Answer
(a) $a_R = 0.0337~m/s^2$
$a_R = 0.00344~g$
(b) The rotation period would have to be 84.5 minutes.
Work Step by Step
(a) We can find the speed on the equator.
$v = \frac{2\pi r}{t} = \frac{(2\pi) (6.38\times 10^6~m)}{24 \times 3600~s}$
$v = 464~m/s$
We can use the speed to find the radial acceleration.
$a_R = \frac{v^2}{r} = \frac{(464~m/s)^2}{6.38\times 10^6~m}$
$a_R = 0.0337~m/s^2$
$a_R = \frac{0.0337~m/s^2}{9.80~m/s^2} = 0.00344~g$
(b) $a_R = \frac{v^2}{r} = g$
$\frac{(2\pi~r)^2}{t^2~r} = g$
$t^2 = \frac{4\pi^2~r}{g}$
$t = {2\pi}\sqrt{\frac{r}{g}}$
$t = {2\pi}\sqrt{\frac{6.38\times 10^6~m}{9.80~m/s^2}}$
$t = 5069.6~s = 84.5~min$
The rotation period would have to be 84.5 minutes.
