Question

A body is weighed by a spring balance to be $1000$N at the north pole. How much will it weigh(in N) at the equator? Account for the earth's rotation only.

Answer 1

At equator

$g^{\prime }=g-R{\omega }^2$

$=(9.8{)}^2-(6400\times {10}^3){\left(\frac{2\pi }{24\times 3600}\right)}^2$

$=9.766m/s^2$

Now,$mg=1000$

$\therefore m=\frac{1000}{g}$

$w^{\prime }=m{g}^{\prime }=\left(1000\right)\left(\frac{g^{\prime }}{g}\right)$

$=\frac{\left(1000\right)\left(9.766\right)}{\left(9.8\right)}$

$=997$N