An artificial satellite is orbiting at a height of 1800 km from the earth's surface. The earth's radius is 6300 km and `g=10m/s^(2)` on its surface. What is the radial acceleration of the satellite?

Options

A. `6 ms^(-2)`
B. `7 ms^(-2)`
C. `8 ms^(-2)`
D. `9 ms^(-2)`

Answer

A

Explanation

Radial acceleration = centripetal acceleration
` therefore (v^2)/( r) = (GM)/(r^2) = (gR^2)/((R + h)^2) = (10 xx (6300)^2)/((6300 + 1800)^2) = 6ms^(-2)`