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 $\omega 2\text{r}=\dfrac{(2\pi )}{\text{T}}2\text{r}=0.034\text{m}/{{\text{s}}^{2}}$.
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.