If the height of a satellite completing one revolution around the earth in T seconds is ` h_(1)` metres, then what would be the height of a satellite taking ` 2 sqrt2` T seconds for one revolution ?

Explanation

Given (1) Time : T seconds (2) Height : ` h_(1)`
(3) Let us assume the height of the satellite completing one revolution in ` 2sqrt2` T seconds as `h_(2)`
`T = (2pir)/v_(c) i.e, T = ( 2pi (R+h_(1)))/(sqrt((GM)/((R+h_(1))))) therefore T= 2pi sqrt((R-h_(1))^(3)/(GM))`
To be calculated : Height `h_(2) ` in time `2sqrt2` T seconds
` 2sqrt2 T = 2pi sqrt(((R-h_(2))^(3))/(GM))`
From Eqs. (1) and (2)
` T/(2sqrt2T) = 2pi sqrt(((R + h_(1))^3)/(GM))/sqrt((R+h_(2))^(3)/(GM)) " " therefore = sqrt((R+h_(1))^(3))/sqrt((R+h_(2))^(3))`
` 1/2 = (R+h_(2))/((R+h_(2)) therefore R + h_(2) = 2R + 2h_(1) " " therefore h_(2) = R + 2h_(1)`