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# 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) $0.34\dfrac{\text{cm}}{\text{se}{{\text{c}}^{2}}}$

Last updated date: 11th Sep 2024
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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.

$T=24\text{h}=86400\text{s}$
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 $3.4\dfrac{cm}{se{{c}^{2}}}$ .