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Orbital Period Calculator

Enter the required parameters, and the calculator will instantly estimate the orbital period.

Satellite Around Central Sphere

Central Body Density:

kg/m³ ▾

Semi-Major Axis

Binary System:

m ▾

Second Body Mass:

kg ▾

First Body Mass:

kg ▾

Related

Use this orbital period calculator to estimate the time required for a satellite or binary system to complete one full orbit. The calculator uses parameters such as central body density, semi-major axis, and the masses of the orbiting bodies to predict the orbital period.

Orbital Period Definition

"The time required for an object or satellite to complete one full revolution around another larger object is called the orbital period."

This concept applies to celestial bodies such as planets orbiting stars, moons orbiting planets, or satellites orbiting Earth. The orbital period is governed by Kepler’s Laws of Planetary Motion.

Orbital Period Image

Types of Orbits

1. Low Earth Orbit (LEO):

This orbit lies close to Earth's surface. In this case, the orbital period can be calculated using the mean density of the central body.

2. Binary Star System:

In a binary system, two bodies of comparable masses orbit around their common center of mass.

Orbital Period Equations

Low Earth Orbit (LEO):

\(T = \sqrt{\dfrac{3\pi}{G \cdot \rho}}\)

\(T\) = Orbital period
\(G\) = Universal gravitational constant
\(\rho\) = Mean density of the central body

Binary Star System:

\(T_b = 2\pi \sqrt{\dfrac{a^3}{G (M_1 + M_2)}}\)

\(a\) = Semi-major axis
\(M_1, M_2\) = Masses of the two bodies
\(G\) = Universal gravitational constant

How to Calculate Orbital Period

For Low Earth Orbit:

Note the mean density of the central body.
Multiply 3 by π.
Multiply gravitational constant \(G\) by density \(ρ\).
Substitute values into \(T = \sqrt{\dfrac{3\pi}{G\rho}}\).

Example 1: Low Earth Orbit (LEO)

Given:

\(\rho = 6.51 \text{ g/cm³}\)

Step 1: Convert to kg/m³

\(\rho = 6.51 \times 1000 \times \left(\dfrac{1}{100}\right)^3 = 6510 \text{ kg/m³}\)

Step 2: Use gravitational constant

\(G = 6.67 \times 10^{-11} \text{ m³·kg⁻¹·s⁻²}\)

Step 3: Apply formula

\(T = \sqrt{\dfrac{3\pi}{(6.67 \times 10^{-11}) \cdot 6510}}\)

After calculation:

\(T ≈ 4662.08 \text{ seconds}\)

Convert to hours:

\(T = \dfrac{4662.08}{3600} ≈ 1.295 \text{ hours}\)

Example 2: Binary Star System

Given:

\(a = 5 \text{ m}\)
\(M_1 = 30 \text{ kg}\)
\(M_2 = 20 \text{ kg}\)

Step 1: Apply formula

\(T_b = 2\pi \sqrt{\dfrac{a^3}{G(M_1 + M_2)}}\)

Step 2: Substitute values

\(T_b = 2\pi \sqrt{\dfrac{5^3}{(6.67 \times 10^{-11})(50)}}\)

\(5^3 = 125\)

\(G(M_1 + M_2) = 3.335 \times 10^{-9}\)

\(\dfrac{125}{3.335 \times 10^{-9}} ≈ 3.746 \times 10^{10}\)

\(\sqrt{3.746 \times 10^{10}} ≈ 193490.7\)

Step 3: Multiply by \(2\pi\)

\(T_b ≈ 1,218,097 \text{ seconds}\)

Convert to hours:

\(T_b ≈ 338.3 \text{ hours}\)

References

Wikipedia: Orbital Period

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