The eccentricity of Earth's orbit is 0.0167. | vedclass

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(B) According to Kepler's second law,the angular momentum of a planet remains constant. The speed of a planet in an elliptical orbit is given by $v =

Vedclass · Accepted Answer

$1.034$

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(B) According to Kepler's second law,the angular momentum of a planet remains constant. The speed of a planet in an elliptical orbit is given by $v = \frac{L}{mr}$,where $L$ is angular momentum,$m$ is mass,and $r$ is the distance from the Sun.At perihelion (closest approach),$r_{\min} = a(1-e)$,so $v_{\max} = \frac{L}{m a(1-e)}$.At aphelion (farthest distance),$r_{\max} = a(1+e)$,so $v_{\min} = \frac{L}{m a(1+e)}$.The ratio of maximum speed to minimum speed is $\frac{v_{\max}}{v_{\min}} = \frac{1+e}{1-e}$.Given $e = 0.0167$,we have $\frac{v_{\max}}{v_{\min}} = \frac{1 + 0.0167}{1 - 0.0167} = \frac{1.0167}{0.9833} \approx 1.034$.

The eccentricity of Earth's orbit is $0.0167$. The ratio of its maximum speed in its orbit to its minimum speed is

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