Questions: Kepler's Laws of Planetary Motion

Question 1

Mars has a semi-major axis of about 1.52 AU. Using Kepler's third law (T² = a³, in AU and years), what is Mars's approximate orbital period?

• A. 1.23 years
• B. 1.52 years
• C. 1.87 years
• D. 3.51 years

Answer: (C)

T² = a³ = (1.52)³ ≈ 3.51, so T = √3.51 ≈ 1.87 years. A common error is to use T = a³ directly (giving 3.51) rather than taking the square root, or to confuse T² = a³ with T = a. Mars's actual orbital period is about 1.88 years, confirming the calculation.

Question 2

The Sun is located at the geometric center of Earth's elliptical orbit.

• T. True
• F. False

Answer: False

This is a very common misconception. According to Kepler's first law, the Sun occupies one of the two *foci* of the ellipse, not the geometric center. The center is the midpoint between the two foci. For Earth's nearly circular orbit the difference is small, but for highly elliptical orbits (like comets) the focus is dramatically offset from the center.

Question 3

What does Kepler's second law (equal areas in equal times) imply about a planet's speed at different points in its orbit?

Model answer: A planet moves faster when closer to the Sun (near perihelion) and slower when farther away (near aphelion). To sweep equal areas in equal time intervals, the radius vector must be shorter at perihelion and compensated by a greater speed.

If the planet-Sun distance r is small (perihelion), the thin, wide triangle swept in a given time must have a large base (velocity) to keep the area constant. At aphelion, the large r means the planet moves slowly to sweep the same area. This is a geometric expression of angular momentum conservation.