The eccentricity of earth’s orbit is 0.0167. The ratio of its maximum speed in its orbit to its minimum speed is

The correct option is 1.033.

Step-by-step Explanation:

According to Kepler's second law of planetary motion, the areal velocity of a planet revolving around the Sun in an elliptical orbit remains constant. This is a direct consequence of the conservation of angular momentum. Since no external torque acts on the system, the angular momentum of the planet about the Sun is conserved.

The angular momentum ($L$) of the Earth at any point in its orbit is given by:
$L=m·v·r$
where:
• $m$ is the mass of the Earth,
• $v$ is the orbital speed of the Earth, and
• $r$ is the distance of the Earth from the Sun.

At the closest point to the Sun (perihelion), the distance is minimum ($r_{\min }$) and the speed is maximum ($v_{\max }$). At the farthest point from the Sun (aphelion), the distance is maximum ($r_{\max }$) and the speed is minimum ($v_{\min }$).

By the conservation of angular momentum:
$m·v_{\max }·r_{\min }=m·v_{\min }·r_{\max }$
Simplifying this relation, we find the ratio of maximum speed to minimum speed:
$\frac{v_{\max }}{v_{\min }}=\frac{r_{\max }}{r_{\min }}$

For an elliptical orbit with semi-major axis $a$ and eccentricity $e$, the minimum and maximum distances are defined as:
$r_{\min }=a(1-e)$
and
$r_{\max }=a(1+e)$

Substituting these expressions into the speed ratio formula:
$\frac{v_{\max }}{v_{\min }}=\frac{a(1+e)}{a(1-e)}=\frac{1+e}{1-e}$

Given that the eccentricity of the Earth's orbit is $e=0.0167$, we substitute this value into the equation:
$\frac{v_{\max }}{v_{\min }}=\frac{1+0.0167}{1-0.0167}$
$\frac{v_{\max }}{v_{\min }}=\frac{1.0167}{0.9833}\approx 1.03397$

Rounding this ratio to three decimal places yields approximately 1.033.