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(B) The total mechanical energy of a satellite in an elliptical orbit is given by $E = -\frac{GMm}{2a}$,where $M$ is the mass of the planet,$m$ is the

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$\sqrt{\frac{3GM}{a}}$

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(B) The total mechanical energy of a satellite in an elliptical orbit is given by $E = -\frac{GMm}{2a}$,where $M$ is the mass of the planet,$m$ is the mass of the satellite,and $a$ is the semi-major axis.According to the law of conservation of energy,the total energy at any point in the orbit is the sum of kinetic energy and potential energy:$E = K + U = \frac{1}{2}mv^2 - \frac{GMm}{r}$At a distance $r = a/2$ from the planet,the equation becomes:$-\frac{GMm}{2a} = \frac{1}{2}mv^2 - \frac{GMm}{a/2}$$-\frac{GMm}{2a} = \frac{1}{2}mv^2 - \frac{2GMm}{a}$Rearranging to solve for $v^2$:$\frac{1}{2}mv^2 = \frac{2GMm}{a} - \frac{GMm}{2a}$$\frac{1}{2}mv^2 = \frac{GMm}{a} \left( 2 - \frac{1}{2} \right) = \frac{GMm}{a} \left( \frac{3}{2} \right)$$v^2 = \frac{3GM}{a}$$v = \sqrt{\frac{3GM}{a}}$

$A$ satellite is revolving around a planet of mass $M$ in an elliptical orbit of semi-major axis $a$. What is the speed of the satellite when it is at a distance $a/2$ from the planet?

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