example of Kepler's second law with Earth's orbit

1 Problem Statement

Consider an example of Kepler’s Second Law, the Law of Equal Areas, which states: A line joining a planet to the Sun sweeps out equal areas in equal intervals of time. This law reflects a planet’s faster motion near the Sun (perihelion) and slower motion farther away (aphelion).

Earth orbits the Sun in an ellipse with semi-major axis

a = 149.6 × 106 km

and eccentricity

e = 0.0167.

The orbital period is

T = 365.25 days.

Calculate the area swept out by Earth’s radius vector in a 10-day interval when Earth is:

at perihelion (closest to the Sun);
at aphelion (farthest from the Sun).

Then verify Kepler’s Second Law by comparing the two areas. Assume the Sun is at one focus of the ellipse.

2 Solution

2.1 Step 1: Kepler’s Second Law

Kepler’s Second Law states that the areal velocity is constant:

dA- = constant. dt

Therefore, equal time intervals must correspond to equal swept areas, regardless of where the planet is in its orbit.

2.2 Step 2: Orbital Parameters

The focal distance is

6 6 c = ae = (149.6 × 10 )(0.0167 ) ≈ 2.498 × 10 km.

Hence,

rp = a − c ≈ 147.102 × 106 km,

and

r = a + c ≈ 152.098 × 106 km. a

The semi-minor axis is

√ -----2 b = a 1 − e .

Thus,

6∘ -----------2- 6 b = (149.6 × 10 ) 1 − (0.0167 ) ≈ 149.579 × 10 km.

2.3 Step 3: Total Orbital Area

The area enclosed by the ellipse is

Aorbit = πab.

Substituting the values above,

Aorbit ≈ π (149.6 × 106 )(149.579 × 106 ) ≈ 7.030 × 1016 km2.

The constant areal velocity is therefore

dA- Aorbit 7.030-×-1016 14 2 dt = T ≈ 365.25 ≈ 1.925 × 10 km ∕day.

2.4 Step 4: Area Swept in 10 Days

For any 10-day interval,

dA- ΔA = dt Δt.

Therefore,

ΔA ≈ (1.925 × 1014)(10) ≈ 1.925 × 1015 km2.

This value applies both near perihelion and near aphelion.

2.5 Step 5: Angular-Momentum Check

Areal velocity is related to angular momentum by

dA- 1- 2 ˙ -L-- dt = 2 r 𝜃 = 2m ,

which is constant for motion under a central force.

The orbital speed can also be estimated from the vis-viva equation,

∘ -------------- ( 2 1) v = GM --− -- . r a

Using

GM = 1.327 × 1011 km3 ∕s2,

one obtains approximately

vp ≈ 30.29 km ∕s

at perihelion and

v ≈ 29.29 km ∕s a

at aphelion.

The larger speed at perihelion compensates for the smaller orbital radius, while the smaller speed at aphelion compensates for the larger radius, so the areal velocity remains constant.

2.6 Step 6: Compare Areas

Perihelion: $ΔA ≈ 1.925 × 1015 km2.$
Aphelion: $ΔA ≈ 1.925 × 1015 km2.$

The areas are equal, as required by Kepler’s Second Law.

2.7 Step 7: Real-World Interpretation

Near perihelion Earth moves at roughly

30.3 km ∕s,

while near aphelion it moves at roughly

29.3 km ∕s.

Despite this change in orbital speed, equal time intervals correspond to equal swept areas.

3 Answer Summary

Perihelion area in 10 days: approximately 1.925 × 10¹⁵ km².
Aphelion area in 10 days: approximately 1.925 × 10¹⁵ km².
Verification: equal areas are swept in equal times, as Kepler’s Second Law requires.

This example was originally generated by Grok, an AI developed by xAI, on February 28, 2025.

References

[1] C. D. Murray and S. F. Dermott, Solar System Dynamics. Cambridge University Press (1999).

[2] P. K. Seidelmann (ed.), Explanatory Supplement to the Astronomical Almanac. University Science Books (1992).

[3] K. R. Lang, The Cambridge Guide to the Solar System, 2nd ed. Cambridge University Press (2011).

[4] NASA JPL, “Planetary Fact Sheet,” Solar System Dynamics (2025).

[5] J. Kepler, Harmonices Mundi (1619); English translation by E. J. Aiton et al., American Philosophical Society (1997).

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