The orbit of the planet mercury around the sun is in an elliptical shape with the sun at a focus. The semi-major axis is of length 36 million miles and the eccentricity of the orbit is 0.206. Find
- how close does mercury get to the sun;
- the greatest possible distance between mercury and sun.
Let the equation of the elliptical orbit be: $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$
Center of the ellipse is at $C \left(0,0\right)$.
Let the sun be at focus $F_1$.
Vertices of the ellipse are $A_1$ and $A_2$.
Semi-major axis of the ellipse $= CA_1 = CA_2 = a = 36$ million miles
Eccentricity of the elliptical orbit $= e = 0.206$
Now, $CF_1 = a \times e = 36 \times 0.206 = 7.416$ million miles
-
When mercury is at point $A_1$, it is closest to the sun.
$\therefore$ $\;$ Required distance $= F_1 A_1$
Now, $F_1 A_1 = CA_1 - CF_1 = 36 - 7.416 = 28.584$ million miles
-
When mercury is at the point $A_2$, it is farthest from the sun.
$\therefore$ $\;$ Required distance $= F_1 A_2$
Now, $F_1 A_2 = CF_1 + CA_2 = 7.416 + 36 = 43.416$ million miles