Can u derive the formula for orbital velocity with proper explanation?

1 Answer
NJ
Mar 31, 2018

#=> v = sqrt((GM)/r)#

Explanation:

Orbital velocity is the speed and direction of an object in orbit.

To derive orbital velocity, we concern ourselves with two concepts:

  • Gravitational force
  • Centripetal force

Gravitational force is important, because it is the force that allows orbiting to exist. A central body exerts a gravitational force on the orbiting body to keep it in its orbit.

Centripetal force is also important, as this is the force responsible for circular motion. For deriving a simple equation for orbital velocity, we will assume uniform circular motion. In the case of an orbiting body, the centripetal force is the gravitational force.

This "is" usually equates to an "=". So we should start by saying that the gravitational force #F_g# is equal to the centripetal force #F_c#:

#=>F_g = F_c#

Now, we need to know how these forces are defined. Gravitational force is based on the masses of both objects and acts as the inverse of the square of the distance:

#=>F_g = (GMm)/(r^2) " "#
where #G# is the gravitational constant, #M# is the mass of the object applying the force, #m# is the mass of the orbiting body, and #r# is the distance between the two objects.

Centripetal force is defined as mass times acceleration, where we are considering centripetal acceleration #a_c#:

#=>F_c = ma_c = (mv^2)/r " "#
where #m# is the mass of the object undergoing uniform circular motion, #v# is its speed, and #r# is the radius of the circle defined by its motion.

Now we equate:

#=>F_g = F_c#

#=> (GMm)/(r^2) = (mv^2)/r#

We want to find #v#:

#=> (GMcancelm)/(r^(cancel2) )= (cancelmv^2)/cancelr#

#=> (GM)/r = v^2#

Hence:

#=> v = +- sqrt((GM)/r)#

If you're only concerned with magnitude, just take the positive result:

#=> v = sqrt((GM)/r)#

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