What is the direction of areal velocity of the earth around the sun?

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Q. What is the direction of areal velocity of the earth around the sun?

Telangana TsbieShort· 2mImportance★★★★★est

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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 …

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Areal velocity is a vector whose direction is given by the angular velocity vector of the orbiting body — for Earth around the Sun, that direction is perpendicular to the orbital plane, pointing out of the page (or northward, along the axis of the orbit).

The idea of areal velocity comes from Kepler’s second law: a line joining a planet to the Sun sweeps out equal areas in equal times. That sweeping motion has a direction — just like the area swept has an orientation. In physics, any quantity that involves rotation or sweeping has a vector associated with it, and its direction is given by the right-hand rule.

Areal velocity is defined as $\frac{d\vec{A}}{dt}$ , the rate of change of the area vector. The area vector itself is perpendicular to the plane of the orbit, by convention pointing in the direction of the angular velocity $\vec{\omega}$ . So the direction of areal velocity is the same as the direction of $\vec{\omega}$ .

Connect areal velocity to angular momentum. For a planet of mass $m$ orbiting the Sun, the angular momentum $\vec{L} = m (\vec{r} \times \vec{v})$ is conserved. The areal velocity is $\frac{d\vec{A}}{dt} = \frac{1}{2} \vec{r} \times \vec{v}$ . This is directly proportional to the specific angular momentum $\vec{L}/m$ . So the direction of $\frac{d\vec{A}}{dt}$ is exactly the direction of $\vec{L}$ .
Find the direction of $\vec{L}$ for Earth’s orbit. Earth orbits the Sun in a counterclockwise direction when viewed from above the North Pole (i.e., looking down from the north). Using the right-hand rule: curl the fingers of your right hand in the direction of Earth’s motion (counterclockwise), and your thumb points upward — out of the orbital plane, toward the north celestial pole. …

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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:

$v = \frac{v_{\text{esc}}}{2} = \frac{11.2}{2} = 5.6\ \mathrm{km\,s^{-1}}$ …
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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: $$F_c = M \left(\frac{2\pi}{T}\right)^2 R = M \frac{4\pi^2}{T^2} R$$ …
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Q.What is the direction of areal velocity of the earth around the sun?

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