Let ω be the angular velocity of the earth's rotation about its axis. Assume that the acceleration due to gravity on the earth's surface has the same value at the equator and the poles. An object weighed at the equator gives the same reading as a reading taken at a depth d below earth's surface at a pole (d

Question

Let ω be the angular velocity of the earth's rotation about its axis. Assume that the acceleration due to gravity on the earth's surface has the same value at the equator and the poles. An object weighed at the equator gives the same reading as a reading taken at a depth d below earth's surface at a pole (d < < R) . The value of d is (A) (ω2 R2/g) (B) (ω2 R 2/2 g ) (C) (2 ω2 R2/g) (D) (ω 2 R2/4g)

Tardigrade Admin · Accepted Answer

g text eff due to rotation = g text eff due to depth d g-R ω2=g(1-(d/R)) ⇒ R ω2=(g d/R) ⇒ d=(R2 ω2/g)

$\frac{ω ^{2} R ^{2}}{g}$

$\frac{ω ^{2} R ^{2}}{2 g}$

$\frac{2 ω ^{2} R ^{2}}{g}$

$\frac{ω ^{2} R ^{2}}{4 g}$

$g_{eff} d u e$ to rotation $= g_{eff} d u e$ to depth $d$
$g - R ω^{2} = g (1 - \frac{d}{R}) \Rightarrow R ω^{2} = \frac{g d}{R} \Rightarrow d = \frac{R ^{2} ω ^{2}}{g}$

gω2R2​

2gω2R2​

g2ω2R2​

4gω2R2​

geff ​due to rotation =geff ​due to depth d g−Rω2=g(1−Rd​)⇒Rω2=Rgd​⇒d=gR2ω2​

$\frac{ω ^{2} R ^{2}}{g}$

$\frac{ω ^{2} R ^{2}}{2 g}$

$\frac{2 ω ^{2} R ^{2}}{g}$

$\frac{ω ^{2} R ^{2}}{4 g}$

$g_{eff} d u e$ to rotation $= g_{eff} d u e$ to depth $d$
$g - R ω^{2} = g (1 - \frac{d}{R}) \Rightarrow R ω^{2} = \frac{g d}{R} \Rightarrow d = \frac{R ^{2} ω ^{2}}{g}$