Did you know that space missions like GPS satellites, Geo Synchronous satellites, etc. are possible only because of the correct calculations of the orbital velocity? Orbital velocity is one of the most important concepts in Physics. Let us study the orbital velocity formula with relevant examples.

## What is Orbital Velocity?

Orbital Velocity is the velocity at which a body revolves around another body. It is an important concept in the field of astronomy and physics. It is used extensively to launch satellites into orbits and to make sure that they stay in their orbits. Inertia of the moving body tends to make it move on in a straight line, whileÂ gravitationalÂ force tends to pull it down. Thus the path traced will be a balance between these two forces i.e. an elliptical path is formed.

A cannon fired from a mountaintop will throw a projectile farther if its muzzle velocity is increased. The surface of theÂ EarthÂ may be thought of as curving away from the projectile, or satellite, as fast as the latter falls toward it. The orbital velocity is directly proportional to the mass of the body for which it is being calculated and inversely proportional to the radius of the body.

Earth’s orbital velocity near its surface is around eight kilometres (five miles) per second if the air resistance is disregarded. The farther from the centre of attraction a satellite is, the weaker the gravitational force and the less velocity it needs to remain in orbit.

### Orbital Velocity in Space Exploration

The concept of orbital velocity is very important in space exploration. It is used extensively by space agencies to understand how to launch satellites. It helps the scientists to discover the speeds at which the satellites must revolve around a planet or a heavenly body so that it doesn’t collapse into it.

Orbital velocity combined with the escape velocity are the two most important calculated values needed to plan a space mission or a satellite mission. Problems involving these two entities are also extensively asked in competitive exams. Thus it is important to know about these formulas and also the logic behind these formulas so that the problems can be solved easily.

### The Formula:

Let us learn more about its formula here.

For any revolving object, the formula for the orbital velocity is given by,

Where,

- G = gravitational constant with the value 6.673×10
^{(-11)}Â Nâˆ™m^{2}/kg^{2}, - M = mass of the body at center,
- R = radius of orbit.
- In most of the cases M is the weight of the earth.

It’s derivation is explained in the figure below,

## Solved Examples for Orbital Velocity Formula

Q: A satellite launch is made for the study of Jupiter.Determine its velocity around Jupiter.

- Given: Radius of Jupiter R = 70.5Â Ã—Â 106Â m,
- Mass of jupiter M = 1.5Â Ã—Â 1027Â Kg,
- Gravitational constant G = 6.67Â Ã—Â 10-11Â m
^{3}/s^{2}Â kg

Solution:Â Substituting theÂ parameters in the orbital velocity formula, we get

VorbitÂ =Â âˆšGM / R

=Â âˆš6.673Ã—10^{âˆ’11}Ã—1.5Ã—1027Â / 70.5Ã—106

=Â âˆš10.0095 x 1016Â / 70.5 x 106

=Â âˆš0.141 x 1010

3.754 x 109Â m/s.

Q: Calculate the orbital velocity of earth if radius of earth R = 6.5Â Ã—Â 106Â m, mass of earth M = 5.5Â Ã—Â 1024Â kg and Gravitational constant G = 6.67Â Ã—Â 10-11Â m3/s2Â kg.

Solution:Â Given as follows,

R =Â 6.5Â Ã—Â 10^{6}Â m

M = 5.5Â Ã—Â 10^{24}Â kg

G = 6.6Â Ã—Â 10^{-11}Â m^{3}/s^{2}Â kg

The Orbital velocity formula is given by

VorbitÂ =Â âˆšGM / R

=Â âˆš6.67 Ã—10^{âˆ’11}Â Ã— 5.5Ã—10^{24}Â / 6.5 Ã—10^{6}

=Â âˆš36.68 x 10^{13}Â / 6.5 xÂ 10^{6}

=Â 7.5 km/s

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