Question

A $75kg$ man weighs himself at the north pole and at the equator. Which scale reading is higher? By how much? Assume the earth is spherical.

Short answer

The weight at the north pole is larger by $2.5N$.

Step-by-step solution

Given Information 

The mass of the person is $m=75\mathrm{kg}$.

Explanation 

$=75\times 9.8-75\times 6.73\times {10}^6\times \times (7.3\times {10}^{-5})$The earth is rotated on its axis, which is passing through the poles. So on poles object does not have centripetal acceleration by earth. So acceleration is zero.

Therefore, at north pole the apparent weight is:

$\omega =n=mg$

$=75\times 9.8$

localid="1648375745460" $=735N$

On equator,

$\omega =\frac{2\pi }{T}$

$=7.3\times {10}^{-5}rad/s$

Therefore,

localid="1648375916104" $mg-n=\frac{m{v}^2}{R}=mr{\omega }^2$

Apparent weight ${\omega }_2=n=mg-mR{\omega }^2$

$=75\times 9.8-75\times 6.37\times {10}^6\times (7.3\times {10}^{-5})$

$735-732.5$

$=2.5N$

Final Answer 

The weight at the north pole is larger by$2.5N$