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Derive the time period of satellite orbiting the Earth. |
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Answer» Solution :Time period of the satellite: The distance covered by the satellite during one rotation in its orbit is equal to `2pi (R_(E )+H)` and time taken for it is the time period, T. Then, Speed , v =` ("Distancetravelled ")/("Time taken ")= (2pi(R_(E)+h))/T ""...(1)` Speed of the satellite, `V = sqrt((GM_(E))/((R_(E)+h)))` `sqrt((GM_(E))/((R_(E)+h))) = (2pi(R_(E)+h))/T` `T = (2pi)/(sqrt(GM_(E)) )(R_(E)+h)^(3/2)""...(2)` Squaring both sides of the equation (2) we get `T^(2) = (4pi^(2))/(GM_(E)) (R_(E)+h)^(3)`...(2) `(4pi^(2))/(GM_(E)) = `constantsay c `T^(2) =c(R_(E)+h)^(3)` Equation (3) IMPLIES that a satellite orbiting the Earth has the same relation between time and distance as that Kepler.s law of planetary motion. For a satellite orbiting near the surface of the Earth. h is negligible compared to the radius of the Earth.`R_(E)`Then, `T_(2) = (4pi^(2))/(GM_(E))R_(E)^(3)` `T^(2) = (4pi^(2))/(((GM_(E))/R_(E)^(2)))R_(E)RARR T^(2) = (4pi^(2))/g R_(E)` Since ` (GM_(E))/(R_(E)^(2)) = g ` `T = 2pisqrt(R_(E)/g)` By substituting the values of`R_(E) = 6.4 xx10^(6) ` m and g = `9.8 " m s"^(-2)`the orbital time period is obtained `T cong 85`minutes . |
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