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Orbital Period Calculator + Online Solver With Free Steps

The Orbital Period Calculator is a free online tool that calculates how long it takes an entity to complete a revolution.

The orbital period is obtained in a shorter time by just taking the central object density, semi-major axis, 1st body weight, & 2nd body weight.

We will also examine geostationary orbit, low Earth orbit, and geosynchronous orbits, as well as Johannes Kepler and his contributions to determining planet orbits in our planetary system.

What Is an Orbital Period Calculator?

The Orbital Period Calculator is an online tool that calculates the route that a body takes as it moves around another object. As an explanation, consider the annual trajectory that our dear planet takes as it orbits the Sun.

However, not all planets need to orbit the Sun once every 365 days, or one year. If we consider an orbit other than that of the Sun, such as that of the Moon, things become considerably more complex.

The definition of the orbital period must be given at this point, along with an explanation of what it includes.

Fortunately for us, the solution is pretty straightforward: the orbital period is the amount of time required to complete one full rotation of the primary object, or, to put it another way, the time required to complete one orbit. The sidereal era is another name for it.

How To Use an Orbital Period Calculator

You can use the Orbital Period Calculator by following the given detailed guide. You are only required to input the data properly and the calculator will automatically solve it for you.

Following are the steps that must be followed accordingly to get the path or orbit that a body follows in its movement.

Step 1

Enter the semi-major axis and the mass of the body you are orbiting in the appropriate input boxes.

Step 2

The whole step-by-step answer for the orbital period will be provided once you click the “SUBMIT” button to calculate the orbit that a body follows.

How Does an Orbital Period Calculator Work?

The Orbital Period Calculator works by using two different techniques, the first of which is titled satellite around the central body and the second of which is appropriately titled binary system.

In this first section, we’ll concentrate on using the calculator’s top portion to determine the orbital periods of tiny objects in low orbit around Earth.

It will be simple because there are just two different fields to complete in this part. As we previously stated, all you have to know to determine the orbital period of the little satellite revolving around the main body is its density.

This approximation is based on the following fairly straightforward equation:

\[ T = \sqrt{3 \dot \pi / (G \dot \rho)} \]

where ‘T’ is the orbital period, ‘G’ denotes the gravitational constant of the universe, and ‘ $ \rho $’ denotes the average density of the center body.

This straightforward equation can be used to determine the orbital period of any object orbiting any heavenly sphere.

For instance, the Earth has a density of 5.51 $ \frac{g}{cm^3 } $, which corresponds to a period of 1.4063 hours.

It’s vital to keep in mind that this assumption decreases as we get farther from the Earth’s top layer.

When we consider the fact that various satellites have various orbital durations, this becomes very obvious. Geostationary and geosynchronous trajectories are examples. The orbital period of such trajectories is precisely equivalent to:

1 day = 23.934446 hours

The position with regard to the equator distinguishes the geostationary orbit from the geosynchronous orbit.

Because the geostationary orbit is directly above the equator, orbiting satellites in this orbit remain over the aforementioned region of the surface of the Earth.

The geosynchronous orbit, however, can be found anywhere and is not directly mapped to any one location on Earth.

Orbital Period of a Binary Star System

We should now turn our attention to binary star systems. The definition of a binary star, which is a system made up of two stars orbiting each other and having identical sizes, has already been discussed. It’s time to determine their orbital period at this point.

We created the second section of the orbital period calculator with this objective in mind. There are several indicators such as:

  • 1st body mass of star: The mass of the first star M₁,
  • 2nd body mass of star: The mass of the second star M₂,
  • Principal axis: The principal axis of the elliptical orbit with one star as a center of attention is labeled as a.
  • Time Span: Orbital time of the binary star system T$_{binary}$.

Following is the system’s governing orbital period equation:

\[ Tbinary = 2 \cdot \pi \sqrt{\frac{a^3}{G \cdot (M_1+M_2)}} \]

where G is the universal gravitational constant.

This equation can be used in any binary system; it is not just applicable to systems that perfectly fit the description of a binary star.

One such case is the Pluto-Charon system. Even though neither of these objects is a star, they are still binary systems, and we can use our Orbital Period Calculator to determine their orbital period.

Solved Examples

Let’s solve some critical examples to better understand the working and concept of the Orbital Period Calculator.

Example 1

Find the orbit of a satellite in low earth orbit.

Solution

The most frequent orbit for commercial satellites is in low Earth orbit.

Given the severe mass disparity and proximity to the planet’s surface, we may use the first equation to calculate the orbital period:

\[ T= \sqrt{\frac{3\cdot\pi}{G\cdot \rho }} = \sqrt{\frac{3\cdot\pi}{G\cdot 5520}} \]

T =84.3 min

This value is rather near to the bottom limit of the LEO orbits, which is approximately 90 minutes.

Example 2

Find the orbit of the moon.

Solution

The length of the Moon’s orbit around the Earth can also be determined. Enter the following figures in the calculator’s second section:

  • First body mass is equal to one Earth mass and the semi-major axis is 384,748 km.
  • Second body mass is 1/82 of Earth’s mass.

\[ T = 2 \cdot \pi \sqrt{\frac{a^3}{G \cdot (M_1+M_2)}} \]

\[ T = 2 \cdot \pi \sqrt{\frac{(384748)^3}{G \cdot (M_1+M_2)}} \]

T=27 days and 7 hours

The period of the Moon has importance in this way.

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