Kepler's connection — orbits are ellipses (motivation): Question bank

This page assumes only the ideas from the parent topic: an ellipse is the set of points whose two focal distances sum to a constant $2 a$ ; the Sun sits at one focus; $r_{m i n} = a (1 - e)$ , $r_{m a x} = a (1 + e)$ ; eccentricity $e$ measures how off-centre the focus is. Everything here is pure reasoning — no heavy computation (that lives in D3/D4).

First, the picture and the vocabulary these traps rely on

Before any trap makes sense, look at the labelled ellipse below. We draw it the standard way: centred at the origin, with its long (major) axis lying along the horizontal $x$ -axis and its short (minor) axis along the vertical $y$ -axis. This choice is just convenient bookkeeping — you could rotate the whole picture and nothing physical changes.

Figure — Kepler's connection — orbits are ellipses (motivation)

Keep this figure in mind: every time a trap mentions " $(0, b)$ ", "the centre", "the focus", or "the minor axis", it is pointing at a labelled part of this drawing.

True or false — justify

The Sun sits at the centre of a planet's elliptical orbit.

False. The Sun sits at a focus, which is offset from the centre by

c = a e

(see the amber dots in the figure); the geometric centre of the ellipse is empty space. It only looks central for planets because their

e

is tiny.

For a perfectly circular orbit the two foci merge into a single point at the centre.

True. A circle is an ellipse with

e = 0

, so

c = a e = 0

— both foci collapse onto the centre, and every point is the same distance from it.

The semi-major axis

a

equals the planet's average distance to the Sun over a full orbit.

False.

a

is the average of the two extremes only,

a = \frac{1}{2} (r_{m i n} + r_{m a x})

. The planet spends far more of its time out near aphelion (where Kepler's 2nd law says it dawdles) than near perihelion (where it races through), so its time-averaged distance is pulled above

a

At the ends of the minor axis, a planet's distance to the Sun equals exactly

a

True. The top point

(0, b)

marked in the figure is equidistant from both foci, and the two distances sum to

2 a

, so each is exactly

a

. This is a special-point coincidence, not a general average.

A larger orbit (bigger

a

) means the planet moves faster.

False. By Kepler's 3rd law

T^{2} \propto a^{3}

, so a bigger

a

gives a longer period and a longer path — the average speed drops. Distant planets crawl.

Two orbits with the same

a

but different

e

have the same period.

True. Kepler's 3rd law depends only on

a

, not on

e

. A near-circular and a highly elongated orbit of equal semi-major axis complete one loop in the same time.

Eccentricity

e

can be negative if the focus is on the "other side".

False.

e = c / a

is a ratio of two lengths (both non-negative), so

0 \leq e < 1

for an ellipse. "Which side" the focus lies on is just a choice of coordinates, not a sign of

e

e = 1

, the shape is still a (very stretched) ellipse.

False. At

e = 1

the curve is no longer closed — it becomes a parabola. For

e > 1

it opens into a hyperbola. Ellipses require

0 \leq e < 1

strictly.

The planet's speed is greatest at aphelion because it has "fallen" the farthest.

False. Speed is greatest at perihelion (nearest point). Kepler's 2nd law demands equal areas in equal times; with a short radius arm the planet must swing through a bigger angle, so it moves faster there.

Spot the error

"Since

b^{2} = a^{2} (1 - e^{2})

and

e < 1

, we get

b > a

Error:

1 - e^{2} < 1

, so

b^{2} < a^{2}

, giving

b < a

. The semi-minor axis (the short one on the figure) is always the smaller — that's why it's called minor.

r_{m i n} + r_{m a x} = a

, so add the extremes to get the semi-major axis."

Error: The sum is

a (1 - e) + a (1 + e) = 2 a

, which is the full major axis. You must average (divide by 2) to get the semi-major axis

a

"Earth's orbit has

e = 0.0167

, so the Sun is about

1.7%

of the way from centre to edge."

Error: The offset is

c = a e \approx 0.0167 a

, i.e. the Sun is

1.7%

of the semi-major axis from the centre — measured along the major axis, not "toward the edge" in some vague sense. The nearest edge is at distance

a - c = a (1 - e)

"Kepler's 2nd law says the planet moves at constant speed because it sweeps equal areas."

Error: It sweeps equal areas, not equal arc lengths or equal angles. Constant areal rate forces the speed to change — faster near the Sun, slower far away.

T^{2} \propto a^{3}

means doubling the period doubles the orbit size."

Error: Doubling

T

multiplies

a

2^{2/3} \approx 1.59

, not by

2

. The cube-square relationship is nonlinear.

"The perihelion is where the ellipse is 'pointiest' (highest curvature), so the Sun sits there."

Error: The Sun sits at a focus inside the ellipse (an amber dot on the figure), not on the curve. Perihelion is the point of the orbit nearest that focus — it lies on the major axis, on the ellipse, at distance

a (1 - e)

from the Sun.

Why questions

Why must gravity be an inverse-square force (

F \propto 1/ r^{2}

) for orbits to be conics that close up into ellipses?

A deep result (Bertrand's theorem) says only two force laws make every bound orbit close into a perfectly repeating loop: the inverse-square law

1/ r^{2}

and the spring law

\propto r

. For any other exponent the ellipse's long axis slowly rotates each lap, tracing a rosette (petal-like) pattern that never closes — so the clean, fixed ellipse of Kepler's 1st law is special to

1/ r^{2}

Why is the focus (not the centre) the physically meaningful point of an orbital ellipse?

Because the focus is where the attracting mass actually sits — the focus–directrix distance

r

to that point is the real Sun–planet distance, so the abstract geometric focus becomes a physical location.

Why does the planet speed up near the Sun, phrased through angular momentum?

Conservation forces

r^{2} d θ / d t

to stay constant; when

r

shrinks near perihelion,

d θ / d t

must grow, so the planet sweeps faster — exactly the equal-areas statement in disguise.

Why does

e = \frac{r _{m a x} - r _{m i n}}{r _{m a x} + r _{m i n}}

measure "how off-centre" the focus is?

The numerator

r_{m a x} - r_{m i n} = 2 c

is twice the focus offset; the denominator is

2 a

. So

e = c / a

— the offset expressed as a fraction of the semi-major axis. Bigger fraction = more lopsided orbit.

Why does the same eccentricity

e

appear both in the geometry of a conic slice and in an orbit?

Because both describe the same shape — a body under inverse-square force traces a conic, and slicing a cone produces those very curves. One

e

characterises the shape however it arises.

Why is a comet's orbit "highly eccentric" while a planet's is nearly circular?

A comet's perihelion is tiny compared to its aphelion, making

r_{m a x} - r_{m i n}

nearly as large as

r_{m a x} + r_{m i n}

, so

e \to 1

. A planet's two extremes are almost equal, so

e \to 0

Edge cases

What happens to the orbit shape as

e \to 0^{+}

The foci merge to the centre and the ellipse becomes a circle of radius

a

;

r_{m i n}

and

r_{m a x}

both approach

a

, so the "nearest" and "farthest" distinction disappears.

What happens as

e \to 1^{-}

(but still less than 1)?

The ellipse becomes extremely elongated — a cigar.

r_{m i n} = a (1 - e) \to 0

(the planet grazes the Sun) while

r_{m a x} = a (1 + e) \to 2 a

. The orbit is still closed, but barely.

At exactly

e = 1

, what changes qualitatively about the motion?

The orbit stops being closed: it becomes a parabola, the boundary case of a body with just enough energy to escape — it slows to zero speed only at infinity and never returns.

r_{m i n} = a (1 - e)

still valid when

e = 0

Yes — it gives

r_{m i n} = a

, the circle radius, matching

r_{m a x} = a

too. The formula degrades gracefully into the circular case with no contradiction.

Can

r_{m i n} = 0

for a real orbit, and what would it mean physically?

Mathematically

r_{m i n} = 0

needs

e = 1

; physically it means the planet's path passes through the focus (a collision course with the Sun). Real orbits keep

e < 1

and

r_{m i n} > 0

, so the planet clears the star.

What does

b^{2} = a^{2} (1 - e^{2})

predict at the two extremes

e = 0

and

e \to 1

e = 0

b = a

(a circle, both axes equal). As

e \to 1

b \to 0

— the ellipse collapses toward a line segment along the major axis, consistent with the cigar-shape limit.

If the orbit were a hyperbola instead, why can't we speak of a period

T

A hyperbolic path is unbound — the body sweeps in once and leaves forever, never repeating. With no repetition there is no period, so Kepler's 3rd law (

T^{2} \propto a^{3}

) simply does not apply.

Connections

Parent topic (Hinglish)
Ellipse — standard equation and focus
Eccentricity of a conic
Focus–directrix definition of conics
Conic sections as slices of a cone
Parabola — projectile motion connection
Hyperbola — unbound orbits and escape
Angular momentum conservation