People's Education Press High School Physics Compulsory Course Part 2: From Circles to Ellipses – Kepler Unveils the Geometric Beauty of Planetary Motion

For thousands of years, humans have gazed at the stars, striving to find order within chaos. The ancient Greek philosopher Plato declared that celestial bodies must move uniformly along 'perfect circles.' To preserve this philosophical aesthetic,geocentric modelsupporters devised complexepicycles (Epicycle) anddeferents (Deferent) models (Diagram 7.1-5), attempting to explain why planets occasionally exhibitretrograde motion (Retrograde motion) phenomena (Diagram 7.1-4).

The Paradigm Shift from 'Circle' to 'Beauty'

When Copernicus proposedheliocentric model(Diagram 7.1-6), the center of the universe shifted, yet the belief in circular motion still constrained calculation accuracy. It was only through Kepler’s painstaking analysis of Tycho’s observational data that the myth of circular orbits was finally shattered. He concluded that planetary orbits areelliptical, with the Sun located at one focus of the ellipse.

Kepler's Third Law: The Rhythm of the Universe

Kepler not only redefined orbital shapes but also revealed a precise mathematical relationship between the orbital radius $r$ and period $T$ of all planets:$\frac{r^3}{T^2} = k$. In this formula, the proportionality constant $k$ is independent of the planet’s mass and depends solely on the mass of the central body (the Sun). This law weaves all members of the solar system into a single geometric framework.

Simplification in Physical Modeling

When discussing large-scale orbital problems, although planetary orbits are elliptical, for computational convenience we typicallysimplify them as uniform circular motion, where the radius $r$ corresponds to the semi-major axis of the ellipse.

QUESTION 1

Given Mars’ orbital radius is 1.5 AU (astronomical units), what is its orbital period in Earth days according to Kepler’s Third Law?

365 days

Approximately 670 days

Approximately 547 days

88 days

QUESTION 2

If an artificial Earth satellite moves along an elliptical orbit, at which point—closest to Earth (perigee) or farthest (apogee)—is its instantaneous velocity greater?

Perigee

Apogee

Velocities are equal at both points

Cannot be determined

QUESTION 3

According to Kepler’s Third Law, for all planets orbiting the same central body, $\frac{r^3}{T^2} = k$. Which of the following statements about the constant $k$ is correct?

$k$ depends on the planet's mass

$k$ depends on the mass of the central body

$k$ is a universal constant applicable to any celestial system

$k$ has units of m/s

QUESTION 4

Earth’s mass is $6.0 \times 10^{24}\text{ kg}$, and the Earth-Sun distance is $1.5 \times 10^{11}\text{ m}$. If Earth’s revolution is treated as uniform circular motion, approximately how many newtons is the gravitational force exerted by the Sun on Earth?

$3.5 \times 10^{22}\text{ N}$

$3.5 \times 10^{30}\text{ N}$

$6.0 \times 10^{24}\text{ N}$

$9.8 \times 10^{20}\text{ N}$

QUESTION 5

Some say the first cosmic velocity can also be calculated using $v = \sqrt{gR}$ ($g$ is surface gravity, $R$ is Earth’s radius). Do you agree?

Correct

Incorrect; the universal gravitation formula must be used

Deep Dive: From Mars’ Retrograde Motion to Galileo’s Moons

Validating the Universality of Kepler’s Laws Using Real Astronomical Observations

Astronomy is built upon precise observations. Historically, Mars’ retrograde motion relative to background stars puzzled observers, while the discovery of Jupiter’s moons provided strong analogies supporting the heliocentric model. Now, apply your knowledge to solve the following two practical problems.

1. Calculate the time interval between successive oppositions of Jupiter in years? (Jupiter’s orbital radius is 5.2 AU. Hint: Opposition occurs when Earth overtakes an outer planet. Use $1/T_{syn} = |1/T_{earth} - 1/T_{jupiter}|$.)

Answer:
Step 1: First compute Jupiter’s orbital period $T_J$. Using Kepler’s Third Law $\frac{5.2^3}{T_J^2} = \frac{1^3}{1^2}$, we get $T_J = \sqrt{5.2^3} \approx 11.86$ years. Step 2: Compute opposition frequency: $1/T_{syn} = 1 - 1/11.86 \approx 0.9157$. Step 3: Find the interval: $T_{syn} = 1 / 0.9157 \approx 1.09$ years. Thus, Jupiter experiences opposition roughly every 1.09 years.

2. Europa’s mass is $4.8 \times 10^{22} \text{ kg}$, its orbital radius is $6.7 \times 10^8 \text{ m}$; Jupiter’s mass is $1.9 \times 10^{27} \text{ kg}$. Calculate the orbital period of Europa around Jupiter.

Answer:
Use the gravitational force providing centripetal force: $G\frac{Mm}{r^2} = m\frac{4\pi^2}{T^2}r$, leading to $T = \sqrt{\frac{4\pi^2 r^3}{GM}}$. Substitute known values: $G = 6.67 \times 10^{-11}$, $M = 1.9 \times 10^{27}$ kg, $r = 6.7 \times 10^8$ m. Computed: $T \approx \sqrt{1.77 \times 10^{11}} \approx 3.06 \times 10^5 \text{ s}$, approximately 3.55 days.

3. [Extension: Spacetime Concepts] A train moves at speed $v$ relative to the ground (Figure 7.5-4). If a ground observer measures light flashes reaching both front and back walls simultaneously, which wall does someone on the train believe the flash reaches first?

Answer:
Someone on the train observes the flash arriving first at the front wall. Explanation: According to the principle of invariant light speed and relativity of simultaneity, since the carriage moves rightward, the ground observer sees the rear wall moving toward the light while the front wall moves away. If the ground observer records simultaneous arrival, it means the light traveled a shorter distance backward in space. In the train’s reference frame, light speed is constant and both walls are equally distant, so light naturally arrives first at the end with the shorter propagation path—the front wall.