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Derive expression for orbital speed of …

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Derive expression for orbital speed of satellite how it is related to escape velocity

    • Posted by Rahul Kumar 6 years, 8 months ago

    CBSE > Class 11 > Physics
    • 1 answers

    Ayush Tripathi 6 years, 8 months ago

    A/c to law of gravitation, the gravitational force acting on a satellite of mass m ,revolving around the earth which is at height of (Re+h) from earth surface will be F=GMem/(Re+h)^2 The centripetal force required by satellite to revolve around earth is F=mVo^2/(Re+h) In the equilibrium state the centripetal force is just provided by gravitational force Therefore, GMem/(Re+h)^2=mVo^2/(Re+h) => GMe/(Re+h)=Vo^2 => Vo=√GMe/(Re+h) Since, the satellite revolve near the earth so h~0 Vo=√GMe/Re * Re/Re => Vo=√Re * GMe/Re^2 => Vo=√gRe (since g=GMe/Re^2) And here Vo=orbital velocity We know that escape velocity Ve=√2gRe Therefore, Ve=√2 *Vo This is relation b/w orbital and escape velocity

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