Let the value of acceleration due to gravity at poles and equator of earth gp and ge respectively. Assuming the earth to be a sphere of radius R rotating about its axis with angular speed ω then gp-ge is given by:

Question

Let the value of acceleration due to gravity at poles and equator of earth gp and ge respectively. Assuming the earth to be a sphere of radius R rotating about its axis with angular speed ω then gp-ge is given by: (A) (ω2/R) (B) R ω2 (C) R2 ω2 (D) (ω2/R2)

Tardigrade Admin · Accepted Answer

Effective value of acceleration due to gravity

(g_{ϕ})

is given by

g_{ϕ} = g (1 - \frac{ω ^{2} R}{g} cos^{2} ϕ)

where

ω

is angular velocity.
At poles:

ϕ = 9 0^{\circ}

∴ g_{ϕ} = g_{p} = g_{\dots (i)}

At equator

ϕ = 0 g_{ϕ} = g_{e} = g (1 - \frac{ω ^{2} R}{g}) = g - ω^{2} R \dots

(ii)
From Eqs. (i) and (ii), we get

∴ g_{p} - g_{e} = g - (g - ω^{2} R) = ω^{2} R

Q. Let the value of acceleration due to gravity at poles and equator of earth $g_{p}$ and $g_{e}$ respectively. Assuming the earth to be a sphere of radius $R$ rotating about its axis with angular speed $ω$ then $g_{p} - g_{e}$ is given by:

$\frac{ω ^{2}}{R}$

$R ω^{2}$

$R^{2} ω^{2}$

$\frac{ω ^{2}}{R ^{2}}$

Q. Let the value of acceleration due to gravity at poles and equator of earth gp​ and ge​ respectively. Assuming the earth to be a sphere of radius R rotating about its axis with angular speed ω then gp​−ge​ is given by:

Rω2​

Rω2

R2ω2

R2ω2​

Q. Let the value of acceleration due to gravity at poles and equator of earth $g_{p}$ and $g_{e}$ respectively. Assuming the earth to be a sphere of radius $R$ rotating about its axis with angular speed $ω$ then $g_{p} - g_{e}$ is given by:

$\frac{ω ^{2}}{R}$

$R ω^{2}$

$R^{2} ω^{2}$

$\frac{ω ^{2}}{R ^{2}}$