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Gravitational
Force

Explore Newton's Law of Universal Gravitation, understand how mass and distance govern the attractive force between objects, and apply \( F = G\frac{m_1 m_2}{r^2} \) to real-world scenarios.

AP Physics 1

What Is Gravity?

Gravity is one of the four fundamental forces of nature. It is the attractive force that exists between any two objects that have mass. Unlike electromagnetic forces, gravity is always attractive and cannot be shielded or canceled.

Key Terms and Quantities

\( G \)
Gravitational constant: \( 6.674 \times 10^{-11} \; \text{N} \cdot \text{m}^2/\text{kg}^2 \)
\( m_1, m_2 \)
The masses of the two interacting objects, measured in kilograms (kg)
\( r \)
The center-to-center distance between the two objects, measured in meters (m)

Why Does Gravity Matter?

1
Planetary orbits: gravity keeps planets in elliptical orbits around the Sun.
2
Tides: the Moon's gravitational pull causes ocean tides on Earth.
3
Satellites: artificial satellites orbit Earth because gravitational force provides the centripetal acceleration.
4
Everyday life: gravity keeps us on the ground and governs the motion of every thrown, dropped, or launched object.

Newton's Law of Universal Gravitation

\( m_1 \) \( m_2 \) \( r \) \( F \) \( F \)

The force is always attractive, directed along the line connecting the centers of the two masses. Both objects experience the same magnitude of force.

Mutual gravitational forces

The Earth–Moon example illustrates the same law for two bodies: each pulls the other with equal magnitude in the opposite direction. The separation is measured center to center.

Earth and Moon exert equal and opposite gravitational forces, approximately 1.99 × 10²⁰ N, along their center line. Use center-to-center separation 3.84 × 10⁸ m; the depicted bodies are not to scale.
Earth and Moon exert equal and opposite gravitational forces, approximately 1.99 × 10²⁰ N, along their center line. Use center-to-center separation 3.84 × 10⁸ m; the depicted bodies are not to scale.

The Inverse-Square Relationship

Because \( F \propto \frac{1}{r^2} \), doubling the distance between two objects reduces the gravitational force to one-quarter of its original value. Tripling the distance reduces it to one-ninth.

\( r \) (distance) \( F \) (force) \( r \): \( F \) \( 2r \): \( F/4 \) \( 3r \): \( F/9 \)

Torsion balance tests the inverse-square law

A four-sphere torsion balance with independent source supports accompanies a top-view interaction diagram and a corrected force-versus-inverse-distance-squared plot.A torsion balance measures a tiny gravitational torque through a calibrated fiber’s angular deflection. Use center-to-center mass distances and include both source masses.

A tiny gravitational torque needs calibration

A tiny gravitational torque needs calibration. A torsion balance measures a tiny gravitational torque through a calibrated fiber’s angular deflection. Use center-to-center mass distances and include both source masses.A torsion balance measures a tiny gravitational torque through a calibrated fiber’s angular deflection. Use center-to-center mass distances and include both source masses.

Gravitational Field Strength \( g \)

Near any massive body, we define the gravitational field strength as the force per unit mass experienced by a small test mass:

At Earth's surface, \( g \approx 9.8 \; \text{m/s}^2 \). This value decreases as you move farther from Earth's center.

Gravity vs. Electrostatic Force

Gravitational Electrostatic (Coulomb's)
Law \( F = G\frac{m_1 m_2}{r^2} \) \( F = k\frac{q_1 q_2}{r^2} \)
Direction Always attractive Attractive or repulsive
Depends on Mass Charge
Relative strength Very weak Very strong
Range Infinite Infinite

How to Apply the Law

1
Identify the two masses \( m_1 \) and \( m_2 \). Convert to kilograms if needed.
2
Determine \( r \): the center-to-center distance between the objects. Convert to meters.
3
Substitute into \( F = G\frac{m_1 m_2}{r^2} \) with \( G = 6.674 \times 10^{-11} \).
4
Compute and check units. The answer should be in Newtons (N).
5
Interpret direction: the force is always attractive, directed along the line joining the two centers.

Example 1: Two Masses on a Table

Two bowling balls, each with mass \( m = 7.0 \; \text{kg} \), are placed with their centers \( r = 0.50 \; \text{m} \) apart. Find the gravitational force between them.

7.0 kg 7.0 kg 0.50 m attractive

Example 2: Your Weight from Earth

Calculate the gravitational force on a \( 65 \; \text{kg} \) person standing on Earth's surface. Use \( M_E = 5.97 \times 10^{24} \; \text{kg} \) and \( R_E = 6.37 \times 10^6 \; \text{m} \).

Example 3: Doubling the Distance

A satellite orbits Earth at altitude \( h = R_E \) above the surface (so \( r = 2R_E \) from Earth's center). How does the gravitational force on the satellite compare to its weight on the surface?

Superposition of Gravitational Forces

When more than two masses are present, the net gravitational force on any one mass is the vector sum of the individual forces from all other masses:

\( m \) \( M_A \) \( M_B \) \( M_C \) \( \vec{F}_A \) \( \vec{F}_B \) \( \vec{F}_C \)

Calculate each force separately, then add them as vectors (break into components).

Gravity and Circular Orbits

For a satellite in a circular orbit, gravity provides the centripetal force:

Orbital speed
Depends only on the central mass \( M \) and orbital radius \( r \), not the satellite mass.
Larger orbit
Greater \( r \) means slower speed and longer orbital period.

Common Mistakes

✕ Using surface distance
You must use the center-to-center distance \( r \), not the gap between surfaces. For objects on Earth's surface, \( r = R_E \).
✕ Forgetting to square \( r \)
The denominator is \( r^2 \), not \( r \). Missing the square gives a force that is far too large.
✕ Confusing \( g \) and \( G \)
\( G \) is the universal constant (\( 6.674 \times 10^{-11} \)). \( g \) is local field strength (\( \approx 9.8 \; \text{m/s}^2 \) on Earth).

Summary of Key Relationships

Quantity Formula Key Insight
Gravitational force \( F = G\frac{m_1 m_2}{r^2} \) Inversely proportional to \( r^2 \)
Field strength \( g = \frac{GM}{r^2} \) Independent of test mass
Weight \( W = mg \) Varies with location
Orbital speed \( v = \sqrt{\frac{GM}{r}} \) Independent of satellite mass

Question 1 of 5

If the distance between two objects is tripled, by what factor does the gravitational force between them change?

Question 2 of 5

Two objects of mass \( m \) are separated by distance \( d \). If both masses are doubled and the distance is halved, the new force is:

Question 3 of 5

A planet has twice the mass of Earth and twice the radius. What is the gravitational field strength \( g \) on its surface compared to Earth's?

Question 4 of 5

An astronaut in the International Space Station (ISS) feels "weightless." Which statement best explains why?

Question 5 of 5

What is the orbital speed of a satellite orbiting Earth at a distance of \( 2R_E \) from Earth's center? Use \( g = 9.8 \; \text{m/s}^2 \) and \( R_E = 6.37 \times 10^6 \; \text{m} \).