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Obtain an equation of orbital time period ofsatellite revolves around the earth. |
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Answer» Solution :`implies`The period of a satellite is the time it takes to make one full orbit around the earth is known as period or ORBITAL period of satellite. `implies` Orbital period of satellite `v_(0) = sqrt((GM_(E))/((R_(E)+h)))""...(1)` circumference of orbit `implies` Orbital velocity = `("of satellite ")/("period of orbit ")` `:. v_0 =(2pi(R_(E)+h))/T""...(2)` `implies`Comparing equation (1) and (2), `sqrt((GM_(E))/((R_(E)+h)))=(2pi(R_(E)+h))/T` `:. T = (2pi(R_E+h))/(sqrt((GM_E)/((R_E+h))))` `:. T = 2pi sqrt((R_(E)+h)^3/(GM_E))""...(3)` `:. T^(2) =(4pi^2)/(GM_E)(R_E+h)^3""...(4)` `:. T = 2pi sqrt((R_(E)+h)^3/(GM_E))""...(5)` but `(GM_E)/((R_(E)+h)^(2) )=G` `:. T = (2pi )sqrt(((R_(E)+h))/g)""...(6)` `implies` From equation (4) , `T^(2) = Kr^(3) ""....(7)` `:. T^(2) alpha r^3""...(8)` (where `K=(4pi^2)/(GM_E)and ` take `R_E+h=r` ) This is Kepler.s third law for PLANETARY motion. `implies` For a satellite very close to the surface of earth h can be neglected in comparison to `R_E`equation (6) will be, `:. T = 2pisqrt((R_E)/g)""...(9)` `implies`Taking `g = 9.8 m//s^(2) , R_(E) = 6400 xx 10 ^(3) m//s` `T = 2pi sqrt((6.4xx10^6)/(9.8))` `T = 5078 s ~~ 85` minutes |
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