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. `9.13 kms^(-1)`
B. `6.67 kms^(-1)`
C. `7.76 kms^(-1)`
D. `8.56 kms^(-1)`

Answer

C

Explanation

The orbital speed of the satellite is
`v_0 =R sqrt((g)/((R + h))`
where R is the earth.s radius, g is the acceleration due to gravity on earth.s surface and h is the height above the surface of earth.
` therefore v_0 = (6.38 xx 10^6 m) sqrt(( 9.8 ms^(-2))/(6.38 xx 10^6 m + 0.25 xx 10^6 m))`
` = 7.76 xx 10^3 ms^(-1) = 7.76 km s^(-1) `