Solveeit Logo
Login

Question

Physics Question on Escape Speed

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

A

ω2RE\frac{\omega^{2}}{R_{E}}

B

ω2RE2\frac{\omega^{2}}{R^2_{E}}

C

ω2RE2\omega^{2}R_{E}^2

D

ω2RE\omega^{2}R_{E}

Answer

ω2RE\omega^{2}R_{E}

Explanation

Solution

Acceleration due to gravity at a place of latitude λ\lambda due to the rotation of earth is g′=g−REω2cos2λg'= g - R_{E}\omega^{2}cos^{2}\lambda At equator, λ=0∘,cos0∘=1\lambda = 0^{\circ}, cos0^{\circ} = 1 ∴g′=ge=g−REω2\therefore g'= g_{ e}=g - R_{E}\omega^{2} At poles, λ=90∘,cos90∘=0\lambda = 90^{\circ}, cos90^{\circ} = 0 ∴g′=gp=g\therefore g'=g_{p}=g ∴gp−ge=g−(g−REω2)=REω2\therefore g_{p}-g_{e} = g - \left( g - R_{E}\omega^{2}\right) = R_{E}\omega^{2}