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Orbital Period Calculator

Orbital period equals two pi times radius divided by orbital velocity

Solution

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How It Works

Orbital period is the time it takes one full revolution around a circular path: T = 2πr / v. Geometrically, you're traveling the circumference (2πr) at constant orbital speed v, so the period is just distance over speed. Enter any two of period, orbital radius, or orbital velocity and the calculator rearranges algebraically to solve for the third, converting units automatically.

Example Problem

A satellite orbits 7,000 km from the center of Earth at an orbital speed of 7,500 m/s. What is the orbital period?

  1. Identify the formula: T = 2πr / v.
  2. Convert the radius to meters: 7,000 km = 7,000,000 m.
  3. Multiply 2π by r: 2π × 7,000,000 ≈ 43,982,297 m.
  4. Divide by orbital speed: 43,982,297 / 7,500 ≈ 5,864 s.
  5. Convert to minutes: 5,864 / 60 ≈ 97.7 minutes per orbit.

This is close to the actual period of the International Space Station (≈ 92 min at 6,778 km mean radius). For higher orbits the period grows because the orbital speed drops.

Key Concepts

The relation T = 2πr / v is purely kinematic — it follows from circular geometry, not from gravity. For gravitationally bound circular orbits there is an additional dynamic constraint: gravity must supply the centripetal force, giving v = √(GM/r). Substituting into the kinematic period yields Kepler's third law T² = 4π²r³ / (GM), the relation Newton used to confirm universal gravitation. This calculator uses only the kinematic form — it works for any circular trajectory regardless of what supplies the centripetal force.

Applications

  • Aerospace: estimating satellite revisit times from orbital altitude and speed.
  • Astronomy: computing planetary years from orbital radius and orbital speed.
  • Engineering: timing centrifuge cycles by pairing rotor radius with rim speed.
  • Sports broadcasting: estimating velodrome lap times from track length and rider speed.
  • Earth observation: planning ground-track repeat patterns for sun-synchronous satellites.

Common Mistakes

  • Using radius from the Earth's surface instead of from Earth's center. Orbital radius is measured from the center of the parent body.
  • Mixing distance units. If radius is in km and velocity is in m/s, you'll be off by a factor of 1,000. The calculator handles unit conversion automatically when both selectors are set.
  • Using diameter instead of radius. The formula uses r; from diameter D, r = D / 2.
  • Confusing orbital period with rotational period (a day vs. a year). Period here means one full revolution around the orbit, not one rotation about the body's own axis.

Frequently Asked Questions

How do you calculate orbital period?

Use T = 2πr / v, where r is the orbital radius (distance from the center of the parent body) and v is the orbital speed. Multiply 2π by r to get the circumference of the orbit, then divide by speed.

What is the formula for orbital period?

T = 2πr / v. For a gravitationally bound orbit you can also use Kepler's third law T = 2π√(r³ / GM) — both give the same answer when the orbital speed satisfies v = √(GM / r).

How long is one orbit of the International Space Station?

The ISS orbits at about 6,778 km mean radius (≈ 400 km altitude above Earth's surface) at 7,660 m/s. T = 2π(6,778,000) / 7,660 ≈ 5,560 s ≈ 92.6 minutes — about 15.5 orbits per day.

Does orbital period depend on mass?

For the kinematic form T = 2πr / v, no — only on radius and speed. But the orbital speed itself depends on the central body's mass (v = √(GM/r) for a circular gravitational orbit), so indirectly orbital period varies with the gravity of the parent body.

Why is orbital period longer at higher altitude?

Two compounding effects: the orbital circumference (2πr) is larger, and the required orbital speed v = √(GM/r) is smaller. Both push the period up — that's why geostationary satellites at 42,164 km radius take 24 hours per orbit while ISS at 6,778 km takes only 92 minutes.

Reference: Tipler, Paul A. 1995. Physics For Scientists and Engineers. Worth Publishers. 3rd ed.

Worked Examples

Aerospace

How long is one orbit for a satellite at 7,000 km radius moving 7,500 m/s?

  • T = 2πr / v
  • T = 2π(7,000,000) / 7,500
  • T = 43,982,297 / 7,500
  • T ≈ 5,864 s (≈ 97.7 minutes)

Comparable to the ISS (≈ 92 min at 400 km altitude). Higher orbits take longer.

Geostationary

What orbital speed gives a 24-hour period at geostationary radius (42,164 km)?

  • v = 2πr / T
  • v = 2π(42,164,000) / 86,400
  • v = 264,924,376 / 86,400
  • v ≈ 3,066 m/s (≈ 3.07 km/s)

Geostationary satellites match Earth's rotation rate, so they appear stationary over a fixed ground point.

Astronomy

What orbital radius gives a one-Earth-year period (≈ 365.25 days) at 29,780 m/s?

  • r = vT / (2π)
  • r = (29,780)(31,557,600) / (2π)
  • r = 9.398 × 10¹¹ / 6.283
  • r ≈ 1.496 × 10¹¹ m (≈ 1 AU)

Recovers the astronomical unit — Earth's mean orbital radius around the Sun.

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