Kepler Orbit Visualiser

Set semi-major axis and eccentricity to see the orbit shape, area sweep, and period.

Worked example — Kepler Orbit Visualiser

Speed at perihelion	45.5 km/s
Speed at aphelion	19.5 km/s

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Equal areas is angular momentum in disguise 🖖

Kepler found his three laws by studying Tycho Brahe's naked-eye-precision data for years — decades before anyone knew why planets should behave that way. The 'equal areas in equal time' law you can toggle on here isn't really about geometry: it's a direct consequence of angular momentum conservation, since gravity always pulls straight along the line between planet and star and can never twist the orbit sideways. That's also why a planet swings fastest through perihelion and crawls through aphelion — conserving L = r×v forces speed up as r shrinks. Newton's gravity, unknown to Kepler, is what finally explained why the pattern he'd already found geometrically had to be true.

What eccentricity really means 🖖

Eccentricity is the single number this slider controls - it measures how stretched an orbit is. At e = 0 you get a perfect circle; push it toward 1 and the ellipse elongates into a comet's path, with the star sitting off-center at one focus, never the middle. The surprise for most people: Earth's orbit has e = 0.017, so it is nearly a circle. Seasons come from Earth's axial tilt, not from a changing distance to the Sun.

Why the ellipse closes perfectly 🖖

Watch the planet retrace exactly the same ellipse forever - it never drifts into a slowly rotating rosette. That is not automatic. Bertrand's theorem proves only two force laws let bound orbits close after a single loop: inverse-square gravity (1/r²) and the harmonic spring force (∝ r). Any other power law leaves orbits that never quite repeat. The clean, closed ellipse you see is a fingerprint of gravity's exact 1/r² form.

Problems solved in full

An orbit with e = 0.4 around a solar-mass star 5 steps
An orbit has a = 1 AU and e = 0.4 around a solar-mass star. Find the speed at both ends, and then find the quantity that stays the same all the way round.
1. An ellipse with semi-major axis a and eccentricity e reaches closest at a(1−e) and furthest at a(1+e). Those two radii are all the geometry the problem needs.
  
  $r_{p} = a (1 - e) = 0.6 AU, r_{a} = a (1 + e) = 1.4 AU$
2. Vis-viva gives the speed at any radius from the radius and the semi-major axis alone. Nothing about mass of the orbiting body enters, and nothing about where it started.
  
  $v = μ (\frac{2}{r} - \frac{1}{a})$
3. Evaluate at both ends. The visualiser above prints exactly these two numbers for this orbit, and it does so by running this same equation rather than by simulating anything.
  
  $v_{p} = \frac{μ}{a} \cdot \frac{1 + e}{1 - e} = 45.5 km/s, v_{a} = \frac{μ}{a} \cdot \frac{1 - e}{1 + e} = 19.5 km/s$
4. Now take the ratio and watch the star’s mass cancel. The speed ratio depends on eccentricity and on nothing else — not on a, not on the star. A comet with e = 0.99 arrives 199 times faster than it leaves the far end.
  
  $\frac{v _{p}}{v _{a}} = \frac{1 + e}{1 - e} = \frac{1.4}{0.6} = 2.333$
5. That ratio is a conservation law in disguise. Multiply each speed by its radius and the two products are equal: angular momentum per unit mass is constant, so what the orbit gains in distance it must give up in speed.
  
  $r_{p} v_{p} = r_{a} v_{a} ⟹ 0.6 \times 45.5 = 1.4 \times 19.5 = 27.3$
Answer

45.5 km/s at perihelion, 19.5 km/s at aphelion, a ratio of exactly 2.333 = (1+e)/(1−e). Both radius-times-speed products come to 27.3, which is Kepler’s second law written as a number: the “equal areas in equal times” rule is just rv = constant. That is why comets spend almost all their time in the outer part of an orbit and cross the inner system in weeks — a fact about area, not about gravity switching strength.

$\frac{v _{p}}{v _{a}} = \frac{1 + e}{1 - e} (the constant is L = m r v)$
Fraction of the year a planet spends crawling through the far half 6 steps
The panel gives the orbit a period of 1.000 yr and puts perihelion at 0.6000 AU against 1.400 at aphelion. It animates the planet crawling through the far half and never says how long it spends there. Work that out — as a fraction of the year, from the eccentricity alone.
1. Start from what is printed. The two apses sit at a(1−e) and a(1+e), so this orbit reaches more than twice as far out as it comes in.
  
  $T = a^{3/2} = 1.000, r_{p} = a (1 - e) = 1.00 (1 - 0.400) = 0.6000, r_{a} = a (1 + e) = 1.400$
2. Kepler's second law is the whole tool here, and it is a statement about area, not distance: the line from the Sun sweeps equal areas in equal times. So time spent is area swept, and nothing else.
  
  $\frac{d A}{d t} = \frac{L}{2 m} = const ⟹ Δ t \propto A$
3. Cut the ellipse in half along the minor axis. Each half has the same area — but the Sun is not at the centre, it sits at a focus, offset by c = ae. Seen from there, the far half is the half-ellipse plus the triangle between the focus and that cut.
  
  $A_{f} = \frac{1}{2} π ab + \frac{1}{2} (2 b) (c), c = a e$
4. Divide by the whole area and almost everything cancels. Both a and b disappear, and what survives is a startlingly clean result: the fraction is one half plus the eccentricity over π.
  
  $\frac{A _{f}}{π ab} = \frac{1}{2} + \frac{b c}{π ab} = \frac{1}{2} + \frac{e}{π}$
5. Put this orbit's eccentricity in. Nearly 63% of the year is spent in the outer half — the planet is not merely slower out there, it is out there for two-thirds of its life.
  
  $e = 0.400 : \frac{1}{2} + \frac{0.400}{π} = 0.5000 + 0.1273 = 0.6273$
6. Now put Earth's in, which is only 0.0167.
  
  $e = 0.0167 : \frac{1}{2} + 0.00532 = 0.50532, 2 (0.00532) (365.25) = 3.88 d$
Answer

The far half takes ½ + e/π of the period: 62.73% for the orbit on screen, and 50.53% for Earth. That last figure is small but not nothing — it is 3.88 days more of the year spent on the far side of the minor axis than the near one, and the same eccentricity, cut on a different line, is why the calendar's seasons are of unequal length: Earth passes perihelion in early January, so northern winter is the fast half of the orbit and runs about five days shorter than summer. Nothing in the derivation used the period, the mass, or the size of the orbit. Only the shape.

$\frac{A _{f}}{π ab} = \frac{1}{2} + \frac{e}{π} : e = 0.400 \Rightarrow 62.73%, e = 0.0167 \Rightarrow 50.53%$

Learning path

Orbits from two numbers

Leads to Vis-viva

References (2)

Kepler working from Tycho Brahe’s observations, and the laws themselves: J. Kepler, Astronomia Nova, 1609 — translated by W. H. Donahue as New Astronomy, Cambridge University Press, 1992. ISBN 978-0-521-30131-2.
Equal areas as conservation of angular momentum: H. Goldstein, C. Poole and J. Safko, Classical Mechanics, 3rd ed., §3.2. Addison-Wesley, 2002. ISBN 978-0-201-65702-9.

By the 42Math team · Updated 2026-07-03

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

All Astronomy & Space lessons →

Example problems

Earth - Earth: nearly circular orbit, e=0.017
Mars - Mars: noticeable eccentricity, e=0.093
Mercury - Mercury: highest planet eccentricity, e=0.206
Halley's Comet - Halley Comet: highly eccentric, e=0.967