AP EAPCET 24 May 2025 Shift 1 Paper

Step 1: Escape velocity from the Earth's surface$v_{\text{escape}} = \sqrt{\frac{2GM}{R}} = 11.2\,\mathrm{km}\,\mathrm{s}^{-1}$Here $G$ is the gravitational constant, $M$ is the mass of the Earth and $R$ is the radius of the Earth.Step 2: Orbital velocity of a satellite at height $h = R$The orbital velocity at a distance $r$ from the centre of the Earth is $v_{\text{orbit}} = \sqrt{\frac{GM}{r}}$.For a height $h = R$, the orbital radius is $r = R + R = 2R$.So $v_{\text{orbit}} = \sqrt{\frac{GM}{2R}}$.Step 3: Compare the two velocities$\frac{v_{\text{orbit}}}{v_{\text{escape}}} = \frac{\sqrt{\frac{GM}{2R}}}{\sqrt{\frac{2GM}{R}}} = \sqrt{\frac{GM}{2R}\cdot\frac{R}{2GM}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$Hence the orbital velocity is half the escape velocity.Step 4: Substitute the given value$v_{\text{orbit}} = \frac{v_{\text{escape}}}{2} = \frac{11.2}{2} = 5.6\,\mathrm{km}\,\mathrm{s}^{-1}$Therefore the correct option is $5.6\,\mathrm{km}\,\mathrm{s}^{-1}$.