Q.(a) Earth can be thought of as a sphere of radius 6400 km. Any object (or a person) is performing circular motion around the axis of earth due to earth's rotation (period 1 day). What is acceleration of object on the surface of the earth (at equator) towards its centre? what is it at latitude $\theta$? How does these accelerations compare with $g = 9.8$ m/s$^2$?

(a) Centripetal acceleration due to Earth's rotation

Concept: An object at rest on Earth's surface undergoes circular motion about the rotation axis with centripetal acceleration $a_c = \omega^2 r$, where $r$ is the perpendicular distance from the axis.

At the equator:

The radius of the circular path equals Earth's radius $R = 6400 \times 10^3 \, \text{m}$. The angular velocity is $\omega = \frac{2\pi}{T}$ with $T = 24 \times 3600 = 86400 \, \text{s}$.

This is roughly $\frac{0.034}{9.8} \approx \frac{1}{288}$ of $g$, entirely negligible.

At latitude $\theta$:

The radius of the circular path becomes $r = R \cos\theta$ (perpendicular distance from the rotation axis).

At the poles ($\theta = 90°$), this acceleration vanishes.

(b) Centripetal acceleration due to Earth's orbit around the Sun

Concept: Earth's orbital motion around the Sun requires centripetal acceleration directed toward the Sun.

The orbital radius is $r = 1.5 \times 10^{11} \, \text{m}$ and period $T = 365.25 \times 86400 \approx 3.156 \times 10^7 \, \text{s}$. …

Why this matters: circular motion and centripetal acceleration

Whenever an object moves in a circle at constant speed, it accelerates toward the center. This centripetal acceleration arises because velocity is a vector—even if speed is constant, the direction changes continuously. The magnitude is $a = \frac{v^2}{r}$, or equivalently $a = \frac{4\pi^2 r}{T^2}$ when you know the period $T$ instead of speed.

Earth spins on its axis (period $T_{\text{day}} = 1\,\text{day}$) and orbits the Sun (period $T_{\text{year}} = 1\,\text{year}$). Every object on Earth's surface participates in both motions, so experiences both accelerations. Comparing these with $g$ tells us whether they're dynamically significant—spoiler: they're tiny, which is why we usually ignore them.

(a) Acceleration due to Earth's rotation

1. At the equator

An object at the equator traces a circle of radius $R = 6400\,\text{km} = 6.4 \times 10^6\,\text{m}$ once per day. The period is

The centripetal acceleration is

Calculate the denominator:

Numerator:

So

Compared to $g = 9.8\,\text{m/s}^2$:

The rotational acceleration at the equator is about one three-hundredth of $g$—small but measurable.

2. At latitude $\theta$

At latitude $\theta$, an object is not on the equator but displaced northward (or southward). It still rotates once per day, but now the circular path has a smaller radius.

The radius of the circular path at latitude $\theta$ is $r = R\cos\theta$, where $R$ is Earth's radius. (Picture a horizontal slice through Earth at that latitude; the slice is a circle of radius $R\cos\theta$.)

The centripetal acceleration becomes

At the poles ($\theta = 90^\circ$), $\cos 90^\circ = 0$, so the acceleration vanishes—an object at the pole simply spins in place, no circular motion around Earth's axis.

(b) Acceleration due to Earth's orbit around the Sun

Earth orbits the Sun in an (approximately) circular path of radius $r_{\text{orbit}} = 1.5 \times 10^{11}\,\text{m}$ with period $T_{\text{year}} = 1\,\text{year}$.

Convert the period to seconds:

The centripetal acceleration toward the Sun is

Denominator:

Numerator: …

Concept: Compute the Physical Speed First, Then Apply $a=v^2/r$ -- Instead of the Period-Only Formula

Method: Route Through the Meaningful Intermediate Quantity $v=2\pi r/T$

The direct route plugs straight into $a=4\pi^2r/T^2$, which is algebraically fastest but never produces a number you can sanity-check against everyday experience. This method computes the actual linear speed of the rotating/orbiting point first (a physically meaningful, checkable quantity) and only then applies $a=v^2/r$.

Steps -- (a) Earth's rotation

1. Compute the equatorial linear speed directly:

(A familiar, checkable number -- close to the commonly quoted $\approx1670\text{ km/h}$ equatorial spin speed of the Earth.)

2. Apply $a=v^2/R$ using this speed:

3. At latitude $\theta$, the circle radius shrinks to $R\cos\theta$, so the speed shrinks proportionally too:

4. Apply $a=v^2/r$ again, now with $r=R\cos\theta$:

Steps -- (b) Earth's orbit

5. Compute the orbital speed directly:

(Again a familiar, checkable figure -- Earth's orbital speed is commonly quoted as $\approx29.8\text{ km/s}$.) …