02Gravity and Planetary Motion
Apply Newton's law of universal gravitation to gravitational force, orbital speed, escape velocity, and Kepler's laws of planetary motion.
Gravity governs the motion of every planet, moon, and satellite. Newton's law of gravitation connects terrestrial physics to the cosmos.
Essential Question: How does a single mathematical law — Newton's law of universal gravitation — explain the fall of an apple, the orbit of the Moon, and the motion of every planet in the solar system?
Lesson Overview
In 1687 Isaac Newton unified terrestrial and celestial mechanics with one elegant equation: every pair of masses in the universe attracts each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between them. From this single idea we can derive the orbital speed of the International Space Station, predict the period of any planet, calculate the minimum speed needed to escape Earth's gravity, and understand why tides rise and fall. Johannes Kepler had already described planetary orbits empirically decades earlier; Newton showed why those descriptions are true.
Kepler's Three Laws of Planetary Motion
First Law — Law of Ellipses
Every planet orbits the Sun in an ellipse with the Sun at one focus. Circles are a special case of ellipses (eccentricity = 0). Most planetary orbits are nearly circular but technically elliptical.
Second Law — Law of Equal Areas
A line segment joining a planet to the Sun sweeps out equal areas in equal time intervals. Consequence: a planet moves faster when closer to the Sun (perihelion) and slower when farther away (aphelion). This is a direct result of conservation of angular momentum.
Third Law — Law of Periods
The square of a planet's orbital period is proportional to the cube of its semi-major axis: T² ∝ r³. More precisely, T² = 4π²r³ / (GM). This lets us compare orbits of different planets or satellites around the same central body.
Key Equations
Universal Gravitation
F = G m₁m₂ / r²
G = 6.674×10⁻¹¹ N·m²/kg²
Gravitational Field Strength
g = GM / r²
acceleration due to gravity at distance r
Orbital Velocity
v_orb = √(GM / r)
speed for circular orbit at radius r
Kepler's Third Law
T² = 4π²r³ / (GM)
orbital period T at radius r
Escape Velocity
v_esc = √(2GM / r)
minimum speed to escape gravity from r
Worked Examples
Find the gravitational force between Earth (M = 5.97×10²⁴ kg) and the Moon (m = 7.35×10²² kg) when their centres are separated by r = 3.84×10⁸ m.
Formula: F = G m₁m₂ / r²
G = 6.674×10⁻¹¹ N·m²/kg²
Numerator: G × M × m = (6.674×10⁻¹¹)(5.97×10²⁴)(7.35×10²²)
= 6.674×10⁻¹¹ × 4.388×10⁴⁷
= 2.928×10³⁷ N·m²
Denominator: r² = (3.84×10⁸)² = 1.475×10¹⁷ m²
F = 2.928×10³⁷ / 1.475×10¹⁷ ≈ 1.98×10²⁰ N
Find the gravitational field strength g on the surface of Mars. (M_Mars = 6.39×10²³ kg, R_Mars = 3.39×10⁶ m)
Formula: g = GM / r²
Numerator: GM = (6.674×10⁻¹¹)(6.39×10²³)
= 4.265×10¹³ m³/s²
Denominator: R² = (3.39×10⁶)² = 1.149×10¹³ m²
g = 4.265×10¹³ / 1.149×10¹³ ≈ 3.71 m/s²
The International Space Station orbits at r = 6.77×10⁶ m from Earth's centre (M_Earth = 5.97×10²⁴ kg). Find its orbital speed.
Formula: v_orb = √(GM / r)
GM_Earth = (6.674×10⁻¹¹)(5.97×10²⁴) = 3.985×10¹⁴ m³/s²
GM / r = 3.985×10¹⁴ / 6.77×10⁶ = 5.887×10⁷ m²/s²
v_orb = √(5.887×10⁷) ≈ 7,673 m/s
Using Kepler's Third Law, find the orbital period of the ISS at r = 6.77×10⁶ m.
Formula: T² = 4π²r³ / (GM)
r³ = (6.77×10⁶)³ = 3.099×10²⁰ m³
4π²r³ = 4 × 9.8696 × 3.099×10²⁰ = 1.224×10²² m³
GM_Earth = 3.985×10¹⁴ m³/s²
T² = 1.224×10²² / 3.985×10¹⁴ = 3.071×10⁷ s²
T = √(3.071×10⁷) ≈ 5,542 s
Convert: 5,542 s ÷ 60 ≈ 92.4 minutes
Find the escape velocity from Earth's surface. (M = 5.97×10²⁴ kg, R = 6.37×10⁶ m)
Formula: v_esc = √(2GM / r)
GM_Earth = 3.985×10¹⁴ m³/s²
2GM / R = 2 × 3.985×10¹⁴ / 6.37×10⁶
= 7.970×10¹⁴ / 6.37×10⁶ = 1.251×10⁸ m²/s²
v_esc = √(1.251×10⁸) ≈ 11,185 m/s
Guided Problems
Two 80 kg students stand 2.0 m apart. Calculate the gravitational force between them.
Hint: Use F = Gm₁m₂/r². Both masses are 80 kg, r = 2.0 m, G = 6.674×10⁻¹¹ N·m²/kg². The answer will be very small — gravity between everyday objects is negligible.
Jupiter has mass M = 1.90×10²⁷ kg and radius R = 7.15×10⁷ m. Find the gravitational field strength at Jupiter's cloud tops.
Hint: Use g = GM/r². Substitute M = 1.90×10²⁷ kg and r = R = 7.15×10⁷ m. You should get roughly 24.8 m/s² — about 2.5 times Earth's surface gravity.
A communications satellite orbits Earth at an altitude of 500 km above the surface (R_Earth = 6.37×10⁶ m). Find its orbital speed.
Hint: First find r = R_Earth + altitude = 6.37×10⁶ + 5.00×10⁵ = 6.87×10⁶ m. Then use v = √(GM/r) with GM_Earth = 3.985×10¹⁴ m³/s².
Mars orbits the Sun at an average radius of 2.28×10¹¹ m. The Sun's mass is 1.99×10³⁰ kg. Find Mars's orbital period in Earth years. (1 year = 3.156×10⁷ s)
Hint: Use T² = 4π²r³/(GM_Sun). Compute r³, multiply by 4π², divide by GM_Sun, then take the square root to get T in seconds. Divide by 3.156×10⁷ to convert to years.
What is the escape velocity from the Moon's surface? (M_Moon = 7.35×10²² kg, R_Moon = 1.74×10⁶ m)
Hint: Use v_esc = √(2GM/r) with r = R_Moon. GM_Moon = (6.674×10⁻¹¹)(7.35×10²²). Compare your answer to Earth's 11.2 km/s — the Moon's lower gravity makes escape much easier.
Key Vocabulary
Universal Gravitation
Newton's law stating that every mass attracts every other mass with a force F = Gm₁m₂/r², where r is the distance between their centres of mass.
Example: Earth attracts the Moon with the same magnitude force that the Moon attracts Earth.
Gravitational Constant (G)
The universal proportionality constant in Newton's law of gravitation: G = 6.674×10⁻¹¹ N·m²/kg². It was first measured by Henry Cavendish in 1798.
Example: G is the same everywhere in the universe — on Earth, on Mars, or in a distant galaxy.
Gravitational Field Strength (g)
The gravitational force per unit mass at a point in space, given by g = GM/r². It equals the free-fall acceleration at that location.
Example: On Earth's surface g ≈ 9.81 m/s²; on Mars g ≈ 3.71 m/s².
Orbital Velocity
The tangential speed required for an object to maintain a circular orbit at radius r around a mass M: v = √(GM/r). Gravity provides exactly the centripetal force needed.
Example: The ISS orbits at ~7.66 km/s; a lower orbit requires a higher speed.
Escape Velocity
The minimum launch speed needed for an object to escape a gravitational field without further propulsion: v_esc = √(2GM/r). It is √2 times the circular orbital speed at the same radius.
Example: Earth's escape velocity from the surface is ~11.2 km/s.
Kepler's First Law
Planets orbit the Sun in ellipses with the Sun at one focus. The orbit's shape is described by its eccentricity (0 = circle, approaching 1 = very elongated ellipse).
Example: Earth's orbit has eccentricity ≈ 0.017, making it nearly circular.
Kepler's Second Law
A line from the Sun to a planet sweeps equal areas in equal time intervals. This means a planet moves fastest at perihelion (closest approach) and slowest at aphelion (farthest point).
Example: Earth moves ~30.3 km/s at perihelion (January) and ~29.3 km/s at aphelion (July).
Kepler's Third Law
The square of a planet's orbital period is proportional to the cube of its semi-major axis: T² = 4π²r³/(GM). This allows comparison of any two orbits around the same body.
Example: If planet B is 4× farther from the Sun than planet A, planet B's period is 4^(3/2) = 8× longer.
Workbook Check
Interactive Practice — 5 Questions
If the distance between two masses is doubled, the gravitational force between them becomes:
A satellite is moved to an orbit with twice the original radius. Its new orbital speed is:
Which of the following correctly states Kepler's Second Law?
The escape velocity from a planet is 8 km/s. What is the circular orbital speed just above its surface?
A geosynchronous satellite has an orbital period equal to:
Independent Practice
Calculate the gravitational force between two 70 kg people standing 1.5 m apart. (G = 6.67 × 10⁻¹¹ N·m²/kg²)
A satellite orbits Earth at an altitude of 400 km above the surface. Find its orbital speed. (M_E = 5.97 × 10²⁴ kg, R_E = 6.37 × 10⁶ m, G = 6.67 × 10⁻¹¹ N·m²/kg²)
Mars has an orbital period of 1.88 years. Using Kepler's 3rd law and Earth's orbital radius of 1.00 AU, find the orbital radius of Mars in AU.
Calculate the escape velocity from the Moon's surface. (M_Moon = 7.35 × 10²² kg, R_Moon = 1.74 × 10⁶ m, G = 6.67 × 10⁻¹¹ N·m²/kg²)
★ A spacecraft is launched from Earth's surface. (a) Calculate the minimum speed needed to escape Earth's gravity. (b) At what altitude above Earth's surface does the gravitational acceleration equal g/4? (c) How does the orbital period change if a satellite's orbital radius is doubled? (G = 6.67 × 10⁻¹¹, M_E = 5.97 × 10²⁴ kg, R_E = 6.37 × 10⁶ m)
ChallengeCommon Mistakes
Using g = 9.8 m/s² for gravitational acceleration at altitudes far above Earth's surface
g = 9.8 m/s² is only valid near Earth's surface. At altitude h, use g = GM/(R+h)²
Thinking satellites need continuous thrust to stay in orbit
A satellite in circular orbit is in free fall — gravity provides the centripetal force. No thrust is needed to maintain altitude
Confusing Kepler's 2nd law (equal areas) with constant speed
Equal areas in equal times means the planet moves FASTER when closer to the Sun (perihelion) and slower when farther (aphelion)
Forgetting to square the period in Kepler's 3rd law: T² ∝ r³
Kepler's 3rd law: (T₁/T₂)² = (r₁/r₂)³. Both sides must be squared and cubed respectively
Math Tips
Orbital speed: v = √(GM/r). For circular orbit, set gravitational force = centripetal force: GMm/r² = mv²/r → v = √(GM/r)
Escape velocity: v_esc = √(2GM/r) = √2 × v_orbital. It is independent of the object's mass — only the planet's mass and radius matter
Kepler's 3rd law ratio shortcut: if you know Earth's period (1 yr) and orbit (1 AU), then T²/r³ = constant for all planets in the same system
G = 6.674 × 10⁻¹¹ N·m²/kg². For Earth: M_E = 5.97 × 10²⁴ kg, R_E = 6.37 × 10⁶ m. Memorize these for quick orbital calculations