Unit 2 · Lesson 2a

2aNewton's Law of Gravitation

Explore universal gravitation, the inverse-square law, gravitational field strength, and the distinction between mass and weight.

Newton's law of gravitation unified terrestrial and celestial mechanics — the same force that makes an apple fall also keeps the Moon in orbit. It is the foundation of all orbital mechanics and space exploration.

What force keeps the Moon in orbit around Earth — and why does doubling the distance between two objects reduce the gravitational force to one quarter?

Lesson Overview

Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between them: F = Gm₁m₂/r². The gravitational constant G = 6.67 × 10⁻¹¹ N·m²/kg² is extremely small, so gravity is only significant when at least one mass is very large. Near Earth's surface, this law produces the familiar gravitational field strength g = GM/R² ≈ 9.8 m/s², and weight W = mg. Mass is an intrinsic property; weight depends on the local gravitational field.

Key Equations

Universal gravitationF = Gm₁m₂/r²
Gravitational constantG = 6.67×10⁻¹¹ N·m²/kg²
WeightW = mg
Field strengthg = GM/r²
Surface gravityg = GM/R²
Inverse-square lawF ∝ 1/r²

Worked Examples

Example 1

Find the gravitational force between two 1000 kg masses separated by r = 1 m. (G = 6.67 × 10⁻¹¹ N·m²/kg²)

Answer:F = Gm₁m₂/r² = (6.67×10⁻¹¹)(1000)(1000)/(1)² = 6.67×10⁻⁵ N ≈ 0.0000667 N
Example 2

Earth has mass M = 5.97 × 10²⁴ kg and radius R = 6.37 × 10⁶ m. Find the gravitational field strength g at Earth's surface.

Answer:g = GM/R² = (6.67×10⁻¹¹ × 5.97×10²⁴)/(6.37×10⁶)² = 3.98×10¹⁴/4.06×10¹³ ≈ 9.8 m/s²
Example 3

A 70 kg astronaut is at a distance of r = 2R from Earth's center (twice Earth's radius). Find the gravitational force on the astronaut.

Answer:F = GMm/r² = GMm/(2R)² = (1/4)GMm/R² = (1/4)(70)(9.8) = 171.5 N (one quarter of surface weight)
Example 4

The Moon has mass M_moon = 7.35 × 10²² kg and radius R_moon = 1.74 × 10⁶ m. Find the gravitational field strength on the Moon's surface.

Answer:g_moon = GM_moon/R_moon² = (6.67×10⁻¹¹ × 7.35×10²²)/(1.74×10⁶)² = 4.90×10¹²/3.03×10¹² ≈ 1.62 m/s²
Example 5

Two objects are separated by distance r. If the distance is tripled, by what factor does the gravitational force change?

Answer:F ∝ 1/r². New F = Gm₁m₂/(3r)² = (1/9)Gm₁m₂/r². The force decreases by a factor of 9.

Guided Problems

Guided Problem 1

Find the gravitational force between Earth (M = 5.97 × 10²⁴ kg) and the Moon (m = 7.35 × 10²² kg) separated by r = 3.84 × 10⁸ m.

Hint: Use F = Gm₁m₂/r². Substitute all values carefully and use scientific notation.

Guided Problem 2

A 60 kg person stands on a planet with mass M = 4.0 × 10²⁴ kg and radius R = 5.0 × 10⁶ m. Find their weight on this planet.

Hint: First find g = GM/R², then use W = mg.

Guided Problem 3

The gravitational force between two objects is F. If one mass is doubled and the distance is halved, what is the new force in terms of F?

Hint: Write F = Gm₁m₂/r². Substitute 2m₁ for m₁ and r/2 for r. Simplify.

Guided Problem 4

A satellite orbits at r = 3R above Earth's surface (so its distance from Earth's center is 4R). Find the gravitational field strength at that altitude.

Hint: Use g = GM/r² with r = 4R. Compare to surface g = GM/R².

Guided Problem 5

An object weighs 600 N on Earth's surface. What is its mass? What would it weigh on the Moon (g_moon = 1.62 m/s²)?

Hint: Mass = W/g_Earth. Then W_moon = m × g_moon. Mass is the same everywhere; weight changes.

Key Vocabulary

Newton's Law of Universal Gravitation

Every mass attracts every other mass with a force F = Gm₁m₂/r², where G is the gravitational constant, m₁ and m₂ are the masses, and r is the distance between their centers.

Example: Earth and Moon attract each other with F ≈ 2.0 × 10²⁰ N.

Gravitational Constant (G)

The universal constant G = 6.67 × 10⁻¹¹ N·m²/kg². Its small value means gravity is only significant when at least one mass is very large.

Example: Two 1 kg masses 1 m apart attract with only F = 6.67 × 10⁻¹¹ N — far too small to feel.

Inverse-Square Law

Gravitational force decreases with the square of the distance: F ∝ 1/r². Doubling the distance reduces the force to one quarter.

Example: Moving from Earth's surface to twice the radius reduces g from 9.8 to 2.45 m/s².

Gravitational Field Strength (g)

The gravitational force per unit mass at a point in space: g = F/m = GM/r². At Earth's surface, g ≈ 9.8 N/kg = 9.8 m/s².

Example: On the Moon's surface, g ≈ 1.62 m/s² — about 1/6 of Earth's surface gravity.

Weight vs. Mass

Mass (kg) is an intrinsic property of matter — it is the same everywhere. Weight (N) is the gravitational force on an object: W = mg. Weight depends on the local gravitational field.

Example: A 70 kg astronaut has the same mass on Earth and the Moon, but weighs 686 N on Earth and only 113 N on the Moon.

Workbook Check — Interactive Quiz

Interactive Practice — 5 Questions

1

Two objects are separated by distance r. If the distance is doubled, the gravitational force becomes:

2

What is the gravitational field strength at Earth's surface? (M = 5.97×10²⁴ kg, R = 6.37×10⁶ m)

3

A 50 kg person weighs 490 N on Earth. What is their weight on the Moon (g_moon = 1.62 m/s²)?

4

Which quantity is the same for an object on Earth and on the Moon?

5

The gravitational force between two masses is F. If both masses are doubled, the new force is:

Independent Practice

1

Find the gravitational force between two 500 kg masses separated by r = 2 m. (G = 6.67 × 10⁻¹¹ N·m²/kg²)

2

A 75 kg person stands on Earth's surface (g = 9.8 m/s²). Find their weight in newtons.

3

Find the gravitational field strength at a distance of r = 2R from Earth's center (R = 6.37 × 10⁶ m, M = 5.97 × 10²⁴ kg).

4

An object weighs 800 N on Earth. What is its mass? What would it weigh on Mars (g_Mars = 3.72 m/s²)?

5

★ Find the gravitational force between Earth (M = 5.97 × 10²⁴ kg) and the Moon (m = 7.35 × 10²² kg) at r = 3.84 × 10⁸ m. Then find the Moon's centripetal acceleration and compare it to g/3600 (where g = 9.8 m/s²). This was Newton's key verification.

Challenge
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Common Mistakes

Confusing mass (kg) with weight (N)

Mass is an intrinsic property measured in kg. Weight is a force W = mg measured in newtons. They are proportional but not the same thing

Using r as the distance from Earth's surface instead of from Earth's center

In F = Gm₁m₂/r², r is the distance between the centers of the two objects, not from the surface

Thinking gravity only acts between large objects like planets

Gravity acts between ALL masses. It's just too weak to notice between everyday objects because G is so small

Forgetting that the inverse-square law means doubling distance gives 1/4 force, not 1/2

F ∝ 1/r². Double r → r² quadruples → F becomes 1/4. Triple r → F becomes 1/9

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Math Tips

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F = Gm₁m₂/r². The force is symmetric — Earth pulls the Moon with the same force the Moon pulls Earth (Newton's 3rd law).

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Gravitational field strength g = GM/r². At Earth's surface r = R, giving g ≈ 9.8 m/s². At altitude h, use r = R + h.

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Inverse-square law: if r doubles, F becomes (1/2)² = 1/4 of original. If r triples, F becomes 1/9. Always square the distance ratio.

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Weight W = mg is a force (newtons). Mass m is in kg. On any planet: g_planet = GM_planet/R_planet², then W = mg_planet.