A remote-sensing satellite of earth revolves in a circular orbit at a hight of `0.25xx10^(6)m` above the surface of earth. If earth's radius is `6.38xx10^(6)m` and `g=9.8ms^(-2)`, then the orbital speed of the satellite is

Options

A. 6.67 km `s^(-1)`
B. 7.76 km `s^(-1)`
C. 8.56 km `s^(-1)`
D. 9.13 km `s^(-1)`

Answer

B

Explanation

`(GMm)/(r^(2))=(mv^(2))/(r)" "r=R+h`
`v=sqrt((GM)/(r))=sqrt((GMR^(2))/(R^(2)r))=sqrt((g)/(r))R`
`v=(sqrt((9.8)/(.25xx10^(6)+6.38xx10^(6))))xx6.38xx10^(6)`
`sqrt((1.47)/(10^(6)))xx6.38xx10^(6)`
`=7.76xx10^(3)m//s=7.76km//s`