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The orbital velocity of a body is the velocity at which it revolves around the other body. The objects that travel around the Earth in uniform circular motion are said to be in an orbit. The velocity of this orbit depends on the distance between the object under consideration and the center of the earth. This velocity is usually associated with artificial satellites because they can revolve around any particular planet. The Orbital Velocity Formula is used to calculate the orbital velocity of an object if its mass and radius are known.

The equation of the orbital velocity is given by:Â

$V_{\text{orbit}} = \sqrt{\dfrac{GM}{R}}$

In the above equation, $G$ stands for Gravitational Constant, $M$ stands for the mass of the body at the center and $R$ is the radius of the orbit.

The Orbital Velocity Equation is used to find the orbital velocity of a planet if its mass $M$ and radius $R$ are known.Â

The unit used to express Orbital Velocity is meter per second $(m/s)$.

To derive the orbital velocity, we first need to know about the gravitational force and the centripetal force. It is important to know about the gravitational force because it is this force that allows orbiting to occur. This gravitational force is exerted by a central body on the orbiting body to keep it in its orbit. Centripetal force is also important because it is this force that is responsible for circular motion.

For the derivation of the formula, let us take a satellite of mass m which revolves around the Earth in a circular orbit of radius $r$ at a height h from the surface of the Earth. Let $M$ and $R$ be the mass and radius of the Earth respectively, then $r=R+h$.

To revolve the satellite, a centripetal force $\dfrac{mv_{0}^{2}}{r}$ is needed which is provided by the gravitational force of $G{\dfrac{Mm}{r^2}}$ acting between the satellite and the Earth.

Thus, equating the two equations, we get

$\dfrac{mv_{0}^{2}}{r} = G{\dfrac{Mm}{r^2}}$

$v_{0}^{2} = \dfrac{GM}{r} = \dfrac{GM}{R+h}$

Simplifying the above equation, we get

$v_0 = \sqrt{\dfrac{GM}{R+h}}$Â ----------$(1)$

Also, $GM = gR^2$, where $g$ is the acceleration due to gravity.Â

Therefore,

$v_0 = \sqrt{\dfrac{gR^2}{R+h}}$

Simplifying the above equation, we get

$v_0 = R \sqrt{\dfrac{g}{R+h}}$

Let $g^\prime$ be the acceleration due to gravity at a height $h$ from the surface,

$g^\prime = \dfrac{GM}{(R+h)^2}$

On further simplifying, we get

$\dfrac{GM}{(R+h)} = g^\prime(R+h) = g^\prime r$Â ---------$(2)$

Now, substituting $(2)$ in $(1)$, we get

$v_0 = \sqrt{g^\prime r} = \sqrt{g^\prime (R+h)}$

Orbital velocity is a very important and scoring chapter of physics. If studied thoroughly, very good marks can be scored from this part.

FAQ (Frequently Asked Questions)

1. What Is Orbital Velocity?

Orbital velocity refers to the velocity required by the satellites to remain in their orbits. The orbital pathway being either elliptical or circular displays a balance between the inertia of the satellite which helps toÂ make it move in a straight line with the gravity of the planet which in turn pulls the satellite closer and closer to the planet. This orbital velocity is also dependent on the distance between the center of the planet and the satellite revolving around it. This velocity would be higher if the center of attraction is a more massive body at a particular altitude.

2. Explain The Relationship Between Escape Velocity And Orbital Velocity In Terms of The Earth.

Escape velocity is the velocity required for a free, non-propelled object to break free from the gravitational impact of a huge center body whereas the orbital velocity is the velocity that is required by objects to remain in their orbits.

It takes a certain level of velocity for an object to achieve orbit around a planet such as Earth. To break free from such an orbit, an even greater velocity is required.Â

The escape velocity is consistently âˆš2 times the orbital speed. At a distance from the moon to the earth, the orbital speed is 1.022 km/s. The escape velocity consequently at this separation is 1.445 km/s. However, this would need to be coordinated away from the Earth.Â

3. What is Transverse Orbital Speed?

The transverse orbital speed is inversely proportional to the distance to the central body by the law of conservation of angular force or Keplerâ€™s second law. This states that as a body moves around its orbit during a fixed time, the line from the barycenter to the body sweeps a steady zone of the orbital plane, paying little heed to which part of its orbit the body traces during that time. This implies that at a smaller distance, to cover the same distance, the object needs to move faster in the arc. Hence, a body moves slower near its apoapsis than near its periapsis.