UPSC Maths 2025 Paper 1 Q5c — Solution

10 marks · Section B

Question

Prove that the time taken by the Earth to travel over half of its orbit, which is separated by the minor axis and is remote from the Sun, when the Sun is at the focus of the elliptic orbit, is two days more than half of the year. The eccentricity of the orbit is taken as $\dfrac{1}{60}$ .

Technique

Use Kepler’s second law (constant areal velocity): the time over an arc equals the area swept by the radius vector from the focus divided by the constant areal velocity $\dfrac{\pi a b}{T}$ .

Solution

Place the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with centre at the origin and the Sun at the focus $S=(ae,0)$ . The minor axis is the line $x=0$ ; its endpoints are $L=(0,b)$ and $L'=(0,-b)$ . The minor axis cuts the orbit into two halves:

the half $x>0$ , near the Sun;
the half $x<0$ , remote from the Sun (this is the half in question).

By Kepler’s second law the radius vector from $S$ sweeps area at the constant rate $\frac{dA}{dt} = \frac{\text{total area}}{T} = \frac{\pi a b}{T},$ where $T$ is the orbital period (one year).

Area swept over the remote half. As the Earth moves along the far arc from $L=(0,b)$ to $L'=(0,-b)$ , the area swept out by $SL,\,SL'$ is the region bounded by the far arc and the two radii $SL,\,SL'$ . This equals $A_{\text{remote}} = \underbrace{\tfrac{1}{2}\pi a b}_{\text{area of the half-ellipse }x<0} \;+\; \underbrace{(\triangle S L L')}_{\text{triangle with base }LL'},$ because the focus $S$ lies on the near side ( $x>0$ ), so the triangle $SLL'$ must be added to the half-ellipse area to obtain the focal sector.

The triangle $SLL'$ has base $LL' = 2b$ along the $y$ -axis and height $ae$ (the $x$ -distance of $S$ from the minor axis), so $\triangle SLL' = \tfrac{1}{2}(2b)(ae) = abe.$

Hence $A_{\text{remote}} = \frac{\pi a b}{2} + abe.$

Time for the remote half. $t_{\text{remote}} = \frac{A_{\text{remote}}}{\pi a b / T} = T\left(\frac{1}{2} + \frac{e}{\pi}\right) = \frac{T}{2} + \frac{Te}{\pi}.$

So the remote half takes longer than half the year by $\Delta t = t_{\text{remote}} - \frac{T}{2} = \frac{T e}{\pi}.$

Numerical value. With $e = \dfrac{1}{60}$ and $T = 365.25$ days, $\Delta t = \frac{365.25}{\pi}\cdot\frac{1}{60} = \frac{365.25}{60\pi} \approx 1.938 \text{ days} \approx 2 \text{ days.}$

(With $T=365$ days, $\Delta t\approx 1.936$ days; in either case “two days more than half a year,” as required.)

Answer

$\boxed{\;t_{\text{remote}} = \frac{T}{2} + \frac{Te}{\pi},\qquad \Delta t = \frac{Te}{\pi} = \frac{365.25}{60\pi}\approx 1.94 \approx 2\ \text{days}.\;}$