Inputs
Results
| Quantity | Value |
|---|---|
| Altitude above surface | 35,786.000000 km |
| Earth radius | 6,378.137000 km |
| Orbital radius (r) | 42,164.137000 km |
| Semi-major axis (a) | 42,164.137000 km |
| GM μ | 398,600.441800 km³/s² |
| Specific energy | -4.726771 km²/s² (-4,726,771.021069 J/kg) |
| Sidereal day | 86,164.0905 s |
Steps and formula
- Total radius:
r = R_E + hwhereR_Eis Earth radius andhis altitude. - For a circular orbit, semi-major axis
a = r. - Kepler's third law (two-body):
T = 2π √(a³ / μ), whereμ = GMis the standard gravitational parameter. - Angular speed:
ω = 2π / T. Circular speed:v = √(μ / r). Specific orbital energy:ε = -μ / (2a). - Compare
Tto Earth's sidereal day to assess geosynchronous matching.
Assumes point-mass Earth and circular orbit. Oblateness perturbations station-keeping and longitude drift are not modeled.
FAQs
1) Why is 35786 km considered geosynchronous altitude?
At this height the orbital period closely matches the Earth's sidereal rotation making a spacecraft appear fixed over one longitude when inclination and eccentricity are near zero.
2) How close is the computed period to a sidereal day?
The table reports the difference in seconds and percent. Small variations arise from the chosen Earth radius and μ as well as rounding.
3) Which Earth radius should I use?
Use 6378.137 km for equatorial or 6371.0 km for mean spherical depending on convention. The resulting period changes slightly because the orbital radius shifts by the radius choice.
4) Can I model noncircular or inclined orbits?
This tool assumes a circular equatorial path. For inclined or eccentric orbits you would need more advanced dynamics including nodal precession and longitude drift.
5) How do I export results?
Use the Export JSON button. It returns inputs outputs and a permalink so you can log or share exact settings.