If a satellite has to orbit the earth in a circular path every 6 hrs, at what distance from the surface of the earth should be satellite placed (radius of earth , `R_e = 6400 km ` ) ( Assume , `(GM)/(4pi ^2) = 8.0 xx 10^(12) N// m^2 // kg`
Where ,G and M are gravitational constant and mass of earth and `10^(1//3)` = 2.1.

Options

A. 15100 km
B. 8720 km
C. 20600 km
D. 5560 km

Answer

B

Explanation

Given time period of a satellite , T = 6 hr
`= 2.16 xx 10^4 ` s and radius of earth , `R = 6.4 xx 10^6 m`
As, we know that time period of a satellite
`T = (2pi (R +h))/(v_0)` ……..(i)
Where , R +h = height of satellite from centre of earth and `V_0 `= orbital velocity
Then `V_0 = sqrt((GM)/((R+h)))` ..........(ii)
So , from Eqs. (i) and (ii) we get
`(R+h)^3 =(T) ^2 (GM)/(4pi^2) (because (GM)/(4pi^2) =8xx 10^2 , N//m^2 kg` given )
`rArr (R+h)^3 = (2.16 xx 10^4 )^2 xx 8 xx10^(12)`
`= 373.24 xx 10xx 10^(18)`
`rArr (R+h) = [ 373.24 xx 10 xx 10^(18) ] ^(1//3)`
`rArr 6.4 xx 10^6 +h = 7.2 xx 2.1 xx 10^6 = 15. 12 xx 10^6`
`rArr h = 8.72 xx 10^6 = 8720 km`