Question 317: Gravitation

Question

Earth is assumed to be a sphere of radius R. If '$$\mathrm{g}_\phi$$' is value of effective acceleration due to gravity at latitude $$30^{\circ}$$ and '$$g$$' is the value at equator, then the value of $$\left|g-g_\phi\right|$$ is ($$\omega$$ is angular velocity of rotation of earth, $$\cos 30^{\circ}=\frac{\sqrt{3}}{2}$$ )

Accepted Answer

The acceleration due to gravity on the surface of the Earth varies with latitude due to the centrifugal force resulting from the Earth's rotation. The effective acceleration due to gravity at a latitude $$\phi$$, denoted as $$\mathrm{g}_\phi$$, takes into account the centrifugal force and is less than the acceleration due to gravity that would be experienced if the Earth were not rotating (denoted as $$g_0$$). At the equator ($$\phi = 0^{\circ}$$), this effect is maximal because the velocity due to Earth's rotation is maximal. As we go to the poles, the effect of rotation becomes minimal.
The effective acceleration due to gravity at latitude $$\phi$$ is given by:
$$ g_{\phi} = g_{0} - \omega^2 \mathrm{R} \cos^2 \phi $$
Where:
$$g_0$$ is the acceleration due to gravity without the Earth's rotation, at the equator
$$\omega$$ is the angular velocity of the Earth's rotation
$$\mathrm{R}$$ is the radius of the Earth
$$\phi$$ is the latitude
At the equator, the effective acceleration due to gravity, $$ g $$, is:
$$ g = g_{0} - \omega^2 \mathrm{R} $$
because $$ \cos 0^{\circ} = 1 $$.
At latitude $$ 30^{\circ} $$, the effective acceleration due to gravity, $$ \mathrm{g}_\phi $$, using $$ \cos 30^{\circ} = \frac{\sqrt{3}}{2} $$, is:
$$ g_{\phi} = g_{0} - \omega^2 \mathrm{R} \left(\frac{\sqrt{3}}{2}ight)^2 $$
$$ g_{\phi} = g_{0} - \omega^2 \mathrm{R} 	imes \frac{3}{4} $$
To find $$ \left|g-g_\phiight| $$, we subtract the above two equations:
$$ \left|g-g_\phiight| = \left|\left(g_{0} - \omega^2 \mathrm{R}ight) - \left(g_{0} - \omega^2 \mathrm{R} 	imes \frac{3}{4}ight)ight| $$
$$ \left|g-g_\phiight| = \omega^2 \mathrm{R} - \omega^2 \mathrm{R} 	imes \frac{3}{4} $$
$$ \left|g-g_\phiight| = \omega^2 \mathrm{R} \left(1 - \frac{3}{4}ight) $$
$$ \left|g-g_\phiight| = \omega^2 \mathrm{R} 	imes \frac{1}{4} $$
So the value of $$ \left|g-g_\phiight| $$ is given by the option A, which is:
$$ \frac{1}{4} \omega^2 \mathrm{R} $$

Answer

$$\frac{1}{4} \omega^2 \mathrm{R}$$

Answer

$$\frac{3}{4} \omega^2 \mathrm{R}$$

Answer

$$\omega^2 \mathrm{R}$$

Answer

$$\frac{1}{2} \omega^2 \mathrm{R}$$

Gravitation

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