Question

An early major objection to the idea that Earth is spinning on its axis was that Earth would turn so fast at the equator that people would be thrown into space. Given : rearth = 6.37 × 106 m , mearth = 5.98 × 1024 kg , rmoon = 1.74 × 106 m , and g = 9.8 m/s 2 . Show the error in this logic by calculating the speed of a 117 kg person at the equator. Answer in units of m/s.

Answer 1

Explanation:

In order to a person to be thrown into space, the centripetal force has to be greater than his weight. Thus, we have to calculate the magnitudes of this forces:

$F_c=ma_c(1)\\W=mg$

The centripetal acceleration is given by:

$a_c=\frac{v^2}{r}$

The speed is given by the distance traveled ($2\pi r_{earth}$) divided into the time spent in one revolution (the period), that is, one day.

$v=\frac{2\pi r_{earth}}{T}\\v=\frac{2\pi(6.37*10^{6}m)}{86400s}\\v=463.24\frac{m}{s}$

Now, we can calculate the centripetal force:

$F_c=m\frac{v^2}{r}\\F_c=117kg\frac{(463.24\frac{m}{s})^2}{6.37*10^{6}m}\\F_c=3.94N$

Finally, we calculate the weight of the person:

$W=117kg(9.8\frac{m}{s^2})\\W=1146.6N$

Since the weight is much greater than the centripetal force, we can say that people wouldn't be thrown into space.