Question
Define the term orbital speed. Establish a relation for orbital speed of a satellite orbiting very close to the surface of the earth. Find the ratio of this orbital speed and escape speed.
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Answer
Orbital speed: It is the minimum speed required to put the satellite into given orbit around the earth. 
M = mass of earth R = radius of earth m = mass of satellite $v_0$ = orbital velocity of satellite h = height of satellite above surface of earth. r = R + h According to Newton's law of gravitation, $\text{F} = \frac{\text{GMm}}{\text{r}^2}$ Centripetal force $\text{F} = \frac{\text{mv}_{\text{o}}}{\text{r}}$ In equilibrium, gravitational pull provides the required centripetal force. $\frac{\text{mv}^2_{\text{o}}}{\text{r}}=\frac{\text{GMm}}{\text{r}^2}$ $\Rightarrow\text{v}_{\text{o}}=\sqrt{\frac{\text{GM}}{\text{r}}}$ $\text{v}_{\text{o}}=\sqrt{\frac{\text{gR}^2}{\text{r}}}=\sqrt{\frac{\text{gR}^2}{\text{R+h}}}$ $\text{for h }<<\text{R}$ $\text{R}+\text{h}\approx\text{R}$ $\therefore\text{v}_{\text{o}}=\sqrt{\text{gR}}$ $\Rightarrow\text{v}_{\text{e}}=\sqrt{2\text{gR}}$ $\frac{\text{v}_{\text{o}}}{\text{v}_{\text{e}}}=\frac{1}{\sqrt{2}}$
M = mass of earth R = radius of earth m = mass of satellite $v_0$ = orbital velocity of satellite h = height of satellite above surface of earth. r = R + h According to Newton's law of gravitation, $\text{F} = \frac{\text{GMm}}{\text{r}^2}$ Centripetal force $\text{F} = \frac{\text{mv}_{\text{o}}}{\text{r}}$ In equilibrium, gravitational pull provides the required centripetal force. $\frac{\text{mv}^2_{\text{o}}}{\text{r}}=\frac{\text{GMm}}{\text{r}^2}$ $\Rightarrow\text{v}_{\text{o}}=\sqrt{\frac{\text{GM}}{\text{r}}}$ $\text{v}_{\text{o}}=\sqrt{\frac{\text{gR}^2}{\text{r}}}=\sqrt{\frac{\text{gR}^2}{\text{R+h}}}$ $\text{for h }<<\text{R}$ $\text{R}+\text{h}\approx\text{R}$ $\therefore\text{v}_{\text{o}}=\sqrt{\text{gR}}$ $\Rightarrow\text{v}_{\text{e}}=\sqrt{2\text{gR}}$ $\frac{\text{v}_{\text{o}}}{\text{v}_{\text{e}}}=\frac{1}{\sqrt{2}}$
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