Orbital Period & Gravity Field Calculator: Kepler, g(r), v

You used altitude instead of semi-major axis. The formula T = 2π√(a³/μ) requires the distance from Earth's center, not from the surface. For ISS at 400 km altitude, add Earth's radius: a = 6371 + 400 = 6771 km = 6.771×10⁶ m. With μ = 3.986×10¹⁴ m³/s², you get T = 5553 s ≈ 92.6 minutes. Using 400 km directly gives T = 2π√((4×10⁵)³/(3.986×10¹⁴)) ≈ 31.7 s—off by a factor of 175. Always convert altitude to orbital radius first.

It is right! Kepler's third law depends only on semi-major axis, not eccentricity. A circular orbit at a = 10,000 km and an ellipse with periapsis 7000 km, apoapsis 13,000 km both have a = 10,000 km and identical periods. The ellipse moves faster near periapsis and slower near apoapsis, but these effects cancel exactly. This is why transfer orbits work: the ellipse connecting LEO to GEO has a semi-major axis of (6671 + 42164)/2 ≈ 24,418 km, and that alone determines its half-period of ~5.3 hours.

Both are correct for different quantities. The semi-major axis is a = 42,164 km (from Earth's center). The altitude is h = a − R_Earth = 42,164 − 6,378 = 35,786 km (above the equator). The formula T = 2π√(a³/μ) needs 42,164 km. If your professor asked for altitude and you wrote 42,164 km, that's the error. Always label your answer clearly: 'altitude h = 35,786 km' or 'orbital radius a = 42,164 km.' The distinction matters for mission planning—launch guidance needs altitude, orbit mechanics need radius.

Your method is wrong, not your values. G is only known to ~5 significant figures, and M_Earth has similar uncertainty. Multiplying them compounds errors. Use μ = GM directly: Earth's μ = 3.986004418×10¹⁴ m³/s² is known to 10 significant figures from satellite tracking. JPL publishes these values for mission-critical calculations. For homework, μ ≈ 3.986×10¹⁴ m³/s² is fine. For precision work, never compute G×M separately—always look up μ from authoritative sources like IAU or JPL HORIZONS.

Inclination affects where the orbit goes (ground track), not how long it takes. The period T = 2π√(a³/μ) is identical for equatorial (0°), polar (90°), and any inclined orbit with the same semi-major axis. However, inclination does matter for real ISS tracking because Earth's equatorial bulge (J2 effect) causes the orbit to precess differently at different inclinations. A 51.6° orbit precesses westward at about 5°/day, which slightly affects the apparent period relative to ground stations—but the inertial period remains ~92.6 minutes.

The period tells you how often, not when. For overhead passes, you need the full orbital elements (epoch, semi-major axis, eccentricity, inclination, RAAN, argument of periapsis, true anomaly) plus your ground station coordinates. From there, propagate the orbit forward in time and check when the satellite's ground track comes within your visibility cone. Software like GMAT, STK, or even N2YO.com does this automatically. The period is just one input—you also need the initial position (epoch) and orientation (the other five elements).

GPS actually uses a half-sidereal-day period (~11 hours 58 minutes), not exactly 12 hours. A sidereal day is 23h 56m 4s, so half is 43,082 seconds. This gives a = (μT²/(4π²))^(1/3) = 26,560 km from Earth's center, or altitude = 26,560 − 6,378 = 20,182 km. The ~20 km variations between sources come from using different Earth radius values (equatorial vs mean) and whether they account for J2 secular effects. For homework, 20,200 km is acceptable; for navigation, millimeter-level ephemerides matter.

They're identical! Substitute T = 2π√(a³/μ) into v = 2πa/T: v = 2πa/(2π√(a³/μ)) = a/√(a³/μ) = √(μa/a³) = √(μ/a). Both give the same velocity. The vis-viva equation generalizes this to elliptical orbits: v = √(μ(2/r − 1/a)), which reduces to √(μ/a) when r = a (circular orbit). Use whichever form has the variables you know—if you have period, use 2πa/T; if you have μ and radius, use √(μ/r).

Several factors: (1) The Moon's orbit is elliptical (e ≈ 0.055), so the semi-major axis you used might be the current distance, not the mean. Use a = 384,400 km for the semi-major axis. (2) Earth-Moon μ should be the sum: μ_system = μ_Earth + μ_Moon ≈ 4.035×10¹⁴ m³/s², not just Earth's μ—though this correction is only ~1.2%. (3) The 27.3-day 'sidereal month' is measured relative to stars; if your simulation uses synodic months (~29.5 days), that's a different quantity. With a = 384,400 km and μ = 4.035×10¹⁴ m³/s², you get T ≈ 27.28 days.

Atmospheric drag. Even at 400 km, there's enough residual atmosphere (~10⁻¹² kg/m³) to slow satellites by meters per second over weeks. This lowers the orbit, which increases drag further (exponential decay). ISS loses about 2 km/month in altitude and needs periodic reboosts. The Kepler/two-body formula T = 2π√(a³/μ) assumes vacuum—it's exact for the Moon but an approximation for LEO. Below ~300 km, decay is rapid (days to weeks); above ~1000 km, orbits last centuries. Drag depends on solar activity, which heats and expands the atmosphere unpredictably.