Closer to the sun

Because the question involves times on the same orbit we can solve this by using Kepler's second law of planetary motion .

Assume that:

• A equals to half the area of the ellipse (which is, $\frac {cb\pi}{2}$ )
• B equals to the area of the triangle SPQ (which is $\frac {ec \times 2b}{2}$ = ecb ), where S is the point where the Sun is located (one of the focal points), and P and Q is the end points of the minor axis line
• T $_1$ and T $_2$ equals to the time that Earth spends on the right side and the left side of the ellipse, respectively
• A $_1$ and A $_2$ equals to the area swept by Earth on the right side and the left side of the ellipse, respectively

Because Kepler's second law of planetary motion stated that (in mathematical form), $\frac {dA}{dt}$ is constant, we can write an equation:

A $_1$ /T $_1$ = A $_2$ /T $_2$ $\Rightarrow$ T $_1$ :T $_2$ = A $_1$ :A $_2$

$\Rightarrow$ T $_1$ :T $_2$ = (A + B):(A - B) = ( $\frac {cb\pi}{2}$ + ecb):( $\frac {cb\pi}{2}$ - ecb) = ( $\frac {\pi}{2}$ + e):( $\frac {\pi}{2}$ - e)

And by using T $_1$ +T $_2$ = 365 days, we can get that: T $_1$ = 184.441 days, T $_2$ = 180.558 days, and T $_1$ - T $_2$ = 3.88 days

Referring to the diagram, we are basically looking for the difference in the number of days the Earth spends on the left half of the orbit compared to the right half. Since we know that the Earth takes approximately $365.25$ days to revolve around the Sun, we just need to calculate the ratio of the time the Earth spends in each half of its orbit. For each half-orbit, the top quarter of the orbit takes the same length of time as the bottom quarter (by symmetry), so we just need to look at the ratio of the time the Earth spends in the top-left quadrant of its orbit as compared to the top-right quadrant.

Let the points closest and furthest from the Sun (on the major axis) be $X$ and $Z$ respectively. Further, let $Y$ be the point dividing the top-left and top-right quarters of the orbit. The eccentric anomalies of $X$ , $Y$ and $Z$ are $0$ , $\frac{\pi}{2}$ and $\pi$ respectively. Using Kepler's equation $M = E - e \sin E$ , we can calculate the mean anomalies of $X$ , $Y$ and $Z$ as $0$ , $\frac{\pi}{2}-e$ and $\pi$ respectively. Now, we can calculate the time lapsed as the Earth moves from $X$ to $Y$ and from $Y$ to $Z$ using $M_1 - M_0 = n(t_1 - t_0)$ (where $n$ is the mean motion). The times are $\dfrac{\frac{\pi}{2}-e}{n}$ and $\dfrac{\frac{\pi}{2}+e}{n}$ respectively. So the ratio of the times is $\frac{\pi}{2}-e$ to $\frac{\pi}{2}+e$ . Thus, for every $\pi$ days that lapsed, the difference in the number of days spent in each half is $2e$ . Since each orbit takes $365.25$ days, then the difference in the number of days spent in each half is $\frac{365.25}{\pi} \cdot 2e = 3.88$ days.

Kepler's 2nd law says that the rate at which area that a planet sweeps out is a constant $\frac{dA}{dt} = \frac{L}{2M}$ Where M is the mass if the planet, L is the angular momentum of the planet, but we do not need all these constants.

We just need to find out the ratio of the difference in area that the Earth sweeps out in between both semi-ellipses to the total area of the ellipse.

The area of the ellipse is $\pi ab$ and the difference in the two areas is actually $2 \times$ the area of the triangle formed by the sun (which is at the focus of semi-ellipse 1) and the semi-minor axis b.

Hence the ratio is

$\frac{2bc}{ \pi ab} = \frac{2c}{\pi a} = \frac{2e}{\pi}$

e is the eccentricity which is given Multiply this by the number of days in a year which is 365.25, we will get 3.88 days ( 3 s.f.)

We start with Kepler's Second Law of Planetary Motion, which states that the radius vector from the Sun to the revolving planet (in our case it is Earth) sweeps out equal areas in equal amounts of time.

This implies that the time-rate of change of area is constant everywhere on Earth's path of revolution. Consequently, if Earth sweeps out an area $A_0$ in time $t_0$ and an area $A_1$ in time $t_1$ , we have

$\frac{A_0}{t_0} = \frac{A_1}{t_1} \Rightarrow \frac{A_0}{A_1} = \frac{t_0}{t_1}$

Now, suppose we want to find the time $t$ (in days) Earth spends in region 2 (as shown in the diagram given). We can then apply the equation above, which would give us

$\frac{ab+\frac{1}{2} \pi bc}{\pi bc} = \frac{t}{365}$

Substituting

$e=\frac{a}{c} =0.0167$

into the equation above would give us

$t\approx 184.4403$ days.

Now it's just

$t-(365-t) \approx 3.88$ days.

Kepler's equation relates position of an orbiting body to time:

$\frac {2πt}{P} = E - e * sin(E)$

where $P$ is the orbital period ( $3.15569 * 10 ^ 7 s$ for Earth), $E$ is the eccentric anomaly (the angle from perihelion in an idealized circular orbit), and $e$ is the eccentricity (given as $0.0167$ ). This can easily be solved for $t$ in terms of $E$ :

$t(E) = \frac {P}{2π} (E - e * sin(E))$

Solving for $t$ at the relevant angles for this problem:

$t1 = t(E = 0) = 0.00 s$

$t2 = t(E = \frac{π}{2}) = 7805350.31 s$

$t3 = t(E = \frac{3π}{2}) = 23751549.69 s$

$t4 = t(E = 2π) = 31556900.00 s$

The time spent in the first half of the orbit is therefore $t3 - t2$ , and the time spent in the second half is $(t4 - t3) + (t2 - t1)$ . The difference between these two quantities is $335498.77$ seconds, or $3.88$ days.

We know that the areal velocity is constant, then the difference in time is proportional to the difference in area just as the total time is proportional to the total area.

Using some geometry:

Draw a line from the sun to the highest and the lowest point of the ellipse. Now note that if we call the area of the half of the ellipse of "A" and the little triangle that formed from the lines of "B", will have that the area difference is:

K = (A+B) - (A-B) = 2B

The ratio of this difference to the total area is the same as the difference in time to the total time

B = 2bc (in terms of the ellipse parameters)

Total area = πab

T (total time) = 365.25 days

t = T((2bc)/(πab)) t = T(2/π)(c/a) = 3.88 days

Let us first determine the areas $A_{1}$ and $A_{2}$ as shown in the figure below. The total area of an ellipse with semi-major axis $a$ and semi-minor axis $b$ is given by the well known formula $A_{T}=\pi a b.$ The segments OT and OB divide the ellipse in two regions with areas $A_{1}= \frac{A_{T}}{2}- cb$ and $A_{2}=\frac{A_{T}}{2}+ cb.$ If $T_{1}$ and $T_{2}$ are the times that the Earth spends in regions 1 and 2 then from Kepler's second law we have that $\frac{A_{1}}{T_{1}}=\frac{A_{2}}{T_{2}}=\frac{A_{T}}{T_{year}},$ where $T_{year}=365 ~\mbox{days}.$ Now one can easily show that $\Delta{T}=T_{2}-T_{1}=\frac{A_{2}-A_{1}}{A_{T}} T_{year}=\frac{2bc}{\pi a b} T_{year}=\frac{2e}{\pi} T_{year}=3.88 ~\mbox{days}.$