A planet is orbiting the sun in an elliptical | vedclass

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(A) For a planet orbiting the sun in an elliptical orbit,the total mechanical energy $E$ is given by $E = K + U$.Since the planet is bound to the sun,

Vedclass · Accepted Answer

$K < |U|$ always

Vedclass · Answer

(A) For a planet orbiting the sun in an elliptical orbit,the total mechanical energy $E$ is given by $E = K + U$.Since the planet is bound to the sun,the total energy $E$ is negative $(E From the virial theorem for a gravitational potential $V \propto r^{-1}$,the average kinetic energy $\langle K angle$ and average potential energy $\langle U angle$ are related by $\langle K angle = -\frac{1}{2} \langle U angle$.Specifically,at any point in the orbit,the potential energy is $U = -\frac{GMm}{r}$ and the kinetic energy is $K = \frac{GMm}{2r} + 	ext{constant terms depending on the orbit}$.More simply,for a bound orbit,the total energy $E = K + U Therefore,$K < |U|$ always holds true for a bound elliptical orbit.

Column $-I$	Column $-II$
$(A)$ Kinetic energy	$(p)$ $-\frac{GM_Em}{2r}$
$(B)$ Potential energy	$(q)$ $\sqrt{\frac{GM_E}{r}}$
$(C)$ Total energy	$(r)$ $-\frac{GM_Em}{r}$
$(D)$ Orbital velocity	$(s)$ $\frac{GM_Em}{2r}$

$A$ planet is orbiting the sun in an elliptical orbit. Let $U$ denote the potential energy and $K$ denote the kinetic energy of the planet at an arbitrary point on the orbit. Choose the correct statement.

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