A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.

A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angu

Exercises · 7.8

Q. A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant

(a) linear speed,

(b) angular speed,

(d) kinetic energy,

(e) potential energy,

(f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.

Telangana TsbieTextbookSubjective· 3mImportance★★★★★est

24% · 16/67 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 →

In a central gravitational field (the Sun), angular momentum and total mechanical energy are conserved, but linear speed, angular speed, kinetic energy, and potential energy all vary along an elliptical orbit. The comet’s angular momentum is constant because the gravitational force is always radial (no torque).

Why conservation of angular momentum is the key

The Sun’s gravity is a central force — it always points directly toward the Sun. For any central force, the torque about the centre is zero:

\vec{\tau} = \vec{r} \times \vec{F} = \vec{r} \times \left( -\frac{GMm}{r^2} \hat{r} \right) = 0

because $\vec{r}$ and $\hat{r}$ are parallel. Zero torque means angular momentum is conserved. That single fact drives the answers for most of the quantities listed.

Let’s go through each one.

1. Linear speed — Not constant

In an elliptical orbit, the comet’s distance from the Sun changes. Angular momentum conservation gives:

L = m v_\perp r = \text{constant}

where $v_\perp$ is the component of velocity perpendicular to the radius. When $r$ is small (near perihelion), $v_\perp$ must be large; when $r$ is large (near aphelion), $v_\perp$ is small. So linear speed $v$ (which includes both radial and perpendicular components) changes continuously.

Watch out

A common mistake is to think “angular momentum constant ⇒ speed constant”. That’s false — angular momentum depends on $v_\perp r$ , not on $v$ alone. The radial component of velocity also changes, so the total speed varies significantly.

2. Angular speed — Not constant

Angular speed $\omega = d\theta/dt$ is related to angular momentum by:

L = m r^2 \omega

Since $L$ is constant, $\omega \propto 1/r^2$ . As $r$ varies, $\omega$ varies dramatically — the comet sweeps out equal areas in equal times (Kepler’s second law), so it moves fastest when closest to the Sun.

3. Angular momentum — Constant

As argued above, the gravitational force exerts zero torque about the Sun. Therefore:

\vec{L} = \vec{r} \times m\vec{v} = \text{constant vector}

Both magnitude and direction are fixed. This is a fundamental property of any motion under a central force.

\vec{L} = \text{constant} \quad \text{(for any central force)}

4. Kinetic energy — Not constant

Kinetic energy $K = \frac12 m v^2$ . Since speed varies, $K$ varies. It is maximum at perihelion (fastest) and minimum at aphelion (slowest).

5. Potential energy — Not constant

Gravitational potential energy (taking $U=0$ at infinity) is:

$U = -\frac{GMm}{r}$ …

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.

TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
$4R$
›Reveal solutionSolution
The comet’s maximum velocity occurs at its closest approach to Earth, where gravitational potential energy is most negative. Using conservation of energy between infinity and the perigee distance $4R$ , the maximum speed is found to be half the escape speed from Earth’s surface, i.e., $5.6\ \mathrm{km\,s^{-1}}$ .

Concept & Intuition

The comet starts from rest at infinity (relative to Earth), so its total mechanical energy is zero. As it falls toward Earth, gravitational potential energy converts to kinetic energy. The maximum speed occurs at the point of closest approach — here, a distance $4R$ from Earth’s center. The escape speed from Earth’s surface is the speed needed to just reach infinity; here we reverse the logic: a body falling from infinity gains a speed at distance $r$ equal to the escape speed from that distance. Since the escape speed scales as $1/\sqrt{r}$ , the speed at $4R$ is smaller than at $R$ by a factor $\sqrt{R/(4R)} = 1/2$ .

Step-by-step solution

Set up conservation of energy

At infinity, the comet has zero kinetic energy (starts from rest) and zero gravitational potential energy (convention: $U = 0$ at infinity). So total energy $E = 0$ .

Energy at closest approach

At distance $r = 4R$ from Earth’s center, the comet’s kinetic energy is $\frac{1}{2} m v^2$ and its gravitational potential energy is $-\frac{GMm}{r}$ . Conservation of energy gives:

$0 = \frac{1}{2} m v^2 - \frac{GMm}{4R}$
Cancel $m$ :
$\frac{1}{2} v^2 = \frac{GM}{4R}$
So:
$v^2 = \frac{GM}{2R}$

Relate to escape speed from Earth’s surface Escape speed from Earth’s surface (radius $R$ ) is:

$v_{\text{esc}} = \sqrt{\frac{2GM}{R}} = 11.2\ \mathrm{km\,s^{-1}}$
Therefore:
$\frac{GM}{R} = \frac{v_{\text{esc}}^2}{2}$

Substitute into expression for $v^2$

$v^2 = \frac{GM}{2R} = \frac{1}{2} \cdot \frac{GM}{R} = \frac{1}{2} \cdot \frac{v_{\text{esc}}^2}{2} = \frac{v_{\text{esc}}^2}{4}$
Hence:

$v = \frac{v_{\text{esc}}}{2} = \frac{11.2}{2} = 5.6\ \mathrm{km\,s^{-1}}$ …
Start your 14-day free trial to see the full solution →
TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
$R^{\frac{3}{2}}$
›Reveal solutionSolution
Two stars of equal mass orbiting each other experience a gravitational force that provides the necessary centripetal force for their circular motion. By equating these forces and relating angular velocity to the orbital period, we find that the orbital period is proportional to $R^{\frac{3}{2}}$ .

When two celestial bodies orbit each other, they actually orbit their common center of mass. In this problem, we have two stars of equal mass $M$ . This simplifies the situation significantly: their center of mass lies exactly halfway between them.

Each star moves in a circular path around this common center of mass. The force responsible for this circular motion is the gravitational attraction between the two stars. This gravitational force acts as the centripetal force, continuously pulling each star towards the center of mass and keeping it in its orbit.

We will use Newton's Law of Universal Gravitation to find the attractive force between the stars and equate it to the centripetal force required for circular motion, expressed in terms of the orbital period.

Define the system and forces:

We have two stars, each of mass $M$ . They are orbiting in a circle of radius $R$ . Since their masses are equal, the center of mass is exactly midway between them. This means each star orbits at a radius $R$ from the center of mass. Consequently, the distance between the two stars at any instant is $2R$ .

Calculate the gravitational force:

The gravitational force $F_g$ between the two stars is given by Newton's Law of Universal Gravitation:

$F_g = \frac{G M_1 M_2}{d^2}$
Here, $M_1 = M_2 = M$, and the distance between the stars is $d = 2R$. Substituting these values:
$F_g = \frac{G M \cdot M}{(2R)^2} = \frac{G M^2}{4R^2}$

Calculate the centripetal force: For one star to move in a circular orbit of radius $R$ with an angular velocity $\omega$ , the required centripetal force $F_c$ is:

$F_c = M \omega^2 R$

Relate angular velocity to orbital period: The angular velocity $\omega$ is related to the orbital period $T$ (the time taken for one complete orbit) by:

$\omega = \frac{2\pi}{T}$
Substituting this into the expression for centripetal force: $$F_c = M \left(\frac{2\pi}{T}\right)^2 R = M \frac{4\pi^2}{T^2} R$$ …
Start your 14-day free trial to see the full solution →

←Previous question7.7 — Does the escape speed of a body from the earth depend on (a) the mass …→Next question7.9 — Which of the following symptoms is likely to afflict an astronaut in s…

Q.A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant

Why conservation of angular momentum is the key

1. Linear speed — Not constant

2. Angular speed — Not constant

3. Angular momentum — Constant

4. Kinetic energy — Not constant

5. Potential energy — Not constant

🎓Unlock everything free for 14 days

Unlock everything free for 14 days