A stelite is launched into a circular orbit of radius R around the earth .While a second is lunched into an orbit of radius `1.01R` The period of the second satellite is longer than the first one by approximatly

Options

A. `3.0%`
B. `1.5%`
C. `0.7%`
D. `1.0%`

Answer

b

Explanation

From Keplet's third law of motion , the apare of the period of revolation (T) of any place amount the son is dinesity preporitional in the cube of the same enjor axis of it dipltical orbit `(R)`.
`i.e. T^(2) = kR^(2)`
where k is constant
Therefore , `(T_(2))/(T_(1)) = ((R_(2))/(R_(1)))^(3//2) = ((1.01R)/(R))^(3//2)`
`rArr (T_(2))/(T_(1)) = (1 + 0.01)^(3//2) = 1+ (3)/(2) xx 0.01`
Percentage increases in time period is
`(Delta T)/(T) xx 100 ((T_(2))/(T_(1)) - 1) xx 100`
`= (3)/(2) xx 0.01 xx 100`
`= 1.5 %`