Find the radius of the circular orbit of a satellite moving with an angular speed equal to the angular speed of earth's rotation. Mass of the earth`=6xx10^24kg`

Answer

A, B, C, D

Explanation

Angulr speed of earth and the satellite will be same
`(2pi)/T_e=(2pi)/T_s`
or `1/(24xx3600)=1/(sqrt2pisqrt((R+h)^3/gh^2))`
or `12xx3600=3.14 sqrt(((R+h)^3)/(gR^2))`
or `((R+h)^3)/(gR^2)=((12xx3600)62/((3.14)^2)`
`or ((6400+h^3)xx10^9)/(9.8xx(6400)^2xx10^6)=((12xx3600)^2)/((3.14)^2)`
or `((6400+h)xx10^9)/(6272xx0^9)=432xx10^4`
or `(6400+h)^3=6272x432xx10^4`
or `6400+h=(6272xx432xx10^4)^(1/3)-6400`
b. Time taken from north to equator
`=1/2T`
`=1/2xx6.28(((42300+6400)^3)/(10xx(6400)^2xx10^6))`
`=3.14(sqrt((479)^3xx10^6)/((64)^2xx10^11))`
`=3.14 sqrt((497xx497xx497)/(64xx64xx10^5))`
`=6 hours`