Question

What is the magnitude of the centripetal acceleration of an object on Earth’s equator owing to the rotation of Earth?

Answer 1

To solve this question, we could use the formula of centripetal acceleration which is square of velocity divided by radius of the circle in which the object revolves. But here we need to note that we have not been provided with the value of velocity of the object, hence we need to find that first.

Complete answer:
First let us calculate the velocity of the object experienced by it due to the rotation of the Earth when it is placed on the Earth’s equator. The velocity of the object can be found out by the following formula:
$v=\dfrac{2\pi R}{T}$
Here $R$ is the radius of the circle in which the object is revolving, i.e., the radius of the Earth which is $R=6.37\times {{10}^{6}}m$ because the object is placed on the equator of the Earth.
And $T$ is the time taken by Earth for a complete rotation which is:
$\begin{aligned} & T=24\times 60\times 60 \\\ & \Rightarrow T=86400\text{ s} \\\ \end{aligned}$
On substituting the values, in the equation, we get the following equation:

& v=\dfrac{2\pi \times 6.37\times {{10}^{6}}}{86400} \\\ & \Rightarrow v=463\text{ m}{{\text{s}}^{-1}} \\\ \end{aligned}$$ Now, let us calculate the centripetal acceleration. The formula for centripetal acceleration is: ${{a}_{c}}=\dfrac{{{v}^{2}}}{R}$ On substituting the values in the equation, we get the following equation: $\begin{aligned} & {{a}_{c}}=\dfrac{{{\left( 463 \right)}^{2}}}{6.37\times {{10}^{6}}} \\\ & \therefore {{a}_{c}}=0.034\text{ m}{{\text{s}}^{-2}} \\\ \end{aligned}$ Thus, the centripetal acceleration of an object on Earth’s equator owing to the rotation of Earth is $0.034\text{ m}{{\text{s}}^{-2}}$. **Note:** Centripetal acceleration can be defined as the acceleration that an object experiences which is directed towards the center of the circle in which the object is moving. Here, we must take care that all the units used in the solution are of S.I units otherwise we may conclude to wrong answers.