91Ó°ÊÓ

${\text{R}}_{\text{s}}=6.96×{10}^{\text{8}}\text{m}$

Eccentricity $\left(\text{e}\right)=0.0167$

Semi major axis$\left(\text{a}\right)=1.50×{10}^{11}\text{m}$

As the distance between the focus of the ellipse and the center of the ellipse iseccentricity × semi-major axis, calculate the distance between the two foci in meters of earth’s orbit both in meters and in terms of the radius of the sun.

The formula is as follows:

Distance (focus and center of the ellipse) = eccentricity × semi-major axis

From the diagram,

$\begin{array}{c}{\mathrm{Distance}}_{{\mathrm{F}}_1\text{ }\mathrm{and}{\mathrm{F}}_2}=((\mathrm{a}×\mathrm{e})+(\mathrm{a}×\mathrm{e}))\\ =((0.0167×1.50×{10}^{11}\text{ }\mathrm{m})+(0.0167×1.50×{10}^{11}\text{ }\mathrm{m}))\\ =(2.5×{10}^9\text{ }\mathrm{m})+(2.5×{10}^9\text{ }\mathrm{m})\\ =5.0×{10}^9\text{ }\mathrm{m}\end{array}$

Hence, the distance between two foci of the earth’s orbit is in meters ${\text{5.0×10}}^{\text{9}}\text{m}$.

Now,

${\mathrm{Distance}}_{{\mathrm{F}}_1\text{ }\mathrm{and}{\mathrm{F}}_2}=5.0×{10}^9\text{m}$

Multiply and divide to the right-hand side by the radius of the sun${\text{R}}_{\text{s}}$,

$\begin{array}{c}{\mathrm{Distance}}_{{\mathrm{F}}_1\text{ }\mathrm{and}{\mathrm{F}}_2}=5.0×{10}^9\text{m}×\frac{{\mathrm{R}}_{\mathrm{s}}}{{\mathrm{R}}_{\mathrm{s}}}\\ =5.0×{10}^9\text{m}×\frac{{\mathrm{R}}_{\mathrm{s}}}{6.96×{10}^8\text{ }\mathrm{m}}\\ =7.2×{\mathrm{R}}_{\mathrm{s}}\end{array}$

Hence, the distance between two foci of the earth’s orbit in terms of the solar radius${\mathrm{R}}_{\mathrm{s}}$is$7.2×{\mathrm{R}}_{\mathrm{s}}$.

Therefore, using the formula for the distance between the focus of the ellipse and the center of the ellipse, the distance between the two foci of the earth can be found.