Orbital velocity is defined as the minimum velocity a body must maintain to stay in orbit. Due to the inertia of the moving body, the body has a tendency to move on in a straight line. But, the gravitational force tends to pull it down. The orbital path, thus elliptical or circular in nature, represents a balance between gravity and inertia. Orbital velocity is the velocity needed to achieve a balance between gravity’s pull on the body and the inertia of the body’s motion. For a satellite revolving around the Earth, the orbital velocity of the satellite depends on its altitude above Earth. The nearer it is to the Earth, the faster the required orbital velocity. A satellite runs into traces of Earth’s atmosphere, at lower altitudes, which creates drag. This drag causes decay the orbit, eventually making the satellite to fall back into the atmosphere and burn itself up. To derive the orbital velocity, we concern ourselves with the following two concepts: It is important to know the gravitational force 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 the derivation, let us consider a satellite of mass m revolving around the Earth in a circular orbit of radius r at a height h from the surface of the Earth. Suppose M and R are the mass and radius of the Earth respectively, then r = R + h. To revolve the satellite, a centripetal force of \(\begin{array}{l}\frac{mv_{o}^2}{r}\end{array} \) \(\begin{array}{l}G\frac{Mm}{r^2}\end{array} \) Therefore, equating both the equations, we get \(\begin{array}{l}\frac{mv_{0}^2}{r}=G\frac{Mm}{r^2}\end{array} \) \(\begin{array}{l}v_{0}^2=\frac{GM}{r}=\frac{GM}{R+h}\end{array} \) Simplifying the above equation further, we get \(\begin{array}{l}v_{0}=\sqrt{\frac{GM}{R+h}}\end{array} \) But \(\begin{array}{l}GM = gR^2\end{array} \) , where g is the acceleration due to gravity.Therefore, \(\begin{array}{l}V_0=\sqrt{\frac{gR^{2}}{R+h}}\end{array} \) Simplifying the above equation, we get \(\begin{array}{l}v_0=R\sqrt{\frac{g}{R+h}}\end{array} \) Let g’ be the acceleration due to gravity in the ( at a height h from the surface) \(\begin{array}{l}g’=\frac{GM}{(R+h)^2}\end{array} \) Simplifying further, we get \(\begin{array}{l}\frac{GM}{(R+h)}=g'(R+h)=g’r\end{array} \) ……(eqn 2)Substituting (2) in (1), we get \(\begin{array}{l}v_0=\sqrt{g’r}=\sqrt{g'(R+h)}\end{array} \) To know more on other Physics-related concepts, stay tuned to BYJU’S. Related Physics articles:
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Orbital Velocity Formula
Objects that travel in uniform circular motion around the Earth are said to be "in orbit". The velocity of this orbit depends on the distance from the object to the center of the Earth. The velocity has to be just right, so that the distance to the center of the Earth is always the same.The orbital velocity formula contains a constant, G, which is called the "universal gravitational constant". Its value is = 6.673 x 10-11 N∙m2/kg2 .The radius of the Earth is 6.38 x 106 m. v = the orbital velocity of an object (m/s) G = the universal gravitational constant, G = 6.673x10(-11) N∙m2/kg2 mE = the mass of the Earth (5.98 x 1024 kg) r = the distance from the object to the center of the Earth Orbital Velocity Formula Questions: 1) The International Space Station orbits at an altitude of 400 km above the surface of the Earth. What is the space station's orbital velocity? Answer: The orbital velocity depends on the distance from the center of mass of the Earth to the space station. This distance is the sum of the radius of the Earth and the distance from the space station to the surface: r = (6.38 x 106 m) + (400 km) r = 6380000 + 400000 m r = 6780000 m The orbital velocity can be found using the formula: v=7672 m/s The orbital velocity of the International Space Station is 7672 m/s. 2) A satellite is orbiting the Earth with an orbital velocity of 3200 m/s. What is the orbital radius? Answer: The orbital radius can be found by rearranging the orbital velocity formula: The orbital radius for this satellite is 3.897 x 107 m.
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