Question

The angular speed of the Earth’s rotation is $7.3 \times 10^{-5}\text{ rad}\cdot\text{s}^{-1}$. Take the radius of the Earth at the equator to be $6400\text{ km}$. Then the ratio ($a_c / g$) of the magnitude of the centripetal acceleration $a_c$ at a point on the equator to $g$, is of the order

• A. $10^{-3}$
• B. $10^{-5}$
• C. $10^{0}$
• D. $10^{-1}$

Hint:

An order of magnitude estimate can be done quickly by rounding numbers:
$\omega \approx 7 \times 10^{-5}$, so $\omega^2 \approx 5 \times 10^{-9}$.
With $R \approx 6 \times 10^6$, we get $a_c \approx 3 \times 10^{-2} = 0.03$.
Dividing by $g \approx 10$ yields $0.003 = 3 \times 10^{-3}$, confirming the order of magnitude.

Answer 1

The Correct Option is A

Step 1: Picture what is happening.
A point sitting on the equator is being carried around in a big circle as the Earth spins. Because it moves in a circle, it always has a centripetal acceleration pointing toward the axis. We just need to compare that small acceleration with normal gravity $g$.

Step 2: Write the handy formula.
For circular motion the centripetal acceleration is $a_c = \omega^2 R$, where $\omega$ is the spin rate and $R$ is the equatorial radius.

Step 3: List the numbers in clean SI units.
Here $\omega = 7.3 \times 10^{-5}$ rad per second and $R = 6400$ km $= 6.4 \times 10^{6}$ m.

Step 4: Square the spin rate first.
$\omega^2 = (7.3 \times 10^{-5})^2 \approx 5.3 \times 10^{-9}$ per second squared.

Step 5: Multiply to get the acceleration.
$a_c = 5.3 \times 10^{-9} \times 6.4 \times 10^{6} \approx 3.4 \times 10^{-2}$ metre per second squared.

Step 6: Take the ratio with $g$.
Using $g \approx 9.8$, we get \[ \frac{a_c}{g} = \frac{3.4 \times 10^{-2}}{9.8} \approx 3.5 \times 10^{-3}. \] So this number is right around one thousandth.

\[ \boxed{\dfrac{a_c}{g} \sim 10^{-3}} \]