Interactive physics simulator
Kepler's Second Law
Explore the Law of Equal Areas: a planet sweeps out equal areas in equal time intervals. Adjust eccentricity, capture sweeps, and verify angular momentum conservation.
Kepler's Second Law Simulator
Adjust eccentricity and orbit size. Sweep equal-time sectors and verify their areas, or observe angular momentum conservation vectors.
Live Telemetry
- Planet Speed (v)
- --
- Distance to Sun (r)
- --
- Eccentricity (e)
- 0.50
- Swept Area (A)
- --
- Angular Momentum (L/m)
- --
- Semi-Major Axis (a)
- 4.00 AU
- Captured Wedges
- 0 / 8
- Orbital Period (T)
- 8.00 years
Kepler's Law of Equal Areas
Kepler\'s Second Law, also known as the Law of Equal Areas, states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. In simple terms, this means that a planet does not travel at a constant speed along its orbit. It moves faster when it is close to the Sun and slower when it is far away, yet the area swept by the radius line per unit time (dA/dt) remains perfectly constant.
Wedge Geometry
The shapes of the equal-area sectors change depending on the planet\'s position:
- Perihelion (Closest): The distance r is smallest, but the orbital velocity v is highest. The planet travels a long distance along the orbit, forming a short, wide sector.
- Aphelion (Farthest): The distance r is largest, but the orbital velocity v is lowest. The planet travels a short distance along the orbit, forming a long, narrow sector.
- Constant Sweeping Rate: The area rate is dA/dt = ½ r² dθ/dt = constant.
Conservation of Angular Momentum
Because gravity acts along the line connecting the planet and the Sun, there is no external torque. As a result, the angular momentum L = m(r × v) is conserved. At perihelion and aphelion, the velocity is perpendicular to the radius, simplifying to rp · vp = ra · va.
Speeds of Solar System Planets
| Celestial Body | Eccentricity (e) | Perihelion Speed (vp) | Aphelion Speed (va) | Speed Variance |
|---|---|---|---|---|
| Venus | 0.007 | 35.26 km/s | 34.78 km/s | Minimal (1.4%) |
| Earth | 0.017 | 30.29 km/s | 29.29 km/s | Slight (3.4%) |
| Mars | 0.093 | 26.50 km/s | 21.97 km/s | Noticeable (20.6%) |
| Mercury | 0.206 | 58.98 km/s | 38.86 km/s | Significant (51.8%) |
| Halley\'s Comet | 0.967 | 54.60 km/s | 0.91 km/s | Extreme (6,000%) |
Solved Examples
A planet orbits a star with an eccentricity of e = 0.50. At perihelion, it is at a distance of 2.00 AU from the star and travels at a speed of 30.0 km/s. If its distance at aphelion is 6.00 AU, calculate the planet's orbital speed at aphelion.
- Recall Kepler’s Second Law in terms of angular momentum conservation: at perihelion and aphelion, velocity is purely perpendicular to the radius vector, meaning rpvp = rava.
- Identify the given values: perihelion distance rp = 2.00 AU, perihelion speed vp = 30.0 km/s, and aphelion distance ra = 6.00 AU.
- Rearrange the equation to solve for aphelion speed: va = (rp × vp) / ra.
- Substitute the values: va = (2.00 × 30.0) / 6.00 = 60.0 / 6.00 = 10.0 km/s.
- This shows the planet travels exactly three times slower at aphelion compared to perihelion.
Answer: Aphelion speed va = 10.0 km/s
Earth orbits the Sun with a semi-major axis of a = 1.00 AU and a semi-minor axis of b ≈ 1.00 AU. The total area of its orbit is π·a·b ≈ 3.14 AU². Calculate how much area Earth's position vector sweeps out in a 30-day period (assume a 365-day year).
- According to Kepler's Second Law, the rate of area swept (dA/dt) is constant. Therefore, the area swept is directly proportional to time: A = Atotal × (Δt / T).
- State the parameters: Total orbit area Atotal ≈ 3.1416 AU², time interval Δt = 30 days, and orbital period T = 365 days.
- Substitute values into the proportion: A = 3.1416 × (30 / 365).
- Perform calculation: A ≈ 3.1416 × 0.08219 ≈ 0.258 AU².
- This area remains exactly the same for any 30-day period throughout the year.
Answer: Area swept ≈ 0.258 AU²
A research satellite in an elliptical orbit has a perigee distance of 8,000 km from the center of Earth and a speed of 10.00 km/s. Calculate its specific angular momentum (h = L/m). What is its perpendicular speed component v⊥ at a point in the orbit where it is 16,000 km from Earth's center?
- Specific angular momentum is defined as h = L/m = r × v⊥. At perigee, the velocity is entirely perpendicular to the radius vector, so v⊥ = v.
- Calculate specific angular momentum: h = rp × vp = 8,000 km × 10.00 km/s = 80,000 km²/s.
- Since angular momentum is conserved, h remains constant at all points in the orbit. Thus, at distance r = 16,000 km, we have h = r × v⊥.
- Solve for perpendicular velocity: v⊥ = h / r = 80,000 km²/s / 16,000 km = 5.00 km/s.
- This shows that doubling the distance halves the perpendicular velocity component.
Answer: Specific angular momentum h = 80,000 km²/s, Perpendicular speed v⊥ = 5.00 km/s
Common Misconceptions
- Thinking the planet moves at a constant speed along its elliptical path (it changes continuously).
- Believing that the area swept varies depending on the distance from the Sun (the area swept per unit time is strictly constant).
- Confusing angular velocity (ω) with areal velocity (dA/dt). Areal velocity is constant, while angular velocity fluctuates, peaking at perihelion.
- Assuming gravity changes the total angular momentum (since gravity acts along the radius vector, it creates no torque, conserving angular momentum).
Areal Velocity Equation
The rate of sweeping area is constant and related to the orbit dimensions and orbital period:
This shows that the area swept in a time interval Δt is exactly A = (π · a · b / T) · Δt, proving that any two wedges representing the same elapsed time have the same area.
Practice Questions
1. Kepler's Second Law is a direct mathematical consequence of which conservation law?
It is a direct consequence of the conservation of angular momentum. Since gravity is a central force directed towards the center of mass (the Sun), it exerts zero torque on the planet, meaning the angular momentum remains constant.
2. How does the shape of a swept sector change as the planet moves from perihelion to aphelion?
Near perihelion, the planet is close to the Sun (radius r is small), but it moves quickly, so the swept arc is long, resulting in a short and wide sector. Near aphelion, the planet is far away (radius r is large) but moves slowly, so the swept arc is short, resulting in a long and narrow sector. Both sectors contain identical area.
3. In a perfectly circular orbit, what happens to the rate at which area is swept?
In a perfectly circular orbit, the distance r and the speed v are constant. The planet sweeps out area at the exact same angular rate. All equal-time sectors are identical wedges of the circle.
4. What is the relationship between the orbital speeds at perihelion (vp) and aphelion (va), and their respective distances (rp, ra)?
Because angular momentum is conserved, the cross product of radius and velocity is constant. At these vertex points, the angle is 90 degrees, yielding the relationship rp × vp = ra × va. Speed and distance are inversely proportional.
FAQ
Frequently Asked Questions
What is Kepler's Second Law?
Kepler's Second Law (the Law of Equal Areas) states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time as the planet orbits.
Why does a planet speed up when it is closer to the Sun?
As a planet gets closer to the Sun (perihelion), the distance r decreases. To sweep out an equal area in the same amount of time, the planet must travel a longer arc, which requires it to move faster. This is a direct consequence of the conservation of angular momentum.
What is perihelion and aphelion?
Perihelion is the closest point in an orbit to the Sun, where the planet moves at its maximum speed. Aphelion is the farthest point in an orbit from the Sun, where the planet moves at its minimum speed.
How does Kepler's Second Law relate to conservation of angular momentum?
Kepler's Second Law is mathematically equivalent to the conservation of angular momentum (L = r × m·v = constant). Because gravity is a central force acting directly along the line connecting the bodies, it exerts no torque, so angular momentum is conserved.
Is the rate of area swept constant for all orbits?
Yes, for any specific orbit, the rate of area swept (dA/dt) is constant (dA/dt = L / 2m). However, different orbits with different semi-major axes, masses, or energies will have different constant sweep rates.
How does the Kepler's Second Law simulator demonstrate equal areas?
The simulator divides the orbital period into equal time segments (Δt). It shades the sectors (wedges) swept by the position vector during these segments. The area of each shaded sector is calculated and shown to be mathematically equal, whether the sector is near the Sun (short and wide) or far away (long and narrow).
What is specific angular momentum?
Specific angular momentum is the angular momentum per unit mass, calculated as h = L/m = r × v⊥ = r × v × sin(φ), where φ is the angle between the position and velocity vectors. The simulator calculates and displays this value live to show it remains constant throughout the orbit.
What happens to the speed of a planet in a perfectly circular orbit?
In a perfectly circular orbit (eccentricity e = 0$), the distance to the Sun is constant (r = a). Therefore, the speed of the planet is also constant, and all equal-time swept sectors are identical wedges of the circle.
Who formulated Kepler's Second Law?
Johannes Kepler formulated this law in 1602 (published in 1609 in 'Astronomia nova') along with his First Law, using Tycho Brahe's observational data of Mars.
Do satellites orbiting Earth also obey Kepler's Second Law?
Yes, Kepler's laws apply to any system of two bodies orbiting each other due to gravity, including natural and artificial satellites orbiting Earth, moons orbiting planets, and stars orbiting around each other in binary systems.
Can a planet ever escape an orbit according to Kepler's Second Law?
Kepler's Second Law assumes a bound closed orbit (an ellipse). If a body exceeds escape velocity, it moves in an open parabolic or hyperbolic trajectory. Although angular momentum is still conserved and equal areas are swept in equal times, the path is no longer a repeating loop.
Why are the swept sectors shaped differently near perihelion and aphelion?
Near perihelion, the planet is close to the Sun, so the radius r is small, but it moves fast, making the swept arc along the orbit long. This results in a short, wide wedge. Near aphelion, the planet is far from the Sun, so r is large, but it moves slowly, making the swept arc short. This results in a long, narrow wedge. Both wedges contain the exact same total area.