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. `5.86kms^(-1)`
B. `9.13kms^(-1)`
C. `6.67kms^(-1)`
D. `7.76 kms^(-1)`

Answer

D

Explanation

`v_)=sqrt((GM)/r)`
`v_0=sqrt((gR^2)/(R+h))=sqrt((9.8xx(6.38xx10^6)^2)/(6.38xx10^6+0.25xx10^6))`
`=7.76kms^(-1)`.