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| 1. |
Derive expression for orbital speed of satellite how it is related to escape velocity |
| Answer» 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)^2The 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 forceTherefore,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~0Vo=√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=√2gReTherefore, Ve=√2 *VoThis is relation b/w orbital and escape velocity | |