(i) The closest we are to the sun is when we are at the vertex closest to the sun (point (150, 0) in the diagram below), and the furthest we are is when we are at the other vertex (point (−150, 0)).

Focus (−c, 0)
Sun at (c, 0)
−150
150
−150
150
closest
greatest
distance

Ellipse with minor axis of length 8 and vertex at (0,-5).

Note: In the graph, I have exaggerated the shape of the elliptical orbit (I squashed it a bit so it looks more elliptical) and the positions for the 2 foci are shown much further apart than they really are.

In the diagram above, units are in millions of kilometers.

The distance a = OA = 149,597,871 is the length of the semi-major axis of our ellipse.

We have assumed that the sun is at the right-hand focus, at the point (c, 0).

The vertices are at A (149 597 871, 0) and B (-149 597 871, 0).

We need to find c, to tell us where the foci are.

c=a/60

=149597871/60

=2,493,298

So the foci are at the points (-2 493 298, 0) and (2 493 298, 0).

The closest we are to the sun is

a c = 149 597 871 − 2 493 298 = 147 104 573 km

The furthest we are from the sun (distance from B to the sun) is:

OB + c = a + c = 149 597 871 + 2 493 298 = 152 091 169 km.

ii) The foci are 2 × a = 2 × 2\ 493\ 298 = 4\ 986\ 596 km apart.

The radius of the sun is around 1,400,000 km, so the 2nd focus is not all that much further out than the surface of the sun.

Our orbit is almost circular. (The eccentricity is very small at 1/60).

### Graphing the Ellipse

Note: To graph the ellipse above, I needed to find b, as follows:

b=sqrt(a^2-c^2)

=sqrt(149\ 597\ 871^2-2\ 493 \ 298^2)

=149\ 577\ 092

Using the formula for an ellipse,

x^2/a^2+y^2/b^2=1

the required curve is

x^2/(149\ 597\ 871^2)+y^2/(149\ 577\ 092^2)=1`