Physics Formulas

Orbital Speed Formula

Objects which travel in a uniform circular motion around the Earth are said to be “in orbital motion”. The velocity of the orbit depends on the distance from the object to the center of the Earth. The velocity has to be just right so that the distance to the center of the Earth is always the same. Thus orbital speed is an important computation. This topic will explain the concept of it with orbital speed formula. Relevant examples are also given. Let us learn the interesting concept!

Orbital Speed Formula

Concept of Orbital Speed:

The orbital speed of the body which is generally a planet or a natural satellite is the speed at which it orbits around the center of the system. This system is usually around a massive body. The orbital speed of the earth around the sun is 108,000 km per hour.

It is relating the mass of a given planet to the gravitational constant and radius through the formula given. The orbital speed formula contains a constant, G, known as the “universal gravitational constant”. Its value is \(6.673 \times 10^{-11} N m^2 kg^{-2}\). The value of the radius of the Earth is \(6.38 \times 10^6 m\).

Earth’s escape velocity is greater than the required place an Earth satellite in the orbit. Orbital speed is the speed needed to achieve the balance between gravity’s pull on the satellite and the inertia of the satellite’s motion. This is approximately 27,359 km per hour at an altitude of 242 km.

Without gravity, the inertia of the satellite will carry it off into space. Even with gravity, the intended satellite goes too fast and eventually it will fly away. Also, if the satellite goes too slowly, gravity will pull it back to Earth.  Thus at the correct orbital speed, gravity exactly balances the satellite’s inertia.

The orbital speed of the satellite depends on its altitude above the Earth. To maintain an orbit that is 35,786 km above the Earth, a satellite must orbit at the speed of about 11,300 km per hour. That orbital speed and distance permit the satellite to make one revolution in exactly 24 hours.

Because the satellite stays right over the same spot always all the time. So, this kind of orbit is known as “Geostationary Orbits” are ideal for weather satellites.

Thus, the higher the orbit, the longer the satellite can stay in its orbit.  At higher altitudes, where the vacuum of space is nearly complete, there is almost no drag and a satellite like a moon can stay in orbit for centuries.

The formula for the orbital speed:

\(V_{orbital} = \sqrt{G\; M }{R}\)


\(V_{orbital}\) Orbital speed
G gravitational constant,
M mass of planet
R radius

Solved Examples

Q.1: The mass of the Moon is \(8.35 \times 10^{22}\)Kg and the radius are \(2.7 \times 10^6 m\). Then, determine the orbital speed.


Given parameters:

Mass of the Moon, M = \(8.35 \times 10^{22} Kg\)

Radius, \(R = 2.7 \times 10^6 m\)

\(G = 6.673 \times 10^{-11} m^3 kg^{-1} s^{-2}\)

Thus, orbital speed will be computed as,

\(V_{orbital} = \sqrt{G \times M }{R}\)

= \(\sqrt{ 6.673 \times 10^{-11} \times 10^{22} }{ 2.7 \times 10^6 }\)

= \(20.63 \times  10^6 m per sec\).

Thus Orbital speed will be \(20.63 \times  10^6\) m per sec.

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5 responses to “Spring Potential Energy Formula”

  1. Typo Error>
    Speed of Light, C = 299,792,458 m/s in vacuum
    So U s/b C = 3 x 10^8 m/s
    Not that C = 3 x 108 m/s
    to imply C = 324 m/s
    A bullet is faster than 324m/s

  2. Malek safrin says:

    I have realy intrested to to this topic

  3. umer says:

    m=f/a correct this

  4. Kwame David says:

    Interesting studies

  5. Yashdeep tiwari says:

    It is already correct f= ma by second newton formula…

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