Question 1

Why do inner planets orbit faster than outer planets?

Accepted Answer

Kepler's Third Law states that the square of the orbital period is proportional to the cube of the semi-major axis: T² ∝ a³. This arises because planets in smaller orbits feel stronger gravitational pull from the Sun, requiring higher orbital speed to maintain a stable orbit. Mercury at 0.387 AU has period 88 days; Neptune at 30.07 AU has period 165 years — a 165-year difference for only an 80× distance difference.

Question 2

Why are planetary orbits elliptical rather than circular?

Accepted Answer

Newton's law of gravitation (F=GMm/r²) produces conic section orbits — ellipses, parabolas, or hyperbolas — for bound and unbound trajectories respectively. An ellipse results when the total mechanical energy E = KE + PE < 0 (bound orbit). The eccentricity e describes how elongated the ellipse is: e=0 is circular, e=1 is parabolic (escape). Most planetary orbits have low eccentricity (Earth e=0.017, Venus e=0.007) because the solar nebula condensed from a roughly rotating disk, giving nearly circular initial conditions. Mercury (e=0.206) and Mars (e=0.093) are the most elliptical of the inner planets.

Question 3

What are Lagrange points and where are they in the solar system?

Accepted Answer

Lagrange points are five equilibrium positions in a two-body gravitational system where a smaller third body experiences zero net force in the rotating reference frame. For the Sun-Earth system: L1 is between Sun and Earth (~1.5 million km from Earth, location of SOHO and DSCOVR); L2 is on the opposite side (~1.5 million km from Earth, location of the James Webb Space Telescope); L3 is behind the Sun; L4 and L5 are 60° ahead and behind Earth in its orbit (Trojan asteroids occupy similar points in Jupiter's orbit).

Question 4

How accurate is this simulation?

Accepted Answer

The simulation uses real Keplerian orbital elements (semi-major axes, eccentricities, inclinations, and periods from NASA/JPL data). Each planet's position is computed from its mean anomaly M(t) by solving Kepler's equation M = E − e·sin(E) via Newton-Raphson iteration, giving the true anomaly ν and hence 3D position. This approach accurately represents heliocentric positions for millennia. It does not include gravitational perturbations from other planets (which are small but cumulative), so exact positions will diverge from reality over centuries.

🧪 3D Solar System Simulator

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