Question

(a) Suppose that your measured weight at the equator is one-half your measured weight at the pole on a planet whose mass and diameter are equal to those of Earth. What is the rotational period of the planet?

Answer 1

7160.2812 s or 1.988 hours

Step-by-step explanation:

m = Mass of person

R = Radius of Earth =
`6.37* 10^(6)\ m`

g = Acceleration due to gravity = 9.81 m/s²

`\omega` = Angular speed

Force at equator would be

`F_e=m(g-\omega^2R)`

Force at pole

`F_p=mg`

From the question

`F_e=(1)/(2)F_p\\\Rightarrow m(g-\omega^2R)=(1)/(2)F_p\\\Rightarrow \omega=\sqrt{(g)/(2R)}`

Time period is given by

`T=(2\pi)/(\omega)\\\Rightarrow T=2\pi\sqrt{(2R)/(g)}\\\Rightarrow T=2\pi\sqrt{(2* 6.37* 10^6)/(9.81)}\\\Rightarrow T=7160.2812\ s=1.988\ hours`

The rotational period of the planet is 7160.2812 s or 1.988 hours