Newton's Law of Universal Gravitation — AP Physics 1
1. Core Definition of Newton's Law of Universal Gravitation ★★☆☆☆ ⏱ 3 min
Newton's Law of Universal Gravitation is the fundamental physical law describing the attractive long-range force between any two objects that have mass. It applies to all objects everywhere in the universe, not just planets or celestial bodies near Earth. For AP Physics 1, we only work with point masses or uniform spherical masses, so distance between objects is defined as the distance between their centers of mass. This topic contributes 16–18% of your total AP exam score, appearing in both MCQ and FRQ sections.
2. Force Formula and Inverse-Square Proportionality ★★☆☆☆ ⏱ 4 min
Newton's law gives the magnitude of the gravitational force between two masses as:
F_g = G \frac{m_1 m_2}{r^2}
Where $G = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$ is the universal gravitational constant, $m_1$ and $m_2$ are the masses of the two objects, and $r$ is the center-to-center distance. AP Physics 1 almost always tests proportional reasoning for this relationship instead of full calculation with $G$. The inverse-square dependence on $r$ means if distance doubles, force drops to $1/4$ of its original value; if distance triples, force drops to $1/9$.
Exam tip: Always use ratio reasoning for proportionality questions to save time and avoid arithmetic errors with the very small value of $G$.
3. Gravitational Acceleration Variation With Altitude ★★★☆☆ ⏱ 3 min
The familiar $g = 9.8 \text{ m/s}^2$ acceleration due to gravity near Earth's surface is just a special case of Newton's universal gravitation. We can derive the gravitational acceleration at any distance from a planet's center by combining Newton's second law with the universal gravitation formula.
Exam tip: Never use altitude (distance from the surface) directly in the $g(r)$ formula. Always add the planet's radius to altitude to get the total center-to-center distance first.
4. Circular Orbital Motion and Kepler's Third Law ★★★☆☆ ⏱ 4 min
For a small object in uniform circular orbit around a much more massive central object, the only force providing the required centripetal acceleration is gravitational force. Setting gravitational force equal to centripetal force gives the core relationship for orbital motion:
G \frac{Mm}{r^2} = m \frac{v^2}{r}
The mass of the orbiting object $m$ cancels out, giving the formula for orbital speed:
v = \sqrt{\frac{GM}{r}}
Orbital speed depends only on the central mass $M$ and orbital radius $r$, so two satellites of different masses in the same orbit have the same speed. We can derive Kepler's third law for circular orbits by substituting $v = \frac{2\pi r}{T}$ (where $T$ is orbital period) into the orbital speed formula, resulting in:
T^2 = \left(\frac{4\pi^2}{GM}\right) r^3
This means for all objects orbiting the same central mass, the square of the orbital period is proportional to the cube of the orbital radius: $T^2 \propto r^3$.
Exam tip: The $T^2 \propto r^3$ proportionality only holds for objects orbiting the same central mass. If two objects orbit different planets with different masses, the proportionality constant changes.
Common Pitfalls
Why: Students confuse surface distance with the center-to-center distance required by the formula
Why: Students confuse force magnitude with acceleration, forgetting Newton's third law applies to gravity
Why: Students mix up which mass is which when setting up the force equation
Why: Students mix up the power relationship in Kepler's third law with the inverse-square relationship for force
Why: This is a common pop culture misconception that incorrectly equates space with zero gravity