Gravitation

Step 1: Escape velocity from EarthThe escape velocity from the Earth's surface is given by:$v_{\text{escape}} = \sqrt{\frac{2GM}{R}}$Here, $G$ is the gravitational constant, $M$ is Earth's mass, and $R$ is Earth's radius.Step 2: Orbital velocity at height $h=R$The orbital velocity for a satellite at a distance $r$ from the center of the Earth is:$v_{\text{orbit}} = \sqrt{\frac{GM}{r}}$If the satellite is at a height equal to the Earth's radius ( $h=R$ ), the total distance from Earth's center is $r=R+R=2R$.So, the orbital velocity becomes:$v_{\text{orbit}} = \sqrt{\frac{GM}{2R}}$Step 3: Find the relationship between escape velocity and orbital velocityLet's compare the orbital and escape velocities:$\frac{v_{\text{orbit}}}{v_{\text{escape}}} = \frac{\sqrt{\frac{GM}{2R}}}{\sqrt{\frac{2GM}{R}}}$Simplifying this gives:$\frac{v_{\text{orbit}}}{v_{\text{escape}}} = \frac{1}{2}$Step 4: Calculate the final answerThis means the orbital velocity is half the escape velocity. Given that the escape velocity is $11.2\,\mathrm{km}\,\mathrm{s}^{-1}$ :$v_{\text{orbit}} = \frac{11.2}{2} = 5.6\,\mathrm{km}\,\mathrm{s}^{-1}$