🌍 Newton's Law of Universal Gravitation

Part 1 of 7 — Universal Gravitation

Newton's great insight was that the same force that makes an apple fall from a tree also keeps the Moon in orbit around Earth. Every object with mass attracts every other object with mass — this is universal gravitation.

In this lesson you will learn:

• Newton's Law of Universal Gravitation: $F = Gm_1m_2/r^2$
• The gravitational constant $G$
• How gravitational force depends on mass and distance
• Calculating gravitational force between objects

The Law of Universal Gravitation

Every two objects with mass attract each other with a force:

$F = \frac{Gm_1 m_2}{r^2}$

Key Features

1. Inverse-square law: $F \propto 1/r^2$

   • Double the distance → force is $1/4$ as strong
   • Triple the distance → force is $1/9$ as strong

2. Proportional to both masses: $F \propto m_1 m_2$

   • Double one mass → force doubles
   • Double both masses → force quadruples

3. Always attractive: gravity only pulls, never pushes

4. Newton's 3rd Law applies: the force on $m_1$ due to $m_2$ equals the force on $m_2$ due to $m_1$ (same magnitude, opposite direction)

Why Don't We Feel Gravity Between Everyday Objects?

$G$ is incredibly small: $6.67 \times 10^{-11}$. Two 100 kg people standing 1 m apart attract each other with only $F = 6.67 \times 10^{-7}$ N — about the weight of a grain of sand!

Gravitational Force Concepts 🎯

Gravitational Force Calculations 🧮

Use $G = 6.67 \times 10^{-11}$ N·m²/kg².

1. Find the gravitational force between Earth ($5.97 \times 10^{24}$ kg) and the Moon ($7.35 \times 10^{22}$ kg) separated by $3.84 \times 10^8$ m. Express in scientific notation as $X \times 10^{20}$ N. What is $X$ (round to 3 significant figures)?

2. Two 70 kg people stand 2 m apart. What is the gravitational force between them (in N, express in scientific notation, give the coefficient to 3 significant figures, e.g., "8.17" for $8.17 \times 10^{-8}$ N)?

3. At Earth's surface, $r = 6.37 \times 10^6$ m. Calculate $g = GM/r^2$ where $M = 5.97 \times 10^{24}$ kg. What value do you get (in m/s², round to 3 significant figures)?

Proportional Reasoning 🔍

Exit Quiz — Universal Gravitation ✅

🌐 Gravitational Field: $g = GM/r^2$

Part 2 of 7 — Universal Gravitation

The gravitational field describes how massive objects modify the space around them, creating a region where other masses experience a gravitational force. This is the concept behind $g$ — and it's not always $9.8$ m/s²!

In this lesson you will learn:

• The gravitational field concept: $g = GM/r^2$
• How $g$ varies with altitude and location
• Surface gravity on different planets
• The relationship between weight and gravitational field

The Gravitational Field

Definition

The gravitational field $\vec{g}$ at a point in space tells you the gravitational force per unit mass that would be experienced by a test mass placed there:

$\vec{g} = \frac{\vec{F}}{m_{test}} \quad \Rightarrow \quad g = \frac{GM}{r^2}$

• Direction: toward the center of the mass $M$ creating the field
• Units: N/kg (equivalent to m/s²)
• $g$ is a property of space — it exists whether or not a test mass is there

Weight and Gravitational Field

$W = mg$

where $g$ is the local gravitational field strength. On Earth's surface: $g \approx 9.8$ m/s².

How $g$ Varies with Distance

$g(r) = \frac{GM}{r^2}$

At Earth's surface ($r = R_E$): $g_0 = GM_E/R_E^2 \approx 9.8$ m/s²

At altitude $h$ above the surface ($r = R_E + h$):

$g(h) = \frac{GM_E}{(R_E + h)^2} = g_0 \left(\frac{R_E}{R_E + h}\right)^2$

Surface Gravity on Other Bodies

Gravitational Field Concepts 🎯

Gravitational Field Calculations 🧮

Use $G = 6.67 \times 10^{-11}$ N·m²/kg², $g_E = 10$ m/s², $R_E = 6.4 \times 10^6$ m.

1. What is the gravitational field strength at twice Earth's radius from Earth's center (in m/s²)?

2. Mars has mass $6.42 \times 10^{23}$ kg and radius $3.39 \times 10^6$ m. What is its surface gravity (in m/s², round to 3 significant figures)?

3. At what distance from Earth's center (in units of $R_E$, round to 3 significant figures) is $g = 1$ m/s²?

Field Concepts 🔍

Exit Quiz — Gravitational Field ✅

🛰️ Orbital Motion: Gravity as Centripetal Force

Part 3 of 7 — Universal Gravitation

The most beautiful connection in physics: the gravitational force provides the centripetal force for orbital motion. This insight unifies terrestrial and celestial physics.

In this lesson you will learn:

• How gravity acts as centripetal force for orbiting objects
• Derive orbital velocity: $v = \sqrt{GM/r}$
• Why orbital speed is independent of satellite mass
• The relationship between orbital radius and speed

Gravity = Centripetal Force

For a satellite of mass $m$ orbiting a planet of mass $M$ at radius $r$:

$F_{gravity} = F_{centripetal}$

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

Solving for Orbital Velocity

Cancel $m$ (satellite mass doesn't matter!):

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

Key Insights

1. Mass of satellite doesn't matter: A feather and a bowling ball orbit at the same speed at the same radius!

2. Higher orbit = slower speed: $v \propto 1/\sqrt{r}$

   • Doubling orbital radius → speed decreases by factor $\sqrt{2}$

3. Only one speed works for each radius: You can't orbit at any speed you want — the speed is determined by $r$ and $M$

Orbital Period

$T = \frac{2\pi r}{v} = \frac{2\pi r}{\sqrt{GM/r}} = 2\pi\sqrt{\frac{r^3}{GM}}$

Energy in Orbit

$KE = \frac{1}{2}mv^2 = \frac{GMm}{2r}$

$PE = -\frac{GMm}{r}$

$E_{total} = KE + PE = -\frac{GMm}{2r}$

Total energy is negative (bound orbit) and equals half the PE.

Orbital Motion Concepts 🎯

Orbital Calculations 🧮

Use $G = 6.67 \times 10^{-11}$ N·m²/kg², $M_E = 5.97 \times 10^{24}$ kg, $R_E = 6.37 \times 10^6$ m.

1. What is the orbital speed of the ISS at altitude 400 km above Earth's surface (in m/s, round to nearest 100)? Hint: $r = R_E + h$.

2. What is the orbital period of the ISS (in minutes, round to nearest whole number)?

3. What orbital radius gives a period of exactly 24 hours (geosynchronous orbit)? Express as a multiple of $R_E$ (round to 3 significant figures).

Orbital Relationships 🔍

Exit Quiz — Orbital Motion ✅

🪐 Kepler's Third Law: $T^2 \propto r^3$

Part 4 of 7 — Universal Gravitation

Kepler discovered three laws of planetary motion decades before Newton explained them. The Third Law — the relationship between orbital period and radius — is one of the most powerful tools in astrophysics.

In this lesson you will learn:

• Kepler's Third Law: $T^2 \propto r^3$
• How to derive it from Newton's Law of Gravitation
• Using Kepler's Law to compare orbits
• Calculating unknown orbital properties

Deriving Kepler's Third Law

Starting from gravity = centripetal force:

$\frac{GMm}{r^2} = \frac{mv^2}{r} = \frac{m(2\pi r/T)^2}{r} = \frac{4\pi^2 mr}{T^2}$

Cancel $m$ and rearrange:

$T^2 = \frac{4\pi^2}{GM} r^3$

The Key Relationship

$T^2 = \left(\frac{4\pi^2}{GM}\right) r^3$

This is a linear relationship between $T^2$ and $r^3$.

Comparing Two Orbits (Same Central Body)

$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$

This ratio form is incredibly useful — you don't even need to know $G$ or $M$!

Kepler's Three Laws Summary

Important: Same Central Body Only!

$T^2/r^3 = 4\pi^2/(GM)$ depends on $M$ — the mass of the body being orbited. You can only compare orbits around the same central body.

Kepler's Third Law Concepts 🎯

Kepler's Third Law Calculations 🧮

1. Earth orbits the Sun at 1 AU with $T = 1$ year. What is the orbital period of an asteroid at 2.5 AU (in years, round to 3 significant figures)?

2. A moon orbits a planet with period 10 days at radius $r$. Another moon orbits the same planet at radius $2r$. What is its period (in days, round to 3 significant figures)?

3. Jupiter's moon Io orbits at $r_1 = 4.22 \times 10^8$ m with $T_1 = 1.77$ days. Europa orbits at $r_2 = 6.71 \times 10^8$ m. What is Europa's period (in days, round to 3 significant figures)?

Kepler's Law Applications 🔍

Exit Quiz — Kepler's Third Law ✅

🚀 Satellite Problems & Orbital Velocity

Part 5 of 7 — Universal Gravitation

Satellites are a cornerstone of AP Physics 1 gravitation problems. In this lesson, we'll tackle practical satellite problems, escape velocity, and the physics of different orbit types.

In this lesson you will learn:

• Low Earth orbit (LEO) vs. geosynchronous orbit (GEO)
• Escape velocity: $v_{esc} = \sqrt{2GM/r}$
• Energy required to launch satellites
• Transfer orbits and orbit changes

Common Orbit Types

Key Pattern

Higher orbit → slower speed, longer period

Escape Velocity

The minimum speed needed to escape a gravitational field (reach infinite distance with zero final speed):

$\frac{1}{2}mv_{esc}^2 = \frac{GMm}{r}$

$v_{esc} = \sqrt{\frac{2GM}{r}} = \sqrt{2} \cdot v_{orbit}$

Escape Velocity from Earth's Surface

$v_{esc} = \sqrt{2gR_E} = \sqrt{2 \times 10 \times 6.37 \times 10^6} \approx 11{,}200 \text{ m/s} \approx 11.2 \text{ km/s}$

Key Insight

Escape velocity = $\sqrt{2}$ × orbital velocity at the same radius. A satellite in orbit needs to increase its speed by only about 41% to escape!

Satellite Concepts 🎯

Satellite Calculations 🧮

Use $g = 10$ m/s², $R_E = 6.4 \times 10^6$ m, $M_E = 6.0 \times 10^{24}$ kg, $G = 6.67 \times 10^{-11}$.

1. What is the orbital velocity for a satellite at altitude $h = 200$ km above Earth (in m/s, round to nearest 100)?

2. What is the escape velocity from Earth's surface (in km/s, round to 3 significant figures)?

3. The Moon orbits at $r = 3.84 \times 10^8$ m. What is the Moon's orbital speed (in m/s, round to nearest 10)?

Energy in Satellite Problems

Total Energy of a Satellite

$E = -\frac{GMm}{2r}$

• Negative: the satellite is gravitationally bound
• Less negative at higher orbits: higher orbit = more total energy
• To move a satellite to a higher orbit: must ADD energy (fire rockets forward, then forward again)

Energy to Launch a Satellite

From Earth's surface to orbit at radius $r$:

$\Delta E = E_{orbit} - E_{surface} = -\frac{GMm}{2r} - \left(-\frac{GMm}{R_E}\right)$

$\Delta E = GMm\left(\frac{1}{R_E} - \frac{1}{2r}\right)$

Binding Energy

The energy needed to completely remove a satellite from orbit to infinity:

$E_{bind} = |E| = \frac{GMm}{2r} = \frac{1}{2}mv^2$

The binding energy equals the kinetic energy! You need to add exactly one more KE worth of energy to escape.

Satellite Energy Concepts 🔍

Exit Quiz — Satellite Problems ✅

🔧 Problem-Solving Workshop

Part 6 of 7 — Universal Gravitation

Time to tackle complex gravitation problems that combine multiple concepts: Newton's Law, gravitational field, orbital motion, Kepler's Law, and energy. These multi-step problems mirror what you'll see on the AP exam.

In this lesson you will:

• Solve complex multi-step gravitation problems
• Combine gravitation with circular motion concepts
• Practice AP-level free response problems
• Master the "which formula to use" decision process

Formula Decision Guide

Problem-Solving Steps

1. Identify: What are you given? What do you need to find?
2. Choose: Which formula connects your known and unknown quantities?
3. Check units: Make sure everything is in SI units (m, kg, s)
4. Solve: Algebra first, numbers last
5. Verify: Does the answer make physical sense?

Warm-Up Problems 🧮

Use $G = 6.67 \times 10^{-11}$, $M_E = 6.0 \times 10^{24}$ kg, $R_E = 6.4 \times 10^6$ m.

1. A satellite orbits Earth at twice Earth's radius from Earth's center. What is the gravitational field strength at that location (in m/s², round to 3 significant figures)?

2. What is the orbital speed at that altitude (in m/s, round to nearest 100)?

3. What is the orbital period (in hours, round to 3 significant figures)?

Multi-Step Problems 🎯

Challenge Problems 🧮

Use $G = 6.67 \times 10^{-11}$.

1. A planet has surface gravity $g = 20$ m/s² and radius $R = 8 \times 10^6$ m. What is the planet's mass (in kg, express as $X \times 10^{25}$ — give $X$ to 3 significant figures)?

2. Using the planet from problem 1, what is the escape velocity from its surface (in km/s, round to 3 significant figures)?

3. A satellite orbits this planet at altitude $h = R$ (i.e., at $r = 2R$ from center). What is its orbital period (in hours, round to 3 significant figures)?

Problem-Solving Strategy 🔍

Exit Quiz — Problem-Solving Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Universal Gravitation

This final lesson brings together everything from universal gravitation for AP exam preparation. We'll cover the most tested question types, common mistakes, and strategies for both multiple choice and free response.

In this lesson you will:

• Tackle AP-style multiple choice questions
• Practice FRQ-style derivations
• Review the complete gravitation toolkit
• Master proportional reasoning — the #1 AP skill for gravitation

Complete Gravitation Toolkit

Essential Formulas

The Power of Proportional Reasoning

On the AP exam, most gravitation questions test proportional reasoning — not plug-and-chug. Know these:

Common AP Mistakes

Mistake 1: Using altitude instead of orbital radius

• $r$ is measured from the center of the planet
• $r = R_{planet} + h_{altitude}$
• ALWAYS add the planet's radius!

Mistake 2: Confusing $g$ and $G$

• $G = 6.67 \times 10^{-11}$ — universal constant, same everywhere
• $g$ — local gravitational field strength, varies with location

Mistake 3: Thinking astronauts are "outside gravity"

• The ISS at 400 km altitude: $g \approx 8.7$ m/s² (only 11% less than surface!)
• Astronauts float because they're in free fall, not because gravity is absent

Mistake 4: Wrong proportional reasoning with $v$ and $T$

• $v \propto 1/\sqrt{r}$ (NOT $1/r$)
• $T \propto r^{3/2}$ (NOT $r$ or $r^2$)

Mistake 5: Drawing centripetal force separately from gravity

• For orbiting objects, gravity IS the centripetal force
• Don't draw both on a free body diagram

AP-Style Multiple Choice 🎯

FRQ Practice

Sample FRQ: "Weighing Jupiter"

Astronomers observe that Jupiter's moon Io orbits at radius $r$ with period $T$.

(a) Derive an expression for Jupiter's mass in terms of $r$, $T$, and fundamental constants.

Gravity provides centripetal force: $\frac{GM_J m}{r^2} = \frac{4\pi^2 m r}{T^2}$

Cancel $m$: $\frac{GM_J}{r^2} = \frac{4\pi^2 r}{T^2}$

$M_J = \frac{4\pi^2 r^3}{GT^2}$

(b) If Europa orbits at $r_E = 1.59r_{Io}$, find the ratio $T_E/T_{Io}$.

$\frac{T_E^2}{T_{Io}^2} = \frac{r_E^3}{r_{Io}^3} = (1.59)^3 = 4.02$

$T_E/T_{Io} = \sqrt{4.02} = 2.00$

(c) Does the mass of the moon matter? Justify.

No. The moon's mass cancels in the derivation (Step 2 above). Orbital properties depend only on the central body's mass and the orbital radius, not the orbiting body's mass. This is analogous to all objects falling with the same acceleration $g$ near Earth's surface.

AP-Style Calculations 🧮

1. Saturn's moon Titan orbits at $r = 1.22 \times 10^9$ m with $T = 15.95$ days. Calculate Saturn's mass (in kg, express as $X \times 10^{26}$ — give $X$ to 3 significant figures). Use $G = 6.67 \times 10^{-11}$.

2. A planet has mass $5M_E$ and radius $2R_E$. What is the escape velocity from this planet's surface as a multiple of Earth's escape velocity? (round to 3 significant figures)

3. Two satellites orbit Earth. Satellite A has period 6 hours and Satellite B has period 24 hours. What is the ratio $r_B/r_A$ (round to 3 significant figures)?

Conceptual Review 🔍

Final Exit Quiz — Universal Gravitation ✅