Newton's Law of Universal Gravitation

AP Physics 1· AP Physics 1 CED — Circular Motion and Gravitation· 14 min read

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.

📘 Definition

Newton's Law of Universal Gravitation

$F_{g} = G \frac{m _{1} m _{2}}{r ^{2}}$

Every massive object attracts every other massive object with a force directly proportional to the product of their masses and inversely proportional to the square of the center-to-center distance between them.

2. Force Formula and Inverse-Square Proportionality★★☆☆☆⏱ 4 min

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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 $G = 6.67 \times 1 0^{- 11} Nm^{2} / kg^{2}$ 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 $1/9$ .

📐 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 all changes:
$F' = G \frac{(30)(40)}{(6)^2} = \frac{1200}{36} G = \frac{100}{3}G$
3
Take the ratio of new force to original force to eliminate $G$ :
$\frac{F'}{F} = \frac{\frac{100}{3}G}{50G} = \frac{2}{3}$
4
Final result:
$F' = \frac{2}{3}F$

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

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The familiar $g = 9.8 m/s^{2}$ 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.

🔬 Derivation

Goal:

Derive the general formula for gravitational acceleration at distance $r$ from a planet's center

Starting from:

Newton's second law $F = ma$ and the universal gravitation force formula

1
For an object of mass $m$ at distance $r$ from the center of a planet of mass $M$ , set gravitational force equal to mass times acceleration:
2
$F_g = G \frac{Mm}{r^2} = mg$

Result:

The mass of the object $m$ cancels out from both sides, leaving the general formula $g (r) = \frac{GM}{r ^{2}}$ , which is valid for all points outside the planet ( $r \geq R$ , where $R$ is the planet's radius). At 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 total orbital radius from Earth's center:
$r = R_E + h = 6400 + 400 = 6800 \text{ km} = \frac{17}{16}R_E$
2
Write expressions for surface and orbital acceleration:
$g_s = \frac{GM_E}{R_E^2}, \quad g_{orbit} = \frac{GM_E}{r^2}$
3
Substitute $r = \frac{17}{16} R_{E}$ into the orbital acceleration formula:
$g_{orbit} = \frac{GM_E}{(\frac{17}{16}R_E)^2} = g_s \left(\frac{16}{17}\right)^2 = g_s \frac{256}{289}$
4
Simplify to get the final result:
$g_{orbit} \approx 0.89 g_s$

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

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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 $v = \frac{2 π r}{T}$ (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: $T^{2} \propto r^{3}$ .

📐 Worked Example

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

1
Use Kepler's third law proportionality for the same central planet:
$\frac{T_A^2}{r_A^3} = \frac{T_B^2}{r_B^3}$
2
Substitute known values $T_{A} = T$ , $r_{A} = r$ , $r_{B} = 4 r$ :
$\frac{T^2}{r^3} = \frac{T_B^2}{(4r)^3} = \frac{T_B^2}{64 r^3}$
3
Cancel $r^{3}$ from both sides and solve for $T_{B}$ :
$T_B^2 = 64 T^2 \implies T_B = 8T$

✓ Quick check

Test your understanding of core proportional relationships:

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 piece is moved to a new separation distance of $d /2$ from the intact asteroid. What is the new force magnitude?
- $F /2$
- $F$
- $2 F$
- $4 F$
Reveal answer
$2F$ —
Original force: $F = \frac{G m ^{2}}{d ^{2}}$ . New force: $F^{'} = \frac{G ( m /2 ) ( m )}{( d /2 ) ^{2}} = 2 F$ , so this option is correct.

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

Wrong move:

Using altitude $h$ instead of $r = R + h$ as the distance in gravitational formulas

Why:

Students confuse surface distance with the center-to-center distance required by the formula

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:

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 calculation or 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 relationship 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:

This is a common pop culture misconception that incorrectly equates space with zero gravity

Correct move:

Remember weightlessness comes from continuous free fall around Earth, not zero gravity; gravity at the ISS orbit is ~90% of surface gravity

6. Quick Reference Cheatsheet

Category	Formula	Notes
Gravitational force magnitude	$F_{g} = G \frac{m _{1} m _{2}}{r ^{2}}$	For point/uniform spherical masses; $r$ = center-to-center distance; force is always attractive
Universal gravitational constant	$G = 6.67 \times 1 0^{- 11} Nm^{2} / kg^{2}$	Rarely needs plugging in; most problems use proportional reasoning
Gravitational acceleration at distance $r$	$g (r) = \frac{GM}{r ^{2}}$	Valid for $r \geq$ planet radius; $g$ is independent of the object's mass
Surface gravitational acceleration	$g_{s} = \frac{GM}{R ^{2}}$	$R$ = planet radius; near Earth, $g_{s} \approx 9.8 m/s^{2}$
Circular orbital speed	$v = \frac{GM}{r}$	$M$ = central mass; speed is independent of orbiting object mass
Kepler's Third Law (circular orbits)	$T^{2} = (\frac{4 π ^{2}}{GM}) r^{3}$	$T^{2} \propto r^{3}$ only for objects orbiting the same central mass
Newton's third law for gravity	$∣ F_{1 on 2} ∣ = ∣ F_{2 on 1} ∣$	Force magnitudes are equal, regardless of mass difference

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Inverse-square proportional reasoning
2022 · FRQ
Orbital speed proportionality

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, and sets the foundation for understanding field forces in future physics study.