Suppose the Earth's orbit is divided in two by its minor axis as shown in the figure. How much longer in days does the Earth spend in one half than the other? The eccentricity of an elliptical orbit is defined as

e:=a/c
where c is the semi-major axis and a is the distance from the the center of ellipse to a focal point. For the Earth's orbit this number is
e=0.0167.

To determine how much longer in days the Earth spends in one half of its orbit compared to the other, we need to consider the concept of orbital motion and the eccentricity of the Earth's orbit.

The Earth's orbit around the Sun is not a perfect circle but rather an ellipse, with the Sun located at one of the two foci of the ellipse. The eccentricity of an elliptical orbit quantifies how much the shape of the orbit deviates from a perfect circle.

In this case, the eccentricity of the Earth's orbit is given as e = 0.0167. This means that the distance between the center of the orbit and one of the foci is 0.0167 times the length of the semi-major axis.

To visualize this, imagine a horizontal line passing through the center of the ellipse, dividing the orbit into two equal halves (left and right). The minor axis is the line perpendicular to this horizontal line, passing through the center of the ellipse.

Since the eccentricity is relatively small for the Earth's orbit, the effect is minimal. However, one-half of the Earth's orbit will be slightly closer to the Sun than the other half.

To calculate the time difference, we need to know the orbital period of the Earth, which is approximately 365.25 days. This is the time it takes for the Earth to complete one full revolution around the Sun.

Given that the total orbital period is 365.25 days, the Earth spends half of this time (365.25/2 = 182.625 days) in each half of the orbit if the orbit were a perfect circle.

However, due to the eccentricity of the Earth's orbit, one half of the orbit will be slightly longer than the other. To calculate the time difference, we need to determine the angular speed of the Earth along its orbit.

From basic geometry, we know that the angular speed is given by ω = 2π / T, where T is the orbital period. Substituting in the value of T, we find ω = 2π / 365.25.

Now, the time spent in each half of the orbit is directly proportional to the angle swept by the Earth in that half. In other words, the longer the arc length, the longer the time spent.

Using the formula for arc length in polar coordinates, we have:

s = rθ

where s is the arc length, r is the distance from the center of the ellipse to the current position of the Earth, and θ is the angle swept by the Earth.

Since the average distance from the Earth to the Sun is the semi-major axis a (also referred to as r), we can express r in terms of the eccentricity e and the semi-major axis:

r = a(1 + e)

In the first half of the orbit, the angle swept by the Earth is 180 degrees, while in the second half, it is 360 degrees minus the angle swept in the first half.

Therefore, the time difference will be:

Δt = (θ2 - θ1) / ω

where θ1 = 180 degrees and θ2 = 360 degrees - θ1.

Simplifying this equation, we have:

Δt = (θ2 - θ1) / (2π / 365.25).

By substituting the values of θ1 and θ2 and evaluating the equation, we can find the time difference in days that the Earth spends in one half compared to the other.

Please note that the calculation can be quite complex and involve numerical methods. The given eccentricity value of 0.0167 is relatively small, so the time difference will also be relatively small.