Question

The value of $h$ will be

• A. $\frac{{\omega }^2{R}^2}{2g}$
• B. $\frac{{\omega }^2{R}^2}{g}$
• C. $\frac{2{\omega }^2{R}^2}{3g}$
• D. $\frac{2{\omega }^2{R}^2}{g}$

Answer 1

Answer: (A)

Gravity is related to the angular speed is given by
$g^{{}^{\prime }}=g-{\omega }^{2}R$
So the weight at the equator is $m{g}^{{}^{\prime }}=m(g-{\omega }^{2}R)$
Gravity above the earth's surface is given by
$g{}^{\prime }=g(1-\frac{2h}{R})$
Implies the weight at the equator above the earth's surface is $m{g}^{{}^{\prime }}=mg(1-\frac{2h}{R})$
It is given that these two weights are equal,
$\Rightarrow m(g-{\omega }^{2}R)=mg(1-\frac{2h}{R})$
$\Rightarrow g-{\omega }^{2}R=g-\frac{2gh}{R}$