Find out the correct relation the dependance of change in acceleration due to gravity on the angle the latitude due to rotation of earth-dgpropto cosphi dgpropto - cos - 2 -phi dgpropto - cos - 3-2 -phi dgpropto dfrac - 1 - cosphi -

Question

Toppr Admin · Accepted Answer

We know that- the effective gravity g-x2032- at a latitude -x3D5- is given asg-x2032-g-x2212-x3C9-2R-cos-x3D5-2Thus- g-x2212-g-x2032-dg-x3C9-2R-cos-x3D5-2

Find out the correct relation for the dependance of change in acceleration due to gravity on the angle at the latitude due to rotation of earth?
$d g \propto c o s ϕ$
$d g \propto {c o s}^{2} ϕ$
$d g \propto {c o s}^{3 / 2} ϕ$
$d g \propto \frac{1}{c o s ϕ}$

We know that, the effective gravity $g^{'}$ at a latitude $ϕ$ is given as
$g^{'} = g - ω^{2} R {(cos ϕ)}^{2}$
Thus, $g - g^{'} = d g = ω^{2} R {(cos ϕ)}^{2}$

Find out the correct relation for the dependance of change in acceleration due to gravity on the angle at the latitude due to rotation of earth?dg∝cosϕdg∝cos2ϕdg∝cos3/2ϕdg∝1cosϕ

We know that, the effective gravity g′ at a latitude ϕ is given asg′=g−ω2R(cosϕ)2Thus, g−g′=dg=ω2R(cosϕ)2

Find out the correct relation for the dependance of change in acceleration due to gravity on the angle at the latitude due to rotation of earth?
$d g \propto c o s ϕ$
$d g \propto {c o s}^{2} ϕ$
$d g \propto {c o s}^{3 / 2} ϕ$
$d g \propto \frac{1}{c o s ϕ}$

We know that, the effective gravity $g^{'}$ at a latitude $ϕ$ is given as
$g^{'} = g - ω^{2} R {(cos ϕ)}^{2}$
Thus, $g - g^{'} = d g = ω^{2} R {(cos ϕ)}^{2}$