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# The linear velocity of a particle on the equator is nearly (radius of the earth is $4000\;{\text{miles}}$):

Last updated date: 14th Aug 2024
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Hint: The above numerical problem is based on the concept of the time period and rotational speed of the earth. The time period is the duration in which the earth completes the one revolution around its axis. The rotational speed varies with the time period of the earth.

Complete step by step answer:
Given: The radius of the earth is $R = 4000\;{\text{miles}}$.
The expression to calculate the rotational speed of the particle at the equator is given as:
$\omega = \dfrac{{2\pi }}{T}......\left( 1 \right)$
Here, T is the time period of the earth and its value is 24 hr.
Substitute 24 hr for T in the expression (1) to find the rotational speed of the particle on the equator.
$\omega = \dfrac{{2\pi }}{{24\;{\text{hr}}}}$
$\omega = 0.262\;{\text{rad}}/{\text{hr}}$

The expression to calculate the linear velocity of a particle on the equator is given as:
$v = R\omega ......\left( 2 \right)$
Substitute $0.262\;{\text{rad}}/{\text{hr}}$for $\omega$ in the expression (2) to calculate the linear velocity of the particle at the equator.
$v = \left( {4000\;{\text{miles}}} \right)\left( {0.262\;{\text{rad}}/{\text{hr}}} \right)$
$v = 1048\;{\text{miles}}/{\text{hr}}$

Thus, the linear velocity of the particle at the equator is $1048\;{\text{miles}}/{\text{hr}}$.

The effect of the gravity on the particle varies with the location of the particle. If the particle lies on the equator of the earth then the effect of the gravity becomes minimum and becomes maximum at the poles of the earth.

Note: First calculate the time period of the earth to find the rotational speed of the particle. The units of the radius and time period should be in the same measurement system. The linear velocity of the particle varies with the distance from the surface of the earth.