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).

Understanding the gravitational force between the Earth and an object on its surface involves applying Newton’s law of universal gravitation:

F = (G.m_Earth.m_object)/r²

Here’s a breakdown of the calculation:

1. Gravitational Constant: Denoted by G, this value is approximately 6.67430 × 10⁻¹¹ m³.kg⁻¹.s⁻². It’s a fundamental constant in physics that defines the strength of the gravitational force.

2. Masses and Distance: The mass of the Earth (m_Earth) is (6 × 10²⁴ kg), and we’re considering an object with a mass (m_object) of 1 kg placed on the Earth’s surface. The distance (r) between the center of the Earth and the object is equal to the Earth’s radius, which is 6.4 × 10⁶ m

Let’s compute the gravitational force:

F = (G.m_Earth . m_object)/r²

Substituting the values:
F = (6.67430 × 10⁻¹¹. 6 × 10²⁴.1)/(6.4 × 10⁶)²

Calculating:
F ≈ 9.77 N

Therefore, when a 1kg object is situated on the Earth’s surface, the magnitude of the gravitational force between the Earth and the object is approximately 9.77N. This force represents the attraction between the Earth and the object, highlighting the gravitational pull exerted by the Earth on objects near its surface.