × (M × m) / R²

Where:

• F = Gravitational Force
• G = Gravitational Constant (6.67 × 10^-11 N m²/kg²)
• M = Mass of Earth in kg (6 × 10²⁴ kg)
• m = Mass of object (1 kg)
• R = Radius of Earth (6.4 × 10⁶ m)

Substituting these values into the formula:

F = (6.67 × 10^-11) × (6 × 10²⁴) × 1 / (6.4 × 10⁶)²

After calculating the values:

F = 9.77 N

Therefore, what is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? 

The answer is 9.77 N. 

The magnitude of gravitational force experienced by a 1 kg object due to the Earth’s gravitational pull is 9.77 N.

To summarize, the magnitude of gravitational force between the Earth and a 1 kg object on its surface is calculated using Newton's formula, where the mass of Earth in kg is 6 × 10²⁴ kg, and the radius of Earth is 6.4 × 10⁶ m. The calculated result shows that the gravitational force is 9.77 N.