Solve the following :


A body is weighed by a spring balance to be $1000 \mathrm{~kg}$ at the north pole. How much will it weigh at the equator? Account for the earth's rotation only.


Let $g_{p}$ be the gravity at the poles , $g_{e}$ be the gravity at the equator

$g_{e}=g_{p}-\omega^{2} R$

$=9.81-\left(7.3 \times 10^{-5}\right)^{2} \times\left(7.3 \times 10^{-5}\right)^{2} \times 6400 \times 10^{3}$

$=9.766 \mathrm{~N} / \mathrm{m}^{2}$

Now, $\mathrm{mg}_{\mathrm{e}}=1 \mathrm{~kg} \times 9.766 \mathrm{~N} / \mathrm{m}^{2}$

$=9.766 \mathrm{~N}$

Thus the body weighs $9.766 \mathrm{~N}$ at the equator.

Leave a comment