Circular & Geostationary Orbit Calculator
Calculate orbital radius, period, velocity, and geostationary synchronization for circular Earth orbits.
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Inputs
Live
Math
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Related
Enter parameters and click Calculate to view results
Formula & Theory
r = Rₑ + h, T = 2π√(r³/μ), v = √(μ/r)This formula is used to calculate antenna parameters for circular & geostationary orbit calculator.
Overview
Use this orbital mechanics calculator to evaluate the radius, period, and velocity of circular satellite orbits. At an altitude of ~35,786 km, the orbital period equals Earth’s sidereal day (23.934 hours), establishing a true geostationary orbit (GEO).
Input Guide
Enter Orbital Altitude exactly in the units shown by this circular & geostationary orbit. Check the operating band, unit prefix, and decimal position before calculating; these are the inputs used by the formula.
- Orbital Altitude — use km.
Output Guide
The results describe the calculated circular & geostationary orbit values for the inputs you entered. Check each value against the available space, selected components, feed system, and operating conditions before making a final design decision.
How This Calculator Works
The Circular & Geostationary Orbit uses r = Rₑ + h, T = 2π√(r³/μ), v = √(μ/r). Supply Orbital Altitude (km) in the displayed units, then use the calculated values as the first engineering target for this satellite communication design or analysis.
Design Notes
Geostationary orbits require an altitude of ~35,786 km directly over Earth’s equator with zero inclination. At this exact distance, the satellite matches Earth’s rotation speed, remaining fixed over a single spot on the ground.
Build and Tuning Notes
Changing the altitude away from 35,786 km changes the orbit class: altitudes below ~2,000 km represent Low Earth Orbit (LEO), 2,000–35,786 km represent Medium Earth Orbit (MEO), and altitudes above GEO represent High Earth Orbit (HEO).
Frequently Asked Questions
What is the exact altitude for a geostationary orbit?
A geostationary orbit requires a semi-major axis of approximately 42,164 km, which corresponds to an altitude of roughly 35,786 km (22,236 miles) above Earth’s equator.
Why is the geostationary period 23.934 hours instead of 24 hours?
Earth completes one 360° rotation relative to the stars in a sidereal day (23 hours, 56 minutes, 4 seconds = 23.934 hours). A 24-hour solar day includes extra rotation to account for Earth’s orbit around the Sun.
What happens if a satellite is at 35,786 km but inclined?
If the altitude is 35,786 km but the orbit is inclined (not equatorial), it is a geosynchronous orbit (GSO). The satellite will trace a figure-eight pattern (analemma) in the sky each day rather than remaining stationary.
Alex Warren
B.Sc. in Electrical & Electronic Engineering (EEE)
Alex specialises in antenna design and wave propagation. His expertise helps ensure these calculators present practical RF concepts, useful design estimates, and clear engineering guidance for students, HAM operators, and wireless professionals.