2bOrbital Motion and Kepler's Laws
Apply Kepler's three laws, derive orbital velocity and escape velocity, and analyze geostationary orbits.
Kepler's laws and orbital mechanics are the foundation of space exploration — from calculating satellite orbits to planning interplanetary missions. Every GPS satellite and space telescope depends on these principles.
How did Kepler discover the laws of planetary motion from pure observation — and how did Newton later explain them using gravity?
Lesson Overview
Kepler's three laws describe how planets orbit the Sun: (1) orbits are ellipses with the Sun at one focus; (2) a line from the Sun to a planet sweeps equal areas in equal times; (3) the square of the orbital period is proportional to the cube of the semi-major axis (T² ∝ r³). For circular orbits, setting gravitational force equal to centripetal force gives orbital velocity v = √(GM/r) and period T = 2πr/v. Escape velocity v_esc = √(2GM/R) is the minimum speed needed to escape a body's gravity. Geostationary satellites orbit at a fixed altitude where T = 24 hours.
Key Equations
Worked Examples
Find the orbital speed of a satellite at r = 7.0 × 10⁶ m from Earth's center. (G = 6.67 × 10⁻¹¹ N·m²/kg², M = 5.97 × 10²⁴ kg)
Find the orbital period of the satellite in Example 1 (r = 7.0 × 10⁶ m, v ≈ 7543 m/s).
Earth orbits the Sun at r = 1.50 × 10¹¹ m with T = 3.156 × 10⁷ s. Mars orbits at r_Mars = 2.28 × 10¹¹ m. Use Kepler's 3rd law to find Mars's orbital period.
Find the escape velocity from Earth's surface. (G = 6.67 × 10⁻¹¹ N·m²/kg², M = 5.97 × 10²⁴ kg, R = 6.37 × 10⁶ m)
Find the orbital radius of a geostationary satellite (T = 24 h = 86,400 s) around Earth. (G = 6.67 × 10⁻¹¹ N·m²/kg², M = 5.97 × 10²⁴ kg)
Guided Problems
A satellite orbits Earth at r = 8.0 × 10⁶ m. Find its orbital speed and period.
Hint: Use v = √(GM/r) first, then T = 2πr/v. Keep track of units.
Jupiter orbits the Sun at r_J = 7.78 × 10¹¹ m. Earth orbits at r_E = 1.50 × 10¹¹ m with T_E = 1 year. Use Kepler's 3rd law to find Jupiter's orbital period.
Hint: T² ∝ r³, so T_J²/T_E² = (r_J/r_E)³. Solve for T_J.
Find 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). Substitute Moon's mass and radius.
A satellite in a circular orbit has orbital speed v = 6000 m/s. Find its orbital radius r. (G = 6.67 × 10⁻¹¹, M_Earth = 5.97 × 10²⁴ kg)
Hint: Rearrange v = √(GM/r) to get r = GM/v².
State Kepler's second law and explain what it implies about a planet's speed when it is closest to the Sun (perihelion) versus farthest (aphelion).
Hint: Equal areas in equal times means the planet must move faster when closer to the Sun to sweep the same area.
Key Vocabulary
Kepler's First Law
Each planet orbits the Sun in an ellipse, with the Sun at one focus. A circle is a special case of an ellipse.
Example: Earth's orbit is nearly circular (eccentricity ≈ 0.017), while Halley's Comet has a highly elongated elliptical orbit.
Kepler's Second Law
A line connecting a planet to the Sun sweeps out equal areas in equal time intervals. This means planets move faster when closer to the Sun.
Example: Earth moves slightly faster in January (perihelion, ~147 million km) than in July (aphelion, ~152 million km).
Kepler's Third Law
The square of a planet's orbital period is proportional to the cube of its semi-major axis: T² ∝ r³. For circular orbits: T² = (4π²/GM)r³.
Example: Mars (r ≈ 1.52 AU) has T ≈ 1.88 years; (1.88)² ≈ (1.52)³ ≈ 3.51 ✓
Orbital Velocity
The speed needed to maintain a circular orbit at radius r: v = √(GM/r). Faster orbits are at lower altitudes; higher orbits are slower.
Example: The ISS at ~400 km altitude orbits at ~7660 m/s and completes an orbit in ~92 minutes.
Escape Velocity
The minimum speed needed to escape a body's gravitational pull: v_esc = √(2GM/R). Note that v_esc = √2 × v_orbital at the surface.
Example: Earth's escape velocity is ~11.2 km/s. The Moon's is only ~2.4 km/s (lower mass and radius).
Geostationary Orbit
A circular orbit at ~42,200 km from Earth's center where the orbital period equals 24 hours. The satellite appears stationary above a fixed point on the equator.
Example: Communication and weather satellites (e.g., GPS, DirecTV) use geostationary orbits.
Workbook Check — Interactive Quiz
Interactive Practice — 5 Questions
A satellite moves to a higher circular orbit. What happens to its orbital speed?
According to Kepler's second law, a planet moves fastest when it is:
Planet A has orbital radius r. Planet B has orbital radius 4r. What is the ratio T_B/T_A?
Escape velocity from Earth is ~11.2 km/s. What is the escape velocity from a planet with the same mass but twice the radius?
A geostationary satellite has an orbital period of:
Independent Practice
Find the orbital speed of a satellite at r = 6.8 × 10⁶ m from Earth's center. (G = 6.67 × 10⁻¹¹, M = 5.97 × 10²⁴ kg)
Earth orbits the Sun at r = 1.50 × 10¹¹ m with T = 365.25 days. Find Earth's orbital speed.
Use Kepler's 3rd law to find the orbital period of Venus (r_V = 1.08 × 10¹¹ m) given Earth's data (r_E = 1.50 × 10¹¹ m, T_E = 1 year).
Find the escape velocity from the Moon's surface. (M_moon = 7.35 × 10²² kg, R_moon = 1.74 × 10⁶ m)
★ A black hole has mass M = 10 × M_Sun (M_Sun = 2.0 × 10³⁰ kg) and radius R = 30 km. Find the escape velocity. What does your answer tell you about light escaping from a black hole?
ChallengeCommon Mistakes
Thinking higher orbits are faster
v = √(GM/r) — higher orbits (larger r) have LOWER orbital speeds. The ISS at 400 km is faster than a GPS satellite at 20,000 km
Confusing orbital velocity with escape velocity
Escape velocity = √2 × orbital velocity at the same radius. v_esc = √(2GM/R) while v_orb = √(GM/R)
Applying Kepler's 3rd law with inconsistent units
T² = (4π²/GM)r³ requires SI units (T in seconds, r in meters). Or use ratios: (T₁/T₂)² = (r₁/r₂)³
Thinking a satellite's mass affects its orbital speed or period
v = √(GM/r) and T = 2πr/v depend only on the central mass M and orbital radius r — not on the satellite's mass
Math Tips
Orbital speed: v = √(GM/r). Derived by setting gravitational force = centripetal force: GMm/r² = mv²/r → v² = GM/r.
Kepler's 3rd law ratio method: (T_A/T_B)² = (r_A/r_B)³. No need to know G or M — just compare two orbits around the same body.
Escape velocity v_esc = √(2GM/R) = √2 × v_orbital. It comes from setting KE = gravitational PE: ½mv² = GMm/R.
Geostationary orbit: set T = 86,400 s in T² = (4π²/GM)r³ and solve for r. The answer (~42,200 km) is the same for any satellite around Earth.