The acceleration due to gravity at the poles and the equator is gp and ge respectively. If the earth is a sphere of radius RE and rotating about its axis with angular speed 0), then gp - ge is given by

Question

The acceleration due to gravity at the poles and the equator is gp and ge respectively. If the earth is a sphere of radius RE and rotating about its axis with angular speed 0), then gp - ge is given by (A) (ω2/RE) (B) (ω2/R2E) (C) ω2RE2 (D) ω2RE

Tardigrade Admin · Accepted Answer

Acceleration due to gravity at a place of latitude

λ

due to the rotation of earth is

g^{'} = g - R_{E} ω^{2} co s^{2} λ

At equator,

λ = 0^{\circ}, cos 0^{\circ} = 1

∴ g^{'} = g_{e} = g - R_{E} ω^{2}

At poles,

λ = 9 0^{\circ}, cos 9 0^{\circ} = 0

∴ g^{'} = g_{p} = g

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

Q. The acceleration due to gravity at the poles and the equator is $g_{p}$ and $g_{e}$ respectively. If the earth is a sphere of radius $R_{E}$ and rotating about its axis with angular speed 0), then $g_{p} - g_{e}$ is given by

$\frac{ω ^{2}}{R _{E}}$

$\frac{ω ^{2}}{R _{E}^{2}}$

$ω^{2} R_{E}^{2}$

$ω^{2} R_{E}$

Q. The acceleration due to gravity at the poles and the equator is gp​ and ge​ respectively. If the earth is a sphere of radius RE​ and rotating about its axis with angular speed 0), then gp​−ge​ is given by

RE​ω2​

RE2​ω2​

ω2RE2​

ω2RE​

Acceleration due to gravity at a place of latitude λ due to the rotation of earth is g′=g−RE​ω2cos2λ At equator, λ=0∘,cos0∘=1 ∴g′=ge​=g−RE​ω2 At poles, λ=90∘,cos90∘=0 ∴g′=gp​=g ∴gp​−ge​=g−(g−RE​ω2)=RE​ω2

Q. The acceleration due to gravity at the poles and the equator is $g_{p}$ and $g_{e}$ respectively. If the earth is a sphere of radius $R_{E}$ and rotating about its axis with angular speed 0), then $g_{p} - g_{e}$ is given by

$\frac{ω ^{2}}{R _{E}}$

$\frac{ω ^{2}}{R _{E}^{2}}$

$ω^{2} R_{E}^{2}$

$ω^{2} R_{E}$