Earth's orbit is an ellipse with eccentricity 0.0167. Thus, earth's distance from the sun and speed as it move

NCERT Exemplar · Q37

Q. Earth's orbit is an ellipse with eccentricity 0.0167. Thus, earth's distance from the sun and speed as it moves around the sun varies from day to day. This means that the length of the solar day is not constant through the year. Assume that earth's spin axis is normal to its orbital plane and find out the length of the shortest and the longest day. A day should be taken from noon to noon. Does this explain variation of length of the day during the year?

Karnataka PUCLong· 5mImportance★★★★★est

99% · 66/67 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Conservation of Angular Momentum

Conservation of Angular Momentum

The Intuition First

Imagine you're sitting on a spinning office chair with your arms stretched out. Someone gives you a gentle push, and you start rotating slowly. Now pull your arms in tight against your chest. What happens? You spin faster. Push your arms back out — you slow down again.

Nothing external pushed you to go faster or slower. The change came from inside — from how you arranged your mass relative to the axis of rotation.

That's the core idea: Angular momentum is a quantity that stays constant for a rotating system unless an external torque acts on it. When you pulled your arms in, you didn't change your angular momentum — you changed your distribution of mass, and your rotation speed had to adjust to keep the total constant.

The Precise Statement

\vec{L} = \text{constant} \quad \text{if} \quad \vec{\tau}_{\text{ext}} = 0

Where:

$\vec{L}$ is the angular momentum of the system
$\vec{\tau}_{\text{ext}}$ is the net external torque acting on the system

In words: The total angular momentum of an isolated system (no external torque) remains constant in both magnitude and direction.

What Is Angular Momentum?

For a point mass $m$ moving with velocity $\vec{v}$ at position $\vec{r}$ from a reference point:

\vec{L} = \vec{r} \times m\vec{v}

For a rigid body rotating about a fixed axis:

L = I\omega

Where:

$I$ = moment of inertia (how mass is distributed relative to the axis)
$\omega$ = angular velocity (how fast it spins)

Note

Moment of inertia $I$ depends on where the mass is, not just how much. Mass far from the axis gives larger $I$ ; mass close to the axis gives smaller $I$ .

Why It Works: The Physics

Newton's second law for rotation says:

\tau_{\text{ext}} = \frac{dL}{dt}

If $\tau_{\text{ext}} = 0$ , then $\frac{dL}{dt} = 0$ , so $L$ is constant.

Since $L = I\omega$ , if $I$ changes (you pull arms in), $\omega$ must change in the opposite way to keep $L$ the same:

I_1 \omega_1 = I_2 \omega_2

Smaller $I$ → larger $\omega$ (spin faster). Larger $I$ → smaller $\omega$ (spin slower).

Real-World Examples

Situation	What happens	Why
Ice skater pulling arms in	Spins faster	$I$ decreases, $\omega$ increases to keep $L$ constant
Diver tucking into a ball	Rotates faster in midair	Same principle — no external torque during flight
Cat falling upside-down	Twists body to land on feet	Changes $I$ of different body parts to rotate without external torque
Planet orbiting the Sun	Speeds up when closer, slows down when farther	Gravitational force is central (torque = 0), so $L$ is constant

Watch out

Angular momentum is a vector. Its direction matters too. If no external torque acts, the axis of rotation stays fixed in space. This is why a spinning gyroscope or a bicycle wheel resists being tilted.

Common Mistake to Avoid …

Start your 14-day free trial to unlock the full solution →

Because Earth moves faster near perihelion and slower near aphelion (Kepler's second law), the extra rotation Earth must make each day to bring the Sun back overhead is slightly larger near perihelion and slightly smaller near aphelion. Working through the numbers, the solar day is about 7.9 seconds longer than average at perihelion and about 7.9 seconds shorter than average at aphelion — a total swing of roughly 16 seconds. This is far too small to explain the large day-length changes we notice through the seasons; those come from Earth's axial tilt, not from orbital eccentricity.

Why a solar day isn't simply Earth's rotation period

A solar day (noon to noon) is the time for the Sun to return to the same position in the sky, as seen from a fixed point on Earth. Because Earth also moves along its orbit while it spins, it must rotate slightly more than one full turn (relative to the distant stars) to bring the Sun back overhead — the extra bit of rotation needed each day equals the angle Earth has swept along its orbit in that time.

If Earth's orbital speed were constant (a circular orbit), that extra angle would be the same every day, and every solar day would be exactly the same length. But Earth's real orbit is a slightly eccentric ellipse ( $e=0.0167$ ), so by Kepler's second law (equal areas in equal times), Earth moves fastest at perihelion (closest to the Sun, around early January) and slowest at aphelion (farthest, around early July). That means the extra angle to sweep each day is bigger near perihelion and smaller near aphelion — so the solar day itself is longer near perihelion and shorter near aphelion.

Setting up the sizes involved

Let $n=2\pi/T_{year}$ be Earth's mean orbital angular speed, and let $\omega_s$ be Earth's (constant) spin rate relative to the stars. Because Earth completes about one extra rotation per year just from orbiting the Sun, $\omega_s$ and $n$ are related by $\omega_s - n \approx 365.25\,n$ (in other words, over the roughly 365.25 solar days in a year, the mean solar day exactly averages out to $D_{mean}=2\pi/(\omega_s-n)=86400$ s).

By conservation of angular momentum ( $r^2\dot\theta=$ constant) with $r_P=a(1-e)$ and $r_A=a(1+e)$ , and keeping only first-order terms in the small eccentricity $e=0.0167$ , the orbital angular speed at the two extremes works out to

\dot\theta_P \approx n(1+2e), \qquad \dot\theta_A \approx n(1-2e)

so the orbital rate swings by $\pm2ne$ around its mean value $n$ .

From orbital-rate swing to solar-day-length swing

The solar day length is $D=2\pi/(\omega_s-\dot\theta)$ . A small deviation $\delta=\dot\theta-n$ in the orbital rate produces a fractional change in $D$ of

\frac{\Delta D}{D_{mean}} \approx \frac{\delta}{\omega_s-n} = \frac{\delta}{365.25\,n}

At perihelion, $\delta_P=2ne$ , so

\frac{\Delta D_P}{D_{mean}} = \frac{2ne}{365.25\,n} = \frac{2e}{365.25} = \frac{2\times0.0167}{365.25} \approx 9.15\times10^{-5}

\Delta D_P \approx 86400\times9.15\times10^{-5} \approx 7.9\text{ s}

By the same calculation with the opposite sign at aphelion, $\Delta D_A\approx-7.9$ s.

The result …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-A1 markMCQ
$0.625 \mathrm{rads}^{-1}$
›Reveal solutionSolution
The bullet embeds at the door’s centre, conserving angular momentum about the hinge. Using the given moment of inertia and final angular speed, the bullet’s speed is found to be 500 m s⁻¹, matching option (D).

Concept & Intuition

This is a classic angular momentum conservation problem. The bullet is moving linearly before impact, but after it sticks to the door, the system rotates as a rigid body. No external torque acts about the hinge (frictionless pivot), so the total angular momentum just before impact equals the total angular momentum just after. The trick: the bullet’s linear momentum must be converted to angular momentum about the hinge, using the perpendicular distance from the hinge to the bullet’s path.

Step-by-step solution

Identify the system and the conserved quantity The bullet + door form an isolated system for rotation about the hinge (no external torque). Therefore, angular momentum about the hinge is conserved:

$L_{\text{before}} = L_{\text{after}}.$

Angular momentum of the bullet before impact The bullet moves horizontally, perpendicular to the door’s face, and strikes at the centre of the door. The distance from the hinge to the centre is half the door’s width:

$r = \frac{L}{2} = \frac{1.0\ \text{m}}{2} = 0.5\ \text{m}.$
The bullet’s linear momentum is $m_b v$ . Its angular momentum about the hinge is
$L_{\text{bullet}} = (\text{momentum}) \times (\text{perpendicular distance}) = m_b v \, r.$
So
$L_{\text{before}} = m_b v \cdot \frac{L}{2}.$

Angular momentum after the bullet embeds After embedding, the bullet and door rotate together with angular velocity $\omega = 0.625\ \text{rad s}^{-1}$ . The total moment of inertia about the hinge is the sum of the door’s and the bullet’s:

Door: $I_{\text{door}} = \frac{M L^2}{3}$ (given).

Bullet: treated as a point mass at distance $r = L/2$ , so $I_{\text{bullet}} = m_b r^2 = m_b (L/2)^2$ . Hence

$I_{\text{total}} = \frac{M L^2}{3} + m_b \left(\frac{L}{2}\right)^2.$
The angular momentum after is
$L_{\text{after}} = I_{\text{total}} \, \omega.$

Plug in the numbers Mass of door $M = 12\ \text{kg}$ , width $L = 1.0\ \text{m}$ , bullet mass $m_b = 10\ \text{g} = 0.010\ \text{kg}$ , $\omega = 0.625\ \text{rad s}^{-1}$ .

$I_{\text{door}} = \frac{12 \times (1.0)^2}{3} = 4\ \text{kg m}^2.$
Start your 14-day free trial to see the full solution →
COMEDK 2026Set 2026-A1 markMCQ
$1.5 \mathrm{~ms}^{-1}$
›Reveal solutionSolution

The man's angular speed relative to the ground is $\omega=v/R=0.5$ rad/s, giving a period $T=2\pi/\omega\approx12.57$ s — option (A).

The man walks with speed $v=1.5\ \text{m s}^{-1}$ relative to the ground along a circle of radius $R=3\ \text{m}$ . His angular speed relative to the ground is

$\omega=\frac{v}{R}=\frac{1.5}{3}=0.5\ \text{rad s}^{-1}$ …
Start your 14-day free trial to see the full solution →
KCET 2025Set D-41 markMCQ
$r_p, v_p, L_p$
›Reveal solutionSolution
Perihelion is the nearest point ( $r_p < r_a$ ); gravity exerts no torque about the Sun so angular momentum is conserved ( $L_p = L_a$ ); and $mv_pr_p = mv_ar_a$ then forces the planet to move faster where it is closer ( $v_p > v_a$ ).

Step 1 — Fix the vocabulary

For an elliptical orbit with the Sun at one focus:

Perihelion (peri- = near, helios = Sun) — the point of closest approach.

Aphelion (apo- = away) — the farthest point.

Therefore, immediately:
$\boxed{r_p < r_a}$
This single fact already eliminates options (A) and (C), both of which claim $r_p > r_a$ .

Step 2 — Angular momentum is conserved (the central-force argument)

The gravitational force on the planet points along the line joining it to the Sun — it is a central force, i.e. $\vec{F} \parallel \vec{r}$ . The torque about the Sun is therefore
$\vec{\tau} = \vec{r} \times \vec{F} = 0 \qquad (\text{since } \vec{F} \text{ is antiparallel to } \vec{r})$
And since
$\vec{\tau} = \frac{d\vec{L}}{dt} = 0 \quad\Longrightarrow\quad \vec{L} = \text{constant throughout the orbit}$
So the angular momentum has the same value at every point of the orbit — in particular:
$\boxed{L_p = L_a}$
This eliminates option (D), which claims $L_a < L_p$ .

This conservation is Kepler's Second Law (the areal-velocity law): $\dfrac{dA}{dt} = \dfrac{L}{2m} = \text{constant}$ — equal areas swept in equal times.

Step 3 — Deduce the speeds from conserved $L$ …
Start your 14-day free trial to see the full solution →
COMEDK 2025Set 2025-A1 markMCQ
$p$
›Reveal solutionSolution
The bullet’s linear momentum becomes angular momentum about the hinge; using conservation of angular momentum and the door’s moment of inertia gives the angular speed as $\omega = \frac{p}{8}$ , so the correct option is (C).

Concept and intuition:

When the bullet embeds in the door, it’s an inelastic collision, but angular momentum about the hinge is conserved because there’s no external torque (frictionless hinge). The bullet’s linear momentum $p$ is converted into angular momentum about the hinge. The door rotates as a rigid body, so we need its moment of inertia about the hinge. The key is to treat the bullet as a point mass at the center of the door, and the door as a uniform rod rotating about one end.

Step-by-step solution:

Set up the angular momentum conservation. Before impact, only the bullet has angular momentum about the hinge. The bullet strikes at the center of the door, which is at a distance $r = 0.5\,\text{m}$ from the hinge (since the door is 1.0 m wide). Its linear momentum is $p$ , so its angular momentum about the hinge is

$L_{\text{bullet}} = r \times p = (0.5)\,p.$
(Direction: into the page, say positive.)

Moment of inertia of the door about the hinge. The door is a uniform rectangular slab, width 1.0 m, mass $M = 12\,\text{kg}$ . For a rod of length $L$ rotating about one end, the moment of inertia is

$I_{\text{door}} = \frac{1}{3} M L^2.$
Here $L = 1.0\,\text{m}$ , so
$I_{\text{door}} = \frac{1}{3} \times 12 \times (1.0)^2 = 4\,\text{kg·m}^2.$

Moment of inertia of the embedded bullet. The bullet becomes a point mass $m$ at distance $r = 0.5\,\text{m}$ from the hinge. Its contribution is

$I_{\text{bullet}} = m r^2 = m (0.5)^2 = 0.25\,m.$
We don’t know $m$ directly, but we know its momentum $p = m v$ . We’ll keep it symbolic.

Total moment of inertia after embedding.

$I_{\text{total}} = I_{\text{door}} + I_{\text{bullet}} = 4 + 0.25\,m.$

Apply conservation of angular momentum. Initial angular momentum = final angular momentum:

$0.5\,p = I_{\text{total}} \, \omega = (4 + 0.25\,m)\,\omega.$
So
$\omega = \frac{0.5\,p}{4 + 0.25\,m}.$

Eliminate $m$ using $p = m v$ . We need another relation. The bullet’s speed $v$ is not given, but we can express $m$ in terms of $p$ and $v$ . However, we don’t have $v$ . Wait — is there a hidden assumption? Actually, the problem gives only $p$ and the door’s properties. This suggests the answer should be independent of $m$ and $v$ individually. Let’s check:

$\omega = \frac{0.5\,p}{4 + 0.25\,m}.$
This still contains $m$ . But note: the bullet’s mass is not specified, so the answer must be in terms of $p$ only. That means the denominator must simplify to a pure number. How?

The bullet’s momentum $p$ is given, but its mass is not. However, the bullet’s mass is typically much smaller than the door’s mass, but here it’s not stated. The only way the answer is a pure number times $p$ is if the bullet’s mass is negligible compared to the door? But then $m \approx 0$ gives $\omega \approx 0.5p/4 = p/8$ . That matches option (C). …
Start your 14-day free trial to see the full solution →
COMEDK 2024Set 2024-E1 markMCQ
$2 \mathrm{~kg}$
›Reveal solutionSolution

Conserve angular momentum about the axis: the particle adds $mvR=1.0$ to the cylinder's $I\omega=0.2$ ; dividing by the new moment of inertia $0.08$ gives $\omega'=15\text{ rad s}^{-1}$ .

The collision has no external torque about the cylinder's axis, so angular momentum about that axis is conserved.

Moment of inertia of the solid cylinder:
$I_{\text{cyl}} = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(2)(0.2)^2 = \tfrac{1}{2}(2)(0.04) = 0.04\ \text{kg·m}^2.$
Initial angular momentum of the cylinder:
$L_{\text{cyl}} = I_{\text{cyl}}\,\omega = 0.04 \times 5 = 0.2\ \text{kg·m}^2\text{s}^{-1}.$
Angular momentum brought in by the particle (moving tangentially at the rim, radius $R$ ):

$L_{\text{p}} = m v R = 1 \times 5 \times 0.2 = 1.0\ \text{kg·m}^2\text{s}^{-1}.$ …
Start your 14-day free trial to see the full solution →
COMEDK 2024Set 2024-M1 markMCQ
$\mathrm{A}$
›Reveal solutionSolution
Areal velocity is half the angular momentum per unit mass, so angular momentum equals twice the areal velocity times the mass: $L = 2MA$ . The correct option is (C).

The key concept here is areal velocity — the rate at which a planet sweeps out area as it orbits. Kepler’s second law tells us that a planet’s radius vector sweeps equal areas in equal times, which is a direct consequence of the conservation of angular momentum. The connection is geometric: the area swept per unit time is directly proportional to the angular momentum per unit mass.

Let’s derive it step by step.

Define areal velocity. Areal velocity, $A$ , is the area swept out by the planet’s radius vector per unit time. In polar coordinates $(r, \theta)$ , the infinitesimal area swept in time $dt$ is

$dA = \frac{1}{2} r^2 \, d\theta.$
Hence
$A = \frac{dA}{dt} = \frac{1}{2} r^2 \frac{d\theta}{dt} = \frac{1}{2} r^2 \dot{\theta}.$

Recall angular momentum. For a planet of mass $M$ moving in a plane, the angular momentum about the central body is

$L = M r^2 \dot{\theta}.$
This is because the velocity component perpendicular to the radius is $r\dot{\theta}$ , so $L = M r (r\dot{\theta}) = M r^2 \dot{\theta}$ .

Relate the two. Compare the expressions:
Start your 14-day free trial to see the full solution →
KCET 2023Set A-31 markMCQ
$Q. When a planet revolves around the Sun, in general, for the planet (A) linear momentum and aerial velocity are constant. (B) kinetic and potential energy of the planet are constant. (C) angular momentum about the Sun and aerial velocity of the planet are constant. (D) linear momentum and linear velocity are constant.$
›Reveal solutionSolution

The Sun's gravity is a central force $\Rightarrow$ zero torque about the Sun $\Rightarrow$ angular momentum, and hence areal velocity, are conserved.

Step 1 — Why angular momentum is conserved.

The gravitational pull on the planet is directed along the line joining it to the Sun, i.e. antiparallel to $\vec r$ . The torque about the Sun is
$\vec\tau = \vec r \times \vec F = 0 \qquad(\text{since } \vec F \parallel \vec r).$
And $\vec\tau = \dfrac{d\vec L}{dt}$ , so
$\vec L = \text{constant}.$
This is the defining property of any central force.

Step 2 — Areal velocity follows immediately.

In time $dt$ the radius vector sweeps the triangle of area $dA=\frac12|\vec r \times d\vec r|$ , so
$\frac{dA}{dt} = \frac{1}{2}\,|\vec r \times \vec v| = \frac{|\vec L|}{2m}.$
With $L$ constant, the areal velocity is constant — which is exactly Kepler's second law (equal areas in equal times).

Step 3 — Rule out the others. …
Start your 14-day free trial to see the full solution →
COMEDK 2023Set 2023-M1 markMCQ
$M$
›Reveal solutionSolution

Adding the spheres raises the moment of inertia from $MR^2$ to $(M+4m)R^2$ ; conserving angular momentum gives $\omega'=\frac{M}{M+4m}\omega$ .

No external torque acts, so angular momentum is conserved:
$I_i\omega=I_f\omega'.$
Ring about its central axis: $I_i=MR^2$ . Four point masses $m$ at radius $R$ add $4mR^2$ :
$I_f=MR^2+4mR^2=(M+4m)R^2.$
Hence: …
Start your 14-day free trial to see the full solution →
COMEDK 2022Set 20221 markMCQ
$^2$
›Reveal solutionSolution

(4)(16) + (8)(4) = (4 + 8) ω 64 + 32 = 12 ω 96 = 12 ω → ω = 8 rad/s.

Concept: Conservation of angular momentum for two discs coupled about a common axis (no external torque):

I₁ω₁ + I₂ω₂ = (I₁ + I₂) ω.

(4)(16) + (8)(4) = (4 + 8) ω …
Start your 14-day free trial to see the full solution →
COMEDK 2021Set 20211 markMCQ
$Q. Kepler's second law of planetary motion corresponds to (A) conservation of energy (B) conservation of angular momentum (C) conservation of linear momentum (D) conservation of mass$
›Reveal solutionSolution

Areal velocity dA/dt = L/(2m). It is constant precisely because the gravitational force is central, so the torque about the Sun is zero and the ANGULAR MOMENTUM L is conserved.

Concept: Kepler's second law - the radius vector from the Sun to the planet sweeps equal areas in equal times. …
Start your 14-day free trial to see the full solution →
COMEDK 2021Set 2021-B1 markMCQ
$Q. A top of moment of inertia I spins with an angular velocity of 40 rad/s, if the angular velocity changes to half of it s original value the new moment of inertia is (A) I/3 (B) 2I (C) I/2 (D) 3I$
›Reveal solutionSolution

$L=I\omega$ conserved; $\omega\to\omega/2 \Rightarrow I\to 2I$ .

In the absence of external torque, angular momentum is conserved:
$I_1\omega_1 = I_2\omega_2.$
Here $\omega_2 = \omega_1/2$ , so …
Start your 14-day free trial to see the full solution →

←Previous questionQ36 — A satellite is to be placed in equatorial geostationary orbit around e…→Next questionQ38 — A satellite is in an elliptic orbit around the earth with aphelion of …

Why a solar day isn't simply Earth's rotation period

Setting up the sizes involved

From orbital-rate swing to solar-day-length swing

The result …

🎓Unlock everything free for 14 days

Unlock everything free for 14 days