Question

An artificial satellite is revolving in a circular orbit at height of $1200km$ above the surface of the earth. If the radius of the earth is $6400km$ and mass is $6\times {10}^{24}$ kg the orbital velocity $(G=6.67\times {10}^{-11}N{m}^2/k{g}^2)$ is

• A. $7.26$ kms${}^{-1}$
• B. $4.26$ kms${}^{-1}$
• C. $9.26$ kms${}^{-1}$
• D. $2.26$ kms${}^{-1}$

Answer 1

The correct option is A $7.26$ kms${}^{-1}$

Orbital speed is given by, $v=\sqrt{\frac{GM}{R+h}}$

Substituting the given values:

$v=\sqrt{\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}}{(6400+1200)\times {10}^3}}=7256.57$ $m/s$ $=7.26$ $km/s$