Gravitational Force from a Hollow Sphere

Explore how gravitational forces change when mass is removed from a solid sphere—using a lead sphere with a spherical hollow as the example.

You’ll Learn

• Use Newton’s Law of Gravitation for complex mass shapes
• Calculate gravitational force from a full sphere
• Subtract gravitational effects due to missing mass
• Apply the principle of superposition in gravitation
• Compute mass of a spherical cavity using density
• Analyze force from composite and non-uniform objects

Key Concepts Covered

• Gravitational force with a cavity
• Superposition in gravitational fields
• Newton’s Law of Gravitation
• Density and mass of spheres
• Inverse square law
• Subtractive mass method
• Composite mass distributions
• Gravitational field of a hollow sphere

Understanding how mass distribution affects gravitational pull is essential for AP and IB Physics, especially for problems involving cavities inside planets, engineered objects, or varying density structures. This lesson strengthens your grasp of real-world gravitation scenarios and critical thinking in physics.

Prerequisite or Follow-Up Lessons

• Newton’s Law of Universal Gravitation
• Gravitational Potential and Field of Spheres

Full Lesson: Gravitational Force from a Hollow Sphere

We are given a lead sphere with a radius R = 4.00 cm and an original mass M = 2.95 kg. A spherical hollow is carved from this sphere. The hollow’s surface just touches the outer edge of the sphere and passes through its center. A small sphere of mass m = 0.431 kg is placed at a distance d = 9.00 cm from the center of the lead sphere.

The task is to calculate the gravitational force between the hollowed-out lead sphere and the small mass.

Step 1: Full Sphere Force (if no hollow)

If the sphere were still solid, the gravitational force would be:

F₁ = G * M * m / d²

where:

• G = 6.674 × 10⁻¹¹ N·m²/kg²
• M = 2.95 kg
• m = 0.431 kg
• d = 0.0900 m (converted from 9.00 cm)

Step 2: Find the mass of the missing cavity

The radius of the hollow is r = R / 2 = 2.00 cm = 0.0200 m.

Because density is uniform, the mass of the cavity (M꜀) is:

M꜀ = (r³ / R³) * M
= (1/8) * M
= 2.95 / 8 = 0.36875 kg

Step 3: Force due to the cavity

The center of the cavity is located at (d - R/2) = 9.00 cm - 2.00 cm = 7.00 cm = 0.0700 m from the small mass.

F₂ = G * M꜀ * m / (d - R/2)²
= G * (M / 8) * m / (0.0700)²

Step 4: Net Gravitational Force

The total gravitational force is:

F = F₁ - F₂
= G * M * m * [1 / d² - 1 / (8 * (d - R/2)²)]

Plugging in the values:

F = (6.674 × 10⁻¹¹) * 2.95 * 0.431 * [1 / (0.0900)² - 1 / (8 * (0.0700)²)]
F ≈ 8.31 × 10⁻⁹ N

This result shows how a cavity within a sphere reduces the gravitational pull, which we compute by treating the hollow as negative mass and applying the superposition principle.