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Derivation of Orbital Velocity

What is Orbital Velocity?

Last updated date: 26th Mar 2023
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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 further 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 base 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.

The velocity required by a satellite to be in its home orbits is called orbital velocity. So now what about escape velocity. It has got the absolute meaning just like the name escape! Yes, it is the velocity required for releasing down from its primary place and to reach the infinite distance. It can also be called an ideal speed. This is because here we have to ignore the friction offered by the atmosphere. And a close relationship with the mass of a primary body is, the escape velocity increases with an increase in the mass of the primary body. And how it decreases is by the distance from the primary body.

The relation between orbital velocity and escape velocity is given by

$V_{e} = 2V_{0}$

Where $V_{e}$ is escape velocity and $V_{0}$is orbital velocity.

How Escape Velocity Works!

We all know about gravitational force. To reach space, the satellite should be able to overcome this pull. And thus it needs to have a specific velocity called escape velocity to reach the distance it intends to. Lower velocity is needed if the distance from the primary body; in our case, it is the earth, higher the distance. So that is why the communication satellites orbiting the earth don't consume any energy. Because they are at a distance of miles away from the earth. However, commercial aircraft require much energy to stay in the atmosphere for having control over the gravitational pull.

For a rotating body, its escape velocity depends upon the direction it is going. However, apart from theoretical considerations, there is so much we have to decide upon the practical considerations. For example, in real life, it is not that easy to achieve escape velocity instantly. Since the acceleration is already implied and with the hypersonic speeds, there are chances of getting burned as a result of aerodynamic heating. And taking into consideration an elliptical orbit, for getting into the escape velocity for the body the required speed will always vary. Also when the body is closest towards the central body it will be at the greatest at the periapsis.

The Black Hole’s Escape Velocity

We already have a brief idea about black holes. For anything that is near the black hole is taken inside. And the escape velocity for something that enters a black hole is nearly more than the speed of the light.

However, there are black holes that are formed at the center of galaxies ejecting the matter absorbed in the speed of the light. They are called quasars.

Conclusion

Vedantu offers free downloads of Orbital velocity in PDF format. In addition Vedantu offers solved examples, step by step derivations, self study materials such as mock question papers, worksheets  and previous year solved question papers.

FAQs on Derivation of Orbital Velocity

1. Why do satellites need to be launched with escape velocity?

Satellite is an interplanetary mission. Any object that comes under an interplanetary mission is given an escape velocity. If we throw an object from a height we can see that it reaches the ground after some time. However, if you attempt to throw in more heights, students can see a circle forming for the path of stone or the object you throw. So scientists have made a theory out of it, that is, an object launched from a particular height will look forward to staying in circles(orbits) when launched with a particular velocity. And also this velocity is needed for defying the gravitational pull from the earth.

2. Does escape velocity belong to a vector or scalar quantity?

A quantity that is having a magnitude and direction is a vector quantity. While a quantity that has magnitude alone is called a scalar quantity. So here escape velocity is the speed that is needed to be break-free from the earth’s gravity, rather than the velocity. Since speed belongs to the scalar quantity, escape velocity is also a scalar quantity.

This is because escape velocity concentrates on how fast the object can move to be free from the atmospheric pull. There, the direction of the object goes irrelevant. This topic is very well explained in Vedantu, you can register and download free study materials for the site.

3. Calculate the moon’s velocity!

Calculating using the equation

$V_{esc} = \sqrt{\frac{2GM}{r}}$

With the details of the moon below:

$G = 6.67384 \times 10^{-11} Nm^{2} kg^{-2}$

M = 7.34767E22 kg

r = 1737400 m

And calculating this way we get escape velocity as 2375.89m/s

4. Don’t you think escape velocity should be called escape speed?

We can call velocity as speed too as both of these are treated as the same almost. They are synonymous to a level. However, students must know that velocity direction has a particular role. While speed doesn’t consider direction at all. And the number one proof for escape velocity to be escape speed is that it is treated as a scalar quantity rather than as a vector quantity.

5. How do objects are launched when the escape velocity is supposed to be 24000mph and no rockets have acquired this speed so far?

This is one interesting question to be read. In the actual case, the object doesn’t need to reach this 24000 mph to escape the earth’s gravitational field. The escape velocity is calculated by the distance measured from the center of mass of the system. The only force to be applied here is gravity. For rockets, the thrust is applied not just in the initial case, but all along!