Question

The earth, radius $ 6400 \mathrm{km} $ makes one revolution about its own axis in 24 hours. The centripetal acceleration of a point on its equator is nearly.
(A) $ 340\dfrac{\text{cm}}{\text{se}{{\text{c}}^{2}}} $
(B) $ 3.4\dfrac{cm}{se{{c}^{2}}} $
(C) $ 34\dfrac{\text{cm}}{\text{se}{{\text{c}}^{2}}} $
(D) $ 0.34\dfrac{\text{cm}}{\text{se}{{\text{c}}^{2}}} $

Answer 1

Hint
We know that a centripetal force is a force that makes a body follow a curved path. The direction is always orthogonal to the motion of the body and towards the fixed point of the instantaneous centre or curvature of the path. It is a force which is necessary to keep an object moving in a curved path and that is directed inward toward the centre of rotation of a string on the end of which a stone is whirled about exerts centripetal force on the force. Based on this concept we have to answer this question.

Complete step by step answer
The time period of earth's rotation or the time that the earth takes to rotate around its own axis is,
 $ T=24\text{h}=86400\text{s} $
We know that the angular velocity $ \omega=2 \pi / \mathrm{T} $
Radius of the earth is represented as $ \mathrm{r}=6400 \mathrm{km}=640000 \mathrm{m} $
Now, we can equate the centripetal acceleration as $ \omega^2\text{r}={(\dfrac{2\pi }{\text{T}})}^2 2\text{r}=0.034\text{m}/{{\text{s}}^{2}} $ .
Therefore, the total centripetal acceleration of a point on Earth’s equator is $ 3.4\dfrac{cm}{se{{c}^{2}}} $ .
Hence, the correct answer is Option (B).

Note
We know that time period is defined as the time that is taken for one complete cycle of vibration to pass a given point. As the frequency of a wave increases, the time period of the wave decreases. Frequency and time period are in a reciprocal relationship that can be expressed mathematically as T = 1 / f.
The rotation period is calculated using the formula rotation period = rotation speed divided by the planet’s circumference, where the circumference is the pi times the diameter.