Why is this calculation of earth gravitational acceleration incorrect?

[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

 

[PJC01]

Hello Marius,

Thank you for bringing this to my attention. The equation a = {v^2 \over R} is not incorrect, but it is only applicable in certain situations. This equation is derived from the formula for centripetal acceleration, which is a = {v^2 \over r}, where v is the velocity and r is the distance from the center of rotation.

The equation you have used, a = {v^2 \over R}, assumes that the Earth is a perfect sphere with a constant radius of 6378m. However, the Earth is not a perfect sphere and its radius varies at different points on the surface. Additionally, the Earth's rotation causes a centrifugal force that counteracts the gravitational force, resulting in a slightly lower gravitational acceleration at the equator compared to the poles.

To accurately calculate the Earth's gravitational acceleration, we must take into account its non-uniform shape and rotation. This can be done using the formula a = {GM \over r^2}, where G is the gravitational constant, M is the mass of the Earth, and r is the distance from the center of the Earth. This formula takes into account the varying distance from the center of the Earth and the effects of rotation.

It is possible that the equation a = {v^2 \over R} was taught as a simplified version for introductory purposes, but it is important to understand its limitations and use the more accurate formula when necessary. I hope this helps clarify the issue.

Best regards,