Why does centripetal acceleration at Earth's equator give 0.03 m/s²?

[Jonsson]

Hello there,

I was taught:

$$a = {v^2 \over R}$$. I substitute $$v$$ for the speed of the rotating Earth at the equator, and the radius, $$R = 6378m$$. And I get $$a = 0.03 \rm m/s^2$$.

It looks like the equation $$a = {v^2 \over R}$$ may incorrect. Why am I taught this equation in University if it is false?

Thank you for your time.

Kind regards,
Marius

 

[jbriggs444]

That is the correct equation for the centripetal acceleration of an object in uniform circular motion. The numeric result that you have calculated is correct for the centripetal acceleration of a man standing on the equator, although the units are off. It should be in m/s².

But the gravity is not the only force acting on such a man. The ground is also pushing upwards against the soles of his shoes. It is the difference between the upward force of the ground on his shoes and the downward force of gravity on his body that accounts for the residual 0.03 m/sec² acceleration

 

[tiny-tim]

Hello Marius!

v²/r is the centripetal acceleration at the equator

if the Earth had no gravity, it is the tension in the string (divided by m) that would be needed to keep you on the ground at the equator

for that reason, although your weight at the poles is mg, your weight at the equator (as measured by standing on a scale) is less than mg by a very small amount, mv²/r

 

[Jonsson]

Thanks!M