Obtain an equation of orbital time period of satellite revolves around the earth.

Explanation

`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