Answer
(a) The radial acceleration at the equator is $0.0337~m/s^2$
(b) An object's apparent weight at the equator is less than its actual weight.
(c) An object's weight is $0.34\%$ lighter at the equator than the object's actual weight.
(d) At the poles, an object's apparent weight on a bathroom scale is equal to the object's true weight.
Work Step by Step
(a) We can find the angular speed of the Earth as it rotates:
$\omega = \frac{2\pi~rad}{(24)(3600~s)}$
$\omega = 7.27\times 10^{-5}~rad/s$
We can find the radial acceleration at the equator:
$a_r = \omega^2~r$
$a_r = (7.27\times 10^{-5}~rad/s)^2~(6.38\times 10^6~m)$
$a_r = 0.0337~m/s^2$
The radial acceleration at the equator is $0.0337~m/s^2$
(b) We can find the normal force $F_N$ exerted on an object at the equator. Note that the normal force is equal to the object's apparent weight.
$\sum F = m~a_r$
$mg-F_N = m~a_r$
$F_N = mg - m~a_r$
We can see that the normal force is less than the object's weight $mg$. Therefore, an object's apparent weight at the equator is less than its actual weight.
(c) We can find the percentage difference between an object's apparent weight at the equator and the object's actual weight $mg$:
$\frac{(mg)-(mg-m~a_r)}{mg}\times 100\%$
$ = \frac{m~a_r}{mg}\times 100\%$
$ = \frac{a_r}{g}\times 100\%$
$ = \frac{0.0337~m/s^2}{9.80~m/s^2}\times 100\%$
$ = 0.34\%$
An object's weight is $0.34\%$ lighter at the equator than the object's actual weight.
(d) At the poles, the centripetal force is zero because the radius of rotation is zero. Therefore, at the poles, an object's apparent weight on a bathroom scale is equal to the object's true weight.
