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

Answer:

7160.2812 s or 1.988 hours

Explanation:

m = Mass of person

R = Radius of Earth = $6.37\times 10^{6}\ m$

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

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

Time period is given by

$T=\dfrac{2\pi}{\omega}\\\Rightarrow T=2\pi\sqrt{\dfrac{2R}{g}}\\\Rightarrow T=2\pi\sqrt{\dfrac{2\times 6.37\times 10^6}{9.81}}\\\Rightarrow T=7160.2812\ s=1.988\ hours$

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