Newton's Law of Universal Gravitation

Core idea

Overview

The force is always attractive, acting along the line joining the centers of the two masses. This inverse-square relationship means that doubling the distance between the two bodies reduces the gravitational force to one-quarter of its original value. It serves as the foundation for understanding planetary orbits, satellite motion, and the formation of celestial structures.

When to use: Use this equation when calculating the force of gravity between any two massive objects where the separation distance is significantly greater than the radii of the objects.

Why it matters: It explains why planets orbit the Sun, why moons stay in orbit, and how we can calculate the mass of celestial bodies.

Symbols

Variables

F = Gravitational Force, G = Gravitational Constant, M = Mass of first object, m = Mass of second object, r = Distance between centers

F

Gravitational Force

Variable

G

Gravitational Constant

Variable

M

Mass of first object

Variable

m

Mass of second object

Variable

r

Distance between centers

Variable

Walkthrough

Derivation

Derivation of Newton's Law of Universal Gravitation

Newton derived this law by synthesizing Kepler's Third Law of planetary motion with the requirement for centripetal force in circular orbits.

Planetary orbits are approximately circular.
The gravitational force is the sole source of centripetal force for an orbiting body.
The force is proportional to both masses involved (Newton's Third Law symmetry).

1

Centripetal Force Requirement

For an object of mass m moving in a circular orbit of radius r with speed v, a centripetal force is required to maintain the path.

F = \frac{m v ^{2}}{r}

Note: Ensure units are consistent (SI) when using this formula.

2

Relating Orbital Speed and Period

Substitute the definition of speed for a circular orbit (circumference divided by period) into the force equation.

v = \frac{2 π r}{T} ⟹ v^{2} = \frac{4 π ^{2} r ^{2}}{T ^{2}}

Note: T represents the orbital period.

3

Applying Kepler's Third Law

Kepler's Third Law states that the square of the orbital period is proportional to the cube of the radius.

\frac{T ^{2}}{r ^{3}} = k ⟹ T^{2} = k r^{3}

Note: Kepler's Law is empirical; Newton provided the theoretical basis for it.

4

Combining and Simplifying

Substitute T squared into the force equation and simplify to show that F is inversely proportional to r squared, defining G as the constant of proportionality.

F = \frac{m ( 4 π ^{2} r ^{2} / k r ^{3} )}{r} = \frac{4 π ^{2} m}{k r ^{2}} = \frac{GM m}{r ^{2}}

Note: G is the Universal Gravitational Constant.

Result

F = \frac{m ( 4 π ^{2} r ^{2} / k r ^{3} )}{r} = \frac{4 π ^{2} m}{k r ^{2}} = \frac{GM m}{r ^{2}}

Source: AQA/Edexcel A-Level Physics Specification: Gravitational Fields

Free formulas

Rearrangements

Solve for $M$

Make M the subject

M = \frac{F r ^{2}}{G m}

Rearrange the formula to solve for the mass of the primary body.

Difficulty: 3/5

Solve for $m$

Make m the subject

m = \frac{F r ^{2}}{GM}

Rearrange the formula to solve for the mass of the secondary body.

Difficulty: 3/5

Solve for $r$

Make r the subject

r = \frac{GM m}{F}

Rearrange the formula to solve for the distance between the centers of the two masses.

Difficulty: 4/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Why it behaves this way

Intuition

Imagine the force as a 'gravity fountain' emitted by mass M. The field strength spreads out over the surface area of a sphere (4πr²) as it moves away. Because the surface area of a sphere grows with the square of the radius (r²), the concentration of that force must dilute by a factor of 1/r².

F

Gravitational Force

The 'tug' or weight exerted between two objects; the result of their mutual gravitational attraction.

G

Gravitational Constant

The 'strength knob' of the universe; it tells us how much force is produced per unit of mass and distance in our specific universe.

M, m

Masses of the two objects

The 'gravitational charge'; the more matter packed into an object, the stronger its pull on others.

r

Separation distance

How far apart the centers of the two masses are; as this increases, the force drops off drastically due to the inverse-square relationship.

Signs and relationships

1/r²: This represents the inverse-square law, indicating that gravity follows the geometry of 3D space, where intensity spreads over the surface area of a sphere.

One free problem

Practice Problem

Calculate the gravitational force between two 1000 kg masses separated by a distance of 10 meters.

Gravitational Constant6.67e-11

Mass of first object1000

Mass of second object1000

Distance between centers10

Solve for: $F$

Hint: Plug the values into F = GMm/rθ. Remember that rθ is 100.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

In a physics application involving Newton's Law of Universal Gravitation, Newton's Law of Universal Gravitation is used to calculate Gravitational Force from Gravitational Constant, Mass of first object, and Mass of second object. The result matters because it helps check loads, margins, or component sizes before a design is treated as safe.

Study smarter

Tips

Ensure the distance r is measured between the centers of mass of the two objects, not their surfaces.
Use SI units: kilograms for mass and meters for distance to maintain consistency with the Gravitational Constant G.
Remember that the force is mutual; object M exerts the same magnitude of force on m as m exerts on M.

Avoid these traps

Common Mistakes

Forgetting to square the radius (r) in the denominator.
Measuring r from the surface of a planet rather than from its center.
Confusing the gravitational constant G (6.67 × 10^-11) with the acceleration due to gravity g (9.81 m/s²).

Keep going

Related Formulas

Common questions

Frequently Asked Questions

Newton derived this law by synthesizing Kepler's Third Law of planetary motion with the requirement for centripetal force in circular orbits.

Use this equation when calculating the force of gravity between any two massive objects where the separation distance is significantly greater than the radii of the objects.

It explains why planets orbit the Sun, why moons stay in orbit, and how we can calculate the mass of celestial bodies.

Forgetting to square the radius (r) in the denominator. Measuring r from the surface of a planet rather than from its center. Confusing the gravitational constant G (6.67 × 10^-11) with the acceleration due to gravity g (9.81 m/s²).

In a physics application involving Newton's Law of Universal Gravitation, Newton's Law of Universal Gravitation is used to calculate Gravitational Force from Gravitational Constant, Mass of first object, and Mass of second object. The result matters because it helps check loads, margins, or component sizes before a design is treated as safe.

Ensure the distance r is measured between the centers of mass of the two objects, not their surfaces. Use SI units: kilograms for mass and meters for distance to maintain consistency with the Gravitational Constant G. Remember that the force is mutual; object M exerts the same magnitude of force on m as m exerts on M.

References

Sources

Newton, I. (1687). Philosophiæ Naturalis Principia Mathematica.
Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics.
AQA/Edexcel A-Level Physics Specification: Gravitational Fields

Overview

Variables

Derivation

Centripetal Force Requirement

Relating Orbital Speed and Period

Applying Kepler's Third Law

Combining and Simplifying

Rearrangements

Intuition

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

Gravitational Field Strength

Frequently Asked Questions

Sources