Gravitation · Lesson 1 of 7
Planetary Motion and Kepler’s Laws
“Connect observations of planetary motion with elliptical orbits, equal-area motion and the period law.”
• State the three Kepler laws with the correct geometric quantities. • Interpret an ellipse, its foci and its semi-major axis. • Derive the area law from angular-momentum conservation. • Apply period ratios and perihelion–aphelion relations.
From falling objects to planetary motion
A dropped stone and the Moon’s motion seem different: the stone falls to the ground, while the Moon keeps going around Earth. This chapter connects terrestrial falling with celestial motion through one gravitational interaction. Galileo’s investigations of falling and inclined motion helped separate gravitational acceleration from the effects of resistance. Astronomical observations supplied another part of the story. Models progressed from Earth-centred descriptions to Sun-centred descriptions, and careful measurements became the test of those models.
The PDF discusses Ptolemy, Indian astronomical traditions, Aryabhata, Copernicus, Galileo, Tycho Brahe and Kepler as historical context. For the physical argument, the key step is that Brahe’s recorded planetary positions allowed Kepler to identify three empirical regularities. Newton later explained those regularities with a universal force law. Kepler’s laws describe the motion; the gravitational law explains why it occurs.
Elliptical orbits and the meaning of a focus
An ellipse is the curve for which the sum of the distances from two fixed points is constant. Those points are its foci. To reproduce the PDF’s construction, attach the ends of a fixed-length string to two pins, place a pencil against the string and keep the string taut while moving the pencil. The resulting closed curve is an ellipse. Its longest diameter is the major axis, of length 2a, and half of that is the semi-major axis a. The centre is the midpoint of the major axis. A focus is generally not the centre.
Perihelion is the nearest point to the Sun; aphelion is the farthest. The two distances are measured from the occupied focus, so rP + rA = 2a. They are not generally equal to a. An orbital size in Kepler’s period law means a, not a planet’s changing instantaneous distance from the Sun.
Equal areas in equal times
Kepler’s second law says that the Sun-to-planet line sweeps equal areas in equal time intervals. A planet must move faster when close to the Sun because a short radius needs a larger angular sweep to cover the same area. Equal times do not imply equal arc lengths or equal angles. Nor does travelling through half an ellipse necessarily take half a period when the dividing line passes through the centre rather than the occupied focus.
To derive the relation, approximate the small swept region by a triangle with sides r and v dt. Its area is half the cross-product magnitude. Use p = mv to replace r×v by L/m. Gravity points along the Sun–planet joining line, so its torque about the Sun is r×F = 0. Angular momentum is therefore constant, and the area rate is constant. This explanation works for any central force, not only an inverse-square force.
Problem
In source Example 7.1, relate the speeds vP and vA at perihelion and aphelion. Compare the times along the far and near halves BAC and CPB of the source diagram.
- 1.At either end of the major axis, the velocity is perpendicular to the radius from the Sun. Thus LP = mrP vP and LA = mrA vA.
- 2.Conservation gives rP vP = rA vA, so vP/vA = rA/rP. The nearer point has the larger speed.
- 3.The Sun lies on the P side of the centre. The sector SBAC swept around the farther side has larger area than SBPC around the near side.
- 4.Constant area rate therefore makes the far-side traversal BAC take longer than CPB, even though the two arcs are the two centre-divided halves of the ellipse.
The period law and the planetary data
| Planet | a / (10¹⁰ m) | T / year | T²/a³ in 10⁻³⁴ year² m⁻³ |
|---|---|---|---|
| Mercury | 5.79 | 0.240 | 2.95 |
| Venus | 10.8 | 0.615 | 3.00 |
| Earth | 15.0 | 1.00 | 2.96 |
| Mars | 22.8 | 1.88 | 2.98 |
| Jupiter | 77.8 | 11.9 | 3.01 |
| Saturn | 143 | 29.5 | 2.98 |
| Uranus | 287 | 84 | 2.98 |
| Neptune | 450 | 165 | 2.99 |
The approximately common last column supports the law. The numbers are the rounded data in the supplied table, so do not expect identical quotients to every digit. For example, Earth gives 1/(15×10¹⁰)³ ≈ 2.96×10⁻³⁴ year²/m³. Ratios avoid carrying that dimensional constant when comparing two planets around the Sun.
Problem
Use source Exercises 7.3 and 7.14: a hypothetical planet has half Earth’s period; Saturn has a period 29.5 Earth years. Find their orbital-size ratios.
- 1.For the hypothetical planet, a/aE = (T/TE)^(2/3) = (1/2)^(2/3) ≈ 0.630. “Twice as fast” here refers to completing revolutions at twice the rate.
- 2.For Saturn, aS/aE = 29.5^(2/3) ≈ 9.55.
- 3.With aE = 1.50×10⁸ km, aS ≈ 1.43×10⁹ km = 1.43×10¹² m. These are semi-major axes, or radii under the circular approximation.
Check your understanding
Quiz
The Sun is located at which point of a planetary ellipse?
Kepler’s first law places the Sun at one focus.
Equal swept areas correspond to equal
The second law concerns area per unit time.
The area law follows for
A central force has zero torque about the force centre.
A planet with half Earth’s orbital period has a/aE approximately
Use a proportional to T^(2/3).
At perihelion compared with aphelion, speed is
Conservation of mr v at the endpoints gives vP/vA=rA/rP>1.
Key Takeaways
• An ellipse’s focus, centre and semi-major axis have different meanings. • Central-force angular-momentum conservation gives a constant swept-area rate. • For one dominant central mass, T²/a³ is common to its orbiting bodies.
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Universal Gravitation and the Measurement of G