Earth Orbit Calculator — Orbital Speed & Period of Satellites | v = √(GM/r) | ISS, GPS, GEO Presets

What Are Earth's Satellites?

A satellite is a small object that orbits another larger object. The Earth is a satellite because it orbits the Sun. We consider the Moon a satellite because it orbits the Earth. However, most people think of an artificial satellite when they say "satellite." These spacecraft are launched into an orbit around the Earth or any other celestial body.

Sputnik 1 was Earth's first artificial satellite. It was launched into an elliptical low Earth orbit at the height of 939 km. It orbited for three weeks before its batteries ran out.

There are tens of thousands of artificial satellites orbiting the Earth. Some photograph our globe, while others photograph other planets, the Sun, and other celestial bodies. These images aid scientists in studying the Earth, the solar system, and the universe. Other satellites broadcast television and make phone calls all around the world.

How to Calculate the Orbital Speed of a Satellite

According to Nicolaus Copernicus, Earth and the other planets orbit the Sun in circles. He also discovered that as the distance from the Sun increased, so did the orbital periods. Kepler later found these orbits to be ellipses, but the orbits of most planets in the solar system are roughly circular.

Determining a satellite's orbital speed and period is significantly easier in circular orbits. The orbital speed (v) of an Earth satellite at height h above the surface is given by:

v = √(G · M_E / (R_E + h))

• G — gravitational constant = 6.674 × 10⁻¹¹ N·m²/kg²
• M_E — Earth's mass = 5.972 × 10²⁴ kg
• R_E — Earth's radius = 6,371 km
• h — height above Earth's sea level

How Can You Estimate the Period of Rotation for an Earth's Satellite?

The orbital period is the time a satellite takes to complete one full revolution around the Earth. For a circular orbit it is derived directly from the orbital speed:

T = 2π · (R_E + h) / v

Combining both formulas gives Kepler's Third Law for circular orbits:

T = 2π · √((R_E + h)³ / (G · M_E))

Example: How to Use This Earth Orbit Calculator

Suppose you want to find the orbital parameters for the International Space Station (ISS), which orbits at approximately 408 km above Earth's surface.

1. Select Metric as the unit system (or click the ISS preset button).
2. Enter 408 km as the orbital height.
3. Click Calculate.

Results:

• Orbital Speed ≈ 7.67 km/s (≈ 4.77 mi/s)
• Orbital Period ≈ 92.7 minutes (≈ 1.545 hours)
• Orbital Radius ≈ 6,779 km from Earth's center

The ISS completes about 15.5 orbits per day — which is why astronauts aboard it see roughly 16 sunrises and sunsets every 24 hours!

Well-Known Earth Orbits

FAQs

Why does a satellite at a lower orbit move faster?

Because gravitational pull is stronger closer to Earth. To maintain a stable circular orbit, the centripetal acceleration must equal gravitational acceleration. At lower altitudes, gravity is stronger, so the satellite must travel faster to avoid falling toward Earth.

What is geostationary orbit?

A geostationary orbit (GEO) is a special circular orbit at ~35,786 km altitude, where the satellite's orbital period equals exactly 24 hours. Satellites in GEO appear to hover over the same point on the equator — making them ideal for communications and weather monitoring.

What is the difference between metric and imperial (American) unit systems?

In the metric system, height is measured in kilometers (km) and speed in km/s. In the American (imperial) system, height is measured in miles (mi) and speed in miles per second (mi/s). This calculator automatically converts between both systems and shows both values in the results.

Can I calculate the orbit of the Moon with this calculator?

Yes! The Moon orbits Earth at approximately 384,400 km. Simply click the Moon preset button or enter that value manually. The calculator uses the same physics formulas that apply to any Earth satellite, including the Moon.

What formula is used for the orbital period?

This calculator uses T = 2π√((R_E + h)³ / (G · M_E)), which is derived from Newton's law of universal gravitation and is equivalent to Kepler's Third Law applied to circular orbits around Earth.