Question

what is orbital velocity​

Answer 1

Answer:

Orbital Velocity:

⇒ It is the minimum velocity required to put the satellite into its orbit around the earth is known as orbital velocity.

⇒ Orbital velocity of satellite is 7.9 km/s.

Derivation:

$\sf Force\;acting\;on\;gravity\;of\;satellite,\;F=\dfrac{GMem}{(Re+h)^{2}}\;\;\;\;\;.......(1)\\ \\ \\ \sf Centripetal\;force\;for\;circular\;motion,\\ \\ \\ \implies \sf Fe=\dfrac{mv_{o}^{2}}{(Re+h)}\;\;\;\;\;.........(2)\\ \\ \\ \sf For\;circular\;motion,\\ \\ \implies \sf F = Fe\\ \\ \\ \implies \sf \dfrac{GMem}{(Re+h)^{2}}=\dfrac{mv_{o}^{2}}{(Re+h)}\\ \\ \\ \implies v_{o}^{2}=\dfrac{GMe}{Re+h}\\ \\ \\ \implies \sf v_{o}=\sqrt{\dfrac{GMe}{Re+h}}\;\;\;\;........(3)$

$\sf Now,\;we\;know\;that,\\ \\ \\ \implies \sf g=\dfrac{GMe}{r^{2}}\\ \\ \\ \implies \sf GMe = gr^{2}\;\;\;\;........(4)\\ \\ \\ \sf Equation\;(4)\;put\;into\;equation\;(3),\\ \\ \\ \implies \sf v_{o}=\sqrt{\dfrac{gRe^{2}}{Re+h}}\;\;\;\;........(5)\\ \\ \\ \sf If\;satellite\;is\;very\;close\;to\;surface\;of\;earth,\\ \\ \\ \implies \sf v_{o}=\sqrt{\dfrac{gRe^{2}}{Re}}\\ \\ \\ \implies \sf v_{o}=\sqrt{gRe}\\ \\ \\ \sf Where,\;g=9.8\;m/s^{2}\;\&\;Re=6400\;km\\ \\ \\ \implies {\boxed{\sf v_{o}= 7.9\;km/s}}$

Hence Proved!!!