Answer
$\frac{3.4}{1000}$
Work Step by Step
$v = \frac{2\pi r}{T}$..................(1)
$T =(24 ~h)(3600 ~s/h) = 86,400 ~s$
We can use the formula and data above to find the centripetal acceleration at the equator:
$a = \frac{v^2}{r} = \frac{4 \pi^2 r^2}{T^2 r} = \frac{4 \pi^2 r}{T^2}$
$a = \frac{(4\pi^2)(6.38 \times 10^6 ~m)}{(86,400 ~s)^2} = 0.0337~m/s^2$
Let's find the ratio of this centripetal acceleration to the acceleration of gravity.
$\frac{a_c}{g} = \frac{0.0337~m/s^2}{9.80 ~m/s^2} = \frac{3.4}{1000}$
