Out of aphelion and perihelion, where is the speed of the earth more and why ?

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Q. Out of aphelion and perihelion, where is the speed of the earth more and why ?

Yanam BieapShort· 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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The Earth moves faster at perihelion (closest approach to the Sun) than at aphelion (farthest point) because angular momentum is conserved — as the orbital radius decreases, the tangential speed must increase to keep the product $m v r$ constant.

The key to this question is conservation of angular momentum. In the absence of external torque, the angular momentum of a body in orbit remains constant. For the Earth–Sun system, the gravitational force is central (always directed toward the Sun), so it exerts zero torque about the Sun. That means the Earth’s angular momentum about the Sun does not change as it moves along its elliptical orbit.

Angular momentum in orbit For a planet of mass $m$ moving with speed $v$ at a distance $r$ from the Sun, the magnitude of its orbital angular momentum (taking the Sun as the reference point) is

L = m v r \sin\theta

where $\theta$ is the angle between the velocity vector and the radius vector. At the two special points — aphelion (farthest from the Sun) and perihelion (closest) — the velocity is perpendicular to the radius vector, so $\sin\theta = 1$ . Hence at those points:

L = m v r

Conservation of $L$ Since $L$ is constant, we have

m v_{\text{peri}} r_{\text{peri}} = m v_{\text{ap}} r_{\text{ap}}

Cancelling $m$ :

v_{\text{peri}} r_{\text{peri}} = v_{\text{ap}} r_{\text{ap}}

Relating speed and distance Rearranging:

v_{\text{peri}} = v_{\text{ap}} \cdot \frac{r_{\text{ap}}}{r_{\text{peri}}}

Because $r_{\text{ap}} > r_{\text{peri}}$ (aphelion is farther), the ratio $\frac{r_{\text{ap}}}{r_{\text{peri}}} > 1$ . Therefore $v_{\text{peri}} > v_{\text{ap}}$ .

Tip

A quick way to remember: "Closer means faster" — just like a figure skater pulling in their arms spins faster. The same physics governs planetary motion.

Physical intuition …

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Q.Out of aphelion and perihelion, where is the speed of the earth more and why ?

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