# Derivation of Orbital Velocity

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## What is Orbital Velocity?

Orbital velocity refers to the velocity required by satellites (natural or artificial) to remain in their orbits. The orbit or the orbital pathway, whether elliptical or circular, displays a balance between the inertia of the satellite which makes it move in a straight line with the gravity of the planet which pulls the satellite closer to the planet. Now, this orbital velocity depends on the distance between the center of the planet and the satellite revolving around it.

This velocity applies to planets revolving around the sun as much as it applies to natural or artificial satellites revolving around a planet. The orbital velocity would be higher if the center of attraction is a more massive body at a particular altitude, for example, if a satellite is close to the surface of the earth, and there is not much air resistance, the orbital velocity can be as high as 8 km per second. On the other hand, the farther the satellite is from the center of the body of attraction, the weaker will be the force of gravitation due to decreased centripetal force, and therefore, less velocity will be utilized by the satellite to remain in the orbit.

## The Formula for Orbital Velocity

The orbital velocity equation is given by:

$\Rightarrow v = \sqrt{\frac{GM}{r}}$

Where,

R is the radius of the orbit,

M is the mass of the central body of attraction,

G is the gravitational constant.

The orbital velocity can be calculated for any satellite and the consequent planet if the mass and radii are known.

## Orbital Velocity Formula Derivation

The following steps can be followed to derive an expression for the orbital velocity of a satellite revolving in an orbit.

For orbital speed derivation, both the gravitational force and centripetal force are very important. Gravitational force (Fg) is the force exerted by the body at the center to keep the satellite in its orbit. Centripetal force (Fc) justifies the existence of circular motion.

For this derivation, we will assume a uniform circular motion. Therefore, for an object in orbit, both these forces will be equal.

$\Rightarrow Fg = Fc$  - - - - 1

Now, as we know gravitational force depends on the masses of both objects and it's formula is:

$\Rightarrow Fg = \frac{GMm}{r^{2}}$

Where,

G is the gravitational constant, M is the mass of the central body, and m is the mass of the revolving body, and r is the distance between the two bodies.

Also, the centripetal force is related to mass and acceleration, and the formula for which is given by:

$\Rightarrow Fc = \frac{mv^{2}}{r}$

Where,

m is the mass of the satellite in orbit, v is the speed, r is the radius of the orbit.

Now, putting both of these equations in (1),

We get,

$\Rightarrow Fg = Fc$

$\Rightarrow \frac{mv^{2}}{r}$

$\Rightarrow \frac{GM}{r} = v^{2}$

And henceforth,

$\Rightarrow v = \pm \sqrt{\frac{GM}{r}}$

Therefore, orbital velocity derivation is proved.

## Difference Between Orbital Velocity and Escape Velocity

Escape velocity is the base speed required for a free, non-propelled object to escape from the gravitational impact of a huge center body, that is, to move an unlimited distance from it. This velocity is a component of the mass of the body and separation to the focal point of mass of the body.

To find the difference between escape velocity and orbital velocity, let us consider an ideal situation where there is no air resistance and all the bodies are point masses.

An orbital speed would be any speed that retains the point mass into a curved or an elliptical direction. It is any velocity that is less than the escape velocity. There are limitlessly many orbital velocities.

Escape speed on the other hand is the speed that would place the point mass in a parabolic direction. For any point in space in a two-body framework, there is actually one escape speed. This, numerically, compares to the way that all parabolas have the very same mathematical shape- they just vary in scale. In the interim, there are endlessly numerous curved or elliptical shapes.

Escape velocity is the velocity at which the kinetic energy of the object approaches the basic of all likely energy out to infinity, which means the object could get away from the gravity of a body completely. Orbital velocity, on the other hand, depends on the precise orbital trajectory; however, for a roundabout circle, it gets away from the velocity at that separation isolated by the square root of 2. Therefore, the object is not getting away into infinity; it is simply falling in a circular motion around the center body.

FAQs (Frequently Asked Questions)

1. Explain the Relationship of Escape Velocity and Orbital Velocity in Terms of Earth.

Answer: Escape velocity is consistently √2 times the orbital speed. At a distance from the moon to earth, its orbital speed is 1.022 Km/s. The escape velocity at this separation is consequently 1.445 Km/s. However, this would need to be coordinated away from Earth.

At the distance of the earth to the sun, the break speed is more than 41 Km/s.

Were the Moon to get away from Earth, it would even now be caught in a heliocentric circle, were it not to accomplish this latter velocity.

2. What is Meant by Transverse Orbital Speed?

Answer: The transverse orbital speed is conversely relative to the separation to the focal body due to the law of conservation of angular force, or comparably, Kepler's subsequent law. This expresses that as a body moves around its circle during a fixed measure of time, the line from the barycenter to the body clears a steady zone of the orbital plane, paying little heed to which part of its circle the body follows during that time frame. This implies that at a smaller distance, the object needs to move faster in the arc to cover the same distance.

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