Concept 06 / 08Gravitation
Kepler's Three Laws of Planetary Motion and Their Dynamical Origin
Planets move on ellipses with the Sun at a focus, sweep equal areas in equal times because angular momentum is conserved, and obey T squared proportional to a cubed, which follows directly from the inverse-square law.
Kepler extracted three empirical laws from Tycho Brahe's naked-eye observations of Mars between 1609 and 1619, decades before Newton. Newton then showed that all three are consequences of the inverse-square law, which is why they are the sharpest evidence for it.
Law 1, the law of orbits. Every planet moves in an ellipse with the Sun at one focus. The other focus is empty. In polar coordinates centred on the Sun, the orbit is , where is the eccentricity and is the semi-latus rectum. The nearest point (perihelion for the Sun, perigee for the Earth) is at and the farthest (aphelion, apogee) at . Adding these, : the semi-major axis is the arithmetic mean of the extreme distances.
A circle is the special case . Most planetary orbits are nearly circular: the Earth has , so its distance from the Sun varies by only about 3 percent over the year, which is why the seasons cannot be caused by it. Comets have close to 1.
Law 2, the law of areas. The line joining a planet to the Sun sweeps out equal areas in equal intervals of time. In the diagram above the two shaded sectors have equal area, so the planet covers the fat short arc near the Sun in the same time as the long thin arc far away; hence it moves fastest at perihelion.
This law is simply angular momentum conservation in disguise, and it holds for any central force whatsoever, not just an inverse-square one. The proof: the area swept in time by the position vector is the area of the thin triangle, . Therefore
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Since the gravitational force is central, its torque about the Sun, , is zero because is parallel to . So is constant, and the areal velocity is constant. The constancy of the direction of also tells us the orbit is planar.
An immediate corollary: at perihelion and aphelion the velocity is perpendicular to the radius, so , giving
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Law 3, the law of periods. The square of the period is proportional to the cube of the semi-major axis: , with the same constant of proportionality for every body orbiting the same central mass.
Derivation for a circular orbit (this is the derivation JEE expects). For a planet of mass in a circular orbit of radius about a mass , gravity supplies the centripetal force:
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Cancel and rearrange:
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The full result for an ellipse replaces by the semi-major axis : . The constant depends only on the central mass, which is why every planet, asteroid and comet in the Solar System lies on one straight line in a plot of against , and why the Moon and every artificial satellite lie on a different line set by the Earth's mass.
Strictly the two-body result is , where is now the semi-major axis of the relative orbit. For planets so the correction is tiny (for Jupiter it is one part in a thousand), but for binary stars, where the masses are comparable, it is the whole point: measuring and gives the total mass of the system.
Two energy results tie the laws together and are worth memorising, because they turn many hard problems into one-liners.
The total energy of an elliptical orbit depends only on the semi-major axis: . Circular orbits are the case .
The vis-viva equation, obtained by combining that with :
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This single formula gives the speed anywhere on any orbit. At it gives , the circular speed; for (parabolic escape) it gives , the escape speed.
Gotchas. The Sun sits at a focus, not at the centre. The quantity in Law 3 is the semi-major axis, not the average of the distances over time (the time-average of is actually ). Law 2 is about areal speed, which is constant; the linear speed and angular speed are certainly not constant. Kepler's laws apply to any inverse-square two-body system, including satellites around the Earth and electrons in the Bohr model. And the laws fail slightly when other planets perturb the orbit or when general relativity matters, which is how the precession of Mercury's perihelion was discovered.
Key Takeaways:
- Law 1: ellipse, Sun at a focus; , , .
- Law 2: is constant, because a central force exerts no torque; valid for any central force. Corollary .
- Law 3: ; two-body form .
- Energy: (depends only on ). Vis-viva: .
- Angular momentum: , depends on both and ; energy depends only on .
- Common mistakes: putting the Sun at the centre; using instead of ; assuming angular velocity is constant because areal velocity is; forgetting that a body on a given orbit has the same period regardless of its own mass.
Brain-Teaser 1 (a comet's aphelion): A comet passes perihelion at a distance from the Sun with speed . Find its aphelion distance, the ratio of its speeds at the two ends, and its period in terms of .
Solution: First get from the energy, which needs no angular momentum bookkeeping.
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Since , we get , so .
Then .
Speeds: angular momentum conservation gives .
Verify independently with vis-viva at aphelion: . Since , the ratio , so . The two methods agree.
Period: , i.e. about times the period of a circular orbit of radius .
Eccentricity, for completeness: gives .
Brain-Teaser 2 (how long to fall into the Sun): If the Earth were suddenly stopped in its orbit, how long would it take to fall into the Sun? Answer without doing any integral.
Solution: A body dropped radially is not a special case at all, it is a degenerate ellipse of eccentricity , squashed flat. Its aphelion is the starting distance , and its perihelion is essentially zero (the centre of the Sun). So
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Kepler's third law compares this degenerate orbit with the Earth's actual circular orbit of radius and period year:
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The fall to the centre is half of that degenerate orbit (down and back would be the full "orbit"), so the falling time is
days.
A three-line answer to a problem that looks like it needs a nasty integral. The same trick answers "how long does a satellite take to fall to the Earth's centre if stopped", and it is a perennial favourite of JEE Advanced.
Brain-Teaser 3 (the speed at the end of the minor axis): A planet moves on an ellipse of semi-major axis and eccentricity . Show that its speed at either end of the minor axis equals the geometric mean of its perihelion and aphelion speeds, and that it equals the speed of a circular orbit of radius .
Solution: From vis-viva, at perihelion :
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Similarly at aphelion : .
Multiplying, , and since both speeds are positive, , so the geometric mean is .
Now the end of the minor axis. The distance from a focus to that point is exactly , since . Vis-viva there gives
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So , which is both the geometric mean of the extreme speeds and the speed on a circular orbit of radius .
The conceptual payoff: the total energy of the elliptical orbit depends only on , so at the one place where happens to equal the kinetic energy must match that of the circular orbit of radius . Energy knows only about ; the eccentricity is stored entirely in the angular momentum, .
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