Newton's Law of Universal Gravitation — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Newton’s Law of Universal Gravitation force formula, inverse-square proportional reasoning, gravitational acceleration variation with altitude, circular orbital motion, and Kepler’s third law derived from Newtonian gravitation for AP exam problem solving.

You should already know: Newton's second and third laws of motion, centripetal acceleration for uniform circular motion, proportional reasoning for algebraic relationships.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Newton's Law of Universal Gravitation?

Newton's Law of Universal Gravitation is the fundamental physical law describing the attractive long-range force between any two objects that have mass. As the name suggests, it applies to all objects everywhere in the universe — from subatomic particles to entire galaxies — not just planets or celestial bodies near Earth.

In the AP Physics 1 Course and Exam Description (CED), this topic is part of Unit 3: Circular Motion and Gravitation, which contributes 16–18% of the total AP exam score. Universal gravitation questions appear in both multiple-choice (MCQ) and free-response (FRQ) sections, often paired with circular motion concepts. Typically, you can expect 1–2 MCQ questions focused solely on this topic, plus at least one part of a multi-part FRQ that combines gravitation with other Unit 3 concepts.

For AP Physics 1, we only work with point masses or uniform spherical masses, which simplifies calculations because the distance between objects is defined as the distance between their centers of mass.

2. The Universal Gravitation Force Formula and Inverse-Square Proportionality

Newton's law states that the magnitude of the gravitational force between two masses is directly proportional to the product of the two masses and inversely proportional to the square of the distance between their centers of mass. In formula form, this is: $$F_g=G\frac{m_1{m}_2}{r^2}$$ 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 distance between their centers of mass. The force is always attractive along the line connecting the two centers, and it follows Newton's third law: the force $m_1$ exerts on $m_2$ is equal in magnitude and opposite in direction to the force $m_2$ exerts on $m_1$, regardless of any difference in mass.

The key relationship AP Physics 1 tests repeatedly is the inverse-square dependence on $r$: if the distance between two masses doubles, the force drops to 1/4 of its original value; if distance triples, the force drops to 1/9. Nearly all AP gravitation force problems use proportional reasoning rather than full calculation with $G$, so mastering ratio reasoning is critical.

Worked Example

A 10 kg mass and a 20 kg mass separated by 2 meters exert a gravitational force $F$ on each other. What is the new force if the 10 kg mass is changed to 30 kg, the 20 kg mass is changed to 40 kg, and the separation distance is increased to 6 meters, in terms of the original force $F$?

1. Write the original force using the universal gravitation formula: $F=G\frac{(10)(20)}{(2{)}^2}=50G$
2. Write the new force after the changes: $F^{\prime }=G\frac{(30)(40)}{(6{)}^2}=\frac{1200}{36}G=\frac{100}{3}G$
3. Take the ratio of the new force to the original force to eliminate $G$: $\frac{F^{\prime }}{F}=\frac{\frac{100}{3}G}{50G}=\frac{100}{150}=\frac{2}{3}$
4. Final result: $F^{\prime }=\frac{2}{3}F$

Exam tip: AP Physics 1 almost never asks you to calculate a raw force magnitude by plugging in $G$. Always use ratio reasoning for proportionality questions to save time and avoid arithmetic errors.

3. Gravitational Acceleration Variation With Altitude

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.

For an object of mass $m$ at distance $r$ from the center of a planet of mass $M$, the gravitational force on the object is $F_g=G\frac{Mm}{r^2}=ma=mg$. The mass of the object $m$ cancels out from both sides, leaving the general formula for gravitational acceleration: $$g(r)=\frac{GM}{r^2}$$ This formula is valid for all points outside the planet (i.e., $r\ge R$, where $R$ is the planet's radius). At the planet's surface, $r=R$, so the surface gravitational acceleration is $g_s=\frac{GM}{R^2}$. At an altitude $h$ above the surface, the total distance from the center is $r=R+h$, so $g$ decreases as altitude increases.

Worked Example

The International Space Station orbits 400 km above Earth's surface. Earth's radius is approximately 6400 km. What is the gravitational acceleration at the ISS orbital altitude, in terms of the surface gravitational acceleration $g_s$?

1. Calculate the total orbital radius from Earth's center: $r=R_E+h=6400+400=6800\text{ km}=\frac{17}{16}{R}_E$
2. Write the surface and orbital acceleration formulas: $g_s=\frac{G{M}_E}{R_E^2}$, $g_{orbit}=\frac{G{M}_E}{r^2}$
3. Substitute $r=\frac{17}{16}{R}_E$ into the orbital acceleration formula: $g_{orbit}=\frac{G{M}_E}{(\frac{17}{16}{R}_E{)}^2}=g_s{\left(\frac{16}{17}\right)}^2=g_s\frac{256}{289}$
4. Simplify: $g_{orbit}\approx 0.89g_s$, or approximately 8.7 m/s².

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 distance from the center first.

4. Circular Orbital Motion and Kepler's Third Law

For a small object in uniform circular orbit around a much more massive central object, the only force providing centripetal acceleration is gravitational force. Setting gravitational force equal to centripetal force gives us 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 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: $$T^2=\left(\frac{4{\pi }^2}{GM}\right)r^3$$ 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: $T^2\propto r^3$. This is Kepler's third law, derived directly from Newton's universal gravitation.

Worked Example

Two satellites orbit the same planet. Satellite A has orbital radius $r$ and orbital period $T$. Satellite B has orbital radius $4r$. What is the orbital period of Satellite B?

1. Kepler's third law for the same central planet gives the proportionality: $\frac{T_A^2}{r_A^3}=\frac{T_B^2}{r_B^3}$
2. Substitute known values: $T_A=T$, $r_A=r$, $r_B=4r$: $\frac{T^2}{r^3}=\frac{T_B^2}{(4r{)}^3}$
3. Simplify the right-hand side: $\frac{T^2}{r^3}=\frac{T_B^2}{64r^3}$
4. Cancel $r^3$ from both sides and solve for $T_B$: $T_B^2=64T^2\to T_B=8T$

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.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using altitude $h$ (distance above the planet's surface) instead of $r=R+h$ as the distance in gravitational formulas. Why: Students confuse surface distance with the center-to-center distance the formula requires. Correct move: Always add the planet's radius to altitude to get $r$ before plugging into any formula.
• Wrong move: Claiming a more massive object exerts a larger gravitational force on a smaller object than the smaller exerts on the larger. Why: Students confuse force magnitude with acceleration, forgetting Newton's third law applies to gravity. Correct move: Always remember gravitational forces are an action-reaction pair, so they have equal magnitude regardless of mass difference.
• Wrong move: Canceling the central mass $M$ instead of the orbiting/object mass $m$ when deriving $g$ or orbital speed. Why: Students mix up which mass is which when setting up the force equation. Correct move: Label the central mass $M$ and the smaller object/orbiting mass $m$ before starting any derivation.
• Wrong move: Claiming if orbital radius doubles, orbital period doubles, because gravity follows an inverse-square law. Why: Students mix up the power relationship in Kepler's third law with the inverse-square for force. Correct move: Always write the full proportionality $T^2\propto r^3$ explicitly before solving proportional problems.
• Wrong move: Claiming astronauts on the ISS are weightless because there is no gravity in orbit. Why: Pop culture misconception that space equals zero gravity. Correct move: Remember weightlessness comes from continuous free fall, not zero gravity; gravity at the ISS orbit is ~90% of surface gravity.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

Two identical asteroids of mass $m$ are separated by distance $d$ and exert a gravitational force of magnitude $F$ on each other. One asteroid is split into two equal pieces, and one of the pieces is moved to a new separation distance of $d/2$ from the other intact asteroid. What is the new magnitude of the gravitational force between the two objects? A. $F/2$ B. $F$ C. $2F$ D. $4F$

Worked Solution: Start with the original force: $F=\frac{G{m}^2}{d^2}$. After splitting, the mass of the moved piece is $m/2$, the intact asteroid is still mass $m$, and the new separation is $d/2$. Substitute into the force formula: $F^{\prime }=\frac{G(m/2)(m)}{(d/2{)}^2}=\frac{G{m}^2/2}{d^2/4}=2\frac{G{m}^2}{d^2}=2F$. The other options come from incorrect exponent or mass calculations. The correct answer is C.

Question 2 (Free Response)

A dwarf planet Elera has mass $M=2.0\times 10^{22}\text{ kg}$ and radius $R=5.0\times 10^5\text{ m}$. A small moon of mass $m=1.0\times 10^{16}\text{ kg}$ orbits Elera in a circular orbit at an altitude of $h=1.5\times 10^6\text{ m}$ above Elera's surface. Use $G=6.67\times 10^{-11}{\text{ Nm}}^2/{\text{kg}}^2$ for all calculations. (a) Derive an expression for the acceleration due to gravity at the surface of Elera, then calculate its numerical value. (b) What is the orbital speed of the moon? (c) A student claims that if the moon were moved to twice its original orbital radius, its orbital speed would double. Do you agree with this claim? Justify your answer.

Worked Solution: (a) For an object of mass $m_o$ at Elera's surface, $F_g=G\frac{M{m}_o}{R^2}=m_{o}g$. Cancel $m_o$ to get the expression $g=\frac{GM}{R^2}$. Substitute values: $g=\frac{(6.67\times 10^{-11})(2.0\times 10^{22})}{(5.0\times 10^5{)}^2}\approx 5.3{\text{ m/s}}^2$. (b) Total orbital radius $r=R+h=5.0\times 10^5+1.5\times 10^6=2.0\times 10^6\text{ m}$. Orbital speed is $v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\times 10^{-11})(2.0\times 10^{22})}{2.0\times 10^6}}\approx 820\text{ m/s}$. (c) Disagree. Orbital speed follows the relationship $v=\sqrt{\frac{GM}{r}}$, so $v\propto \frac{1}{\sqrt{r}}$. If $r$ doubles, $v$ becomes $\frac{1}{\sqrt{2}}\approx 0.71$ times the original speed, not double. The student confused the proportionality of orbital period with the proportionality of orbital speed.

Question 3 (Application / Real-World Style)

GPS satellites orbit Earth at an altitude of 20,200 km above Earth's surface. Earth's mass is $M_E=5.97\times 10^{24}\text{ kg}$, and Earth's radius is $R_E=6370\text{ km}$. What is the orbital period of a GPS satellite, in hours? What does this result mean for GPS operation?

Worked Solution: First calculate total orbital radius: $r=R_E+h=6370+20200=26570\text{ km}=2.657\times 10^7\text{ m}$. Use Kepler's third law: $T^2=\frac{4{\pi }^2}{G{M}_E}{r}^3$. Substitute values: $T^2\approx 1.86\times 10^9{\text{ s}}^2$, so $T\approx 43100\text{ s}$. Convert to hours: $T=\frac{43100}{3600}\approx 12\text{ hours}$. A 12-hour orbital period means each GPS satellite passes over the same point on Earth's surface twice per day, which provides consistent global coverage for navigation signals.

7. Quick Reference Cheatsheet

8. What's Next

Newton's Law of Universal Gravitation is the foundational prerequisite for all orbital motion topics in Unit 3. Without mastering the inverse-square relationship, proportional reasoning, and how gravity provides centripetal force for orbits, you will not be able to solve more complex problems involving orbital energy, Kepler's full laws, or multi-object gravitational interactions that appear on the AP exam. Beyond Unit 3, the inverse-square relationship you learned here directly translates to Coulomb's law for electrostatics in later units, so mastering proportional reasoning for gravitation will make electrostatics much easier. This topic also sets the foundation for understanding field forces in AP Physics 2.

Next topics to study after this chapter: Centripetal Acceleration and Circular Motion Kepler's Laws of Planetary Motion Conservation of Energy in Orbits Gravitational Fields