Question

At what angular speed should the earth rotate so that a body situated on the equator becomes weightless?
(Acceleration due to gravity $=9.8 m/s^{2}$, radius of the earth $=6400km$)

Answer 1

Hint: The gravity experienced by a body increases from the equator to the poles. This is because the earth is not a perfect circle. It is slightly flattened at the poles, forming an irregularly shaped ellipsoid. Thus, there is a difference in the radius of the earth along the poles and the equator.

Formula used: $W\prime=W-mR_{e}\omega^{2}\cos^{2}\phi$

Complete step-by-step answer:
The radius of the earth along the poles is slightly larger than along the earth. This is due to the irregularity on the earth. Hence, earth is in the shape of an irregular ellipsoid and not a perfect circle. This difference in radius affects the gravitational force on the earth. This in turn affects the weight of the body. The apparent weight of the body on the equator is given by:
$W\prime=W-mR_{e}\omega^{2}\cos^{2}\phi$
$mg\prime=mg-mR_{e}\omega^{2}\cos^{2}\phi$
Where $m$ is the mass of the object, $g$ acceleration due to gravity, $g\prime$ is the acceleration due to gravity at equator, $R_{e}$ is the radius of the earth , $\omega$ is the angular velocity of the earth and $\phi$ is the angle the body forms with the center of the earth.
Weightlessness occurs when the resultant force acting on a body becomes zero.
Since, the body becomes weightless on the equator,
$mg\prime =0$ and $\phi=0^{\circ}$
$0=mg-mR_{e}\omega^{2}cos^{2}0$
Dividing by mass on both the sides
$g-R_{e}\omega^{2}=0$
$g=R_{e}\omega^{2}$
Given that, acceleration due to gravity $g=9.8 m/s^{2}$, radius of the earth $R_{e}=6400km$)
Substituting the values, we get
$\omega^{2}=\dfrac{g}{R_{e}}$
$\omega=\sqrt{(\dfrac{g}{R_{e}})}=\sqrt{(\dfrac{9.8}{6.4\times 10^{6}})}$
$\omega=1.23\times 10^{-3} rad/sec$
Hence, for the body on the equator to become weightless, the earth must spin at $\omega=1.23\times 10^{-3} rad/sec$

Note: The shape of earth is ellipsoidal and not spherical, this causes a difference in the radius of the earth along the equator and the poles, which in turn affect the gravitational force and hence the mass of the body at the equator and the poles. Be careful with the units and the angle $\phi$.