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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 moving body has a tendency to move in a straight path because of its inertia. The gravitational force, on the other hand, tends to drag it down. The orbital route, which is elliptical or circular in shape, is a balance of gravity and inertia. The velocity required to achieve a balance between gravity’s pull on the body and the inertia of the body’s motion is known as orbital velocity. The orbital velocity of a satellite orbiting around the Earth is determined by its altitude above the Earth. The faster the required orbital velocity, the closer it is to the Earth.
At lower altitudes, a satellite collides with traces of Earth’s atmosphere, causing drag. This drag causes the satellite’s orbit to degrade, eventually causing it to fall back into the atmosphere and burn up.
What Is Orbital Velocity?
A natural or artificial satellite’s orbital velocity is the velocity required to keep it in orbit. The moving body’s inertia causes it to move in a straight line, while gravitational force pulls it down. The elliptical or circular orbital path thus shows a balance between gravity and inertia. If the muzzle velocity of a cannon shot from a mountaintop is increased, the projectile will go further. The bullet will never fall to the ground if the velocity is great enough. The Earth’s surface can be imagined curving away from the projectile, or satellite, at the same rate that it falls toward it. The higher the orbital velocity for a given height or distance, the more massive the body is at the center of attraction. If air resistance could be ignored near the Earth’s surface, orbital velocity would be around eight kilometers (five miles) per second. The gravitational force is weaker the farther a satellite is from the center of attraction, and the less velocity it needs to stay in orbit. See also the definition of escape velocity.
Orbital Velocity Equation
The equation of the orbital velocity is given by:
[V_{orbit} = sqrt{frac{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.
The unit used to express Orbital Velocity is meter per second (m/s).
Derive An Expression for Orbital Velocity
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 resolve the satellite, a centripetal force [frac{mv_{0}^{2}}{r}] is needed which is provided by the gravitational force of the G[tfrac{Mm}{r^{2}}] acting between the satellite and the Earth.
Thus, equating the two equations, we got
[frac{mv_{0}^{2}}{r} = Gtfrac{Mm}{r^{2}}]
[v_{0}^{2}= frac{GM}{r}= frac{GM}{R+h}]
Simplifying the above equation, we get
[v_{0}= sqrt{frac{GM}{R+h}}]…………….(1)
Also,GM = g[R^{2}], where g is the acceleration due to gravity
Therefore,
[v_{0}= sqrt{frac{gR^{2}}{R+h}}]
Simplifying the above equation, we get
[v_{0}= Rsqrt{frac{g}{R+h}}]
Let g’ be the acceleration due to gravity at a height h from the surface
[g^{‘}= frac{GM}{(R+h)^{2}}]
Conclusion
Orbital speed is a very important and rewarding chapter in physics. It helps with the understanding of orbital concepts in physics. With thorough research, you can achieve very good results in this area.