# Question #88269

##### 1 Answer
Feb 16, 2017

The annual average distance is ${r}_{M} = \frac{b}{2} = 74788686$ km. The locus of this midpoint M is the parallel-to-orbit ellipse $\frac{\frac{1}{2} b \sqrt{1 - {e}^{2}}}{r} _ M = 1 + e \cos \theta$..

#### Explanation:

The distance between the Sun and the Earth varies from the

perihelion 147098291 km to the aphelion 152098233 km, nearly.

The mean is the semi major axis of the elliptic orbit a =149598262

km.

For these data, the eccentricity of the Earth's

elliptic orbit e =0.01671123, nearly.

The semi minor axis

b = 149577372 km, nearly.

In mathematical exactitude, the annual

average Earth-Sun distance = b =149577372 km.

So, the annual average of the equidistant point M is

${r}_{M} = \frac{b}{2} = 74788686$ km.

The instantaneous distance of the equidistant point M distant ${r}_{M}$ is

1/2((Sun-Earth distance r), at the instant.

The locus of the equidistant point is the ellipse is

$\frac{\frac{1}{2} b \sqrt{1 - {e}^{2}}}{r} _ M = 1 + e \cos \theta$

and this is parallel to the orbit.

Scaled down on an A4 paper, both the ellipses would appear as the

circles

${x}^{2} + {y}^{2} = {b}^{2} \mathmr{and} {x}^{2} + {y}^{2} = {b}^{2} / 4$

graph{(x^2+y^2-2.25)(x^2+y^2-0.5625)=0 [-5, 5, -2.5, 2.5]}