## Gravitational Field Strength: variation from the centre of the Earth

Introduction

The gravitational field strength (g)

The force per unit mass acting on a small test mass placed at a point in the field.

Units: N/kg

The Earth, being a massive object, exerts a gravitational force on any object near it. The magnitude of this force varies with distance from the Earth's center.

Variation of g with Distance

**1. Inside the Earth:**

## Deriving the Gravitational Field Strength Within the Earth

Assumptions made:

- The Earth is a perfect sphere with uniform density.
- The test mass is a point mass.

- If r < R (the Earth's radius), this surface is entirely within the Earth

The volume of the sphere of radius r is (4/3)πr³.

The mass enclosed is ρV, where ρ is the Earth's density.

So, enclosed mass = (4/3)πρr³.

Apply Newton's Law of Gravitation to find the force on 1kg placed on the surface of this imaginary sphere.

F = GM

_{1}M

_{2}

_{2}= 1kg, F = g, d = r

g = 4π(4/3)πρr³/r² = 4π(4/3)πρr

**2. Outside the Earth:**

_{1}M

_{2}/d²

As we move further away from the Earth, the gravitational field strength decreases rapidly.

In theory, the field strength would become zero at an infinite distance from the Earth.

**Key Points:**

The gravitational field strength is not constant within the Earth.

It reaches a maximum value at the boundary between the mantle and core.

Outside the Earth, it follows an inverse square law with distance.

The variation of g with distance can be explained using Newton's Law of Universal Gravitation and the concept of enclosed mass.

The Simulation

The above simulation shows how the gravitational field strength changes with the distance from the centre.

**Worked example**

Calculating Gravitational Field Strength at Different Distances

Given:

Mass of Earth (M) ≈ 5.972 × 10^24 kg

Radius of Earth (R) ≈ 6371 km

Gravitational constant (G) ≈ 6.674 × 10^-11 Nm²/kg²

Formula:

For points outside the Earth: g = GM/r²

1. At 2000 km from the center:

r = 2000 km = 2 × 10^6 m

g = (6.674 × 10^-11 Nm²/kg²) * (5.972 × 10^24 kg) / (2 × 10^6 m)²

g ≈ 9.85 m/s²

2. At 8000 km from the center:

r = 8000 km = 8 × 10^6 m

g = (6.674 × 10^-11 Nm²/kg²) * (5.972 × 10^24 kg) / (8 × 10^6 m)²

g ≈ 0.98 m/s²

Therefore, the gravitational field strength at a distance of 2000 km from the center of the Earth is approximately 9.85 m/s², and at a distance of 8000 km, it is approximately 0.98 m/s².

Conclusion

The gravitational field strength is a fundamental concept in physics. Understanding how it varies with distance is essential for comprehending the motion of objects near the Earth and the behavior of celestial bodies. By combining theoretical knowledge with simulations, students can develop a deeper understanding of this important topic.

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