The weight of an object at the equator is the same as its weight at height h above the pole. If ω is the rotational speed of Earth, then the value of h is

Question

The weight of an object at the equator is the same as its weight at height h above the pole. If ω is the rotational speed of Earth, then the value of h is (A) (ω2 R/2 g) (B) (ω R/2 g) (C) (ω/2 g) (D) (ω2 R2/2 g)

Tardigrade Admin · Accepted Answer

g=gp-ω2 R cos 2 λ At equator, λ=0 ⇒ g e q=gp-ω2 R cos 2 0=gp-ω2 ⋅ R Now, g prime above the pole is g prime=gp(1-(2 h/R)) As ge q=g prime gp-ω2 R =gp(1-(2 h/R)) So, ω2 R=(2 h ⋅ gp/R) ⇒ h=(ω2 ⋅ R2/2 gp)=(ω2 ⋅ R2/2 g)

Q. The weight of an object at the equator is the same as its weight at height $h$ above the pole. If $ω$ is the rotational speed of Earth, then the value of $h$ is

$\frac{ω ^{2} R}{2 g}$

$\frac{ω R}{2 g}$

$\frac{ω}{2 g}$

$\frac{ω ^{2} R ^{2}}{2 g}$

Q. The weight of an object at the equator is the same as its weight at height h above the pole. If ω is the rotational speed of Earth, then the value of h is

2gω2R​

2gωR​

2gω​

2gω2R2​

g=gp​−ω2Rcos2λ At equator, λ=0 ⇒geq​=gp​−ω2Rcos20=gp​−ω2⋅R Now, g′ above the pole is g′=gp​(1−R2h​) As geq​=g′ gp​−ω2R=gp​(1−R2h​) So, ω2R=R2h⋅gp​​ ⇒h=2gp​ω2⋅R2​=2gω2⋅R2​

Q. The weight of an object at the equator is the same as its weight at height $h$ above the pole. If $ω$ is the rotational speed of Earth, then the value of $h$ is

$\frac{ω ^{2} R}{2 g}$

$\frac{ω R}{2 g}$

$\frac{ω}{2 g}$

$\frac{ω ^{2} R ^{2}}{2 g}$