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

Explanation

Given : Height of `1^(st)` satellite = h
Time of rev. of `1^(st)` sateelite = T
Time of rev. of `2^(nd)` satellite = `2sqrt(2)` T
To find : Height of `2^(nd)` satellite `(h_(2))` = ?
Formulae : `(T^(2))/(r^(3)) = k , (2pi(R+h))/(T)`
solution : Case (i) Time = T , r = R ` + h_(1)`
`(T^(2))/((R + h_(1))^(3)) = k`
Case (ii) Time `= 2 sqrt(2)`
` (8T^(2))/((R + h_(2))^(3)) = k`
Form (i) and (ii)
`(T^(2))/((R + h_(1))^(3))` `(8T^(2))/((R+h_(2))^(3))` (by cross multiplying )
`(R + h_(2))^(3) = 8(R + h_(1))^(3)`
Taking cube root on the both side .
`root(3)((R+h_(2))^(3)) = root(3)(8(R+h_(1))^(3))`
` R +h_(2) = 2(R+h_(1))`
`R + h_(2) = 2R + 2h_(1)`
`h_(2) = 2R + 2h_(1) - R`
`h_(2) = R+ 2h_(1)`
`therefore ` Height of the `2^(nd)` satellite will be `R + 2h_(1)`