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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$1.034$

Vedclass · Answer

(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

Similar Questions

The distance of a planet from the sun is $5$ times the distance between the earth and the sun. The time period of the planet is

The motion of satellites in the solar system is an example of which of the following?

Kepler discovered

Two planets move around the sun. The periodic times and the mean radii of the orbits are $T_1, T_2$ and $r_1, r_2$ respectively. The ratio $T_1/T_2$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module