Eli5: is here any noticeable change in our weight based on how close or far away we are from the equator because of centrifugal force from earth’s rotation , since gravity is the same everywhere( at same altitude)?
In: 2
One correction:
>Gravity is the same everywhere
This is not actually true. Gravity towards the poles is actually ALSO stronger because they’re literally closer to the center of the earth. There are also smaller variations all over the place due to local rocks/ice/water.
Depends on what’s your threshold for “noticeable”. You could actually measure the difference on a really high quality scale, because your weight would be about 0.3% lower. That’s about 99.7 kg at the equator compared to 100 at the poles. Your average bathroom scale doesn’t have the sensitivity for it, but it’s not laboratory grade technology.
Not noticeable by a person, but noticeable by scientific instruments.
The formula for centrifugal force is F = mω^2 r, m being mass, ω being angular velocity, r being the distance to the axis of rotation.
Let’s call the weight you experience W, so W = mg – mω^2 r
At a pole, W = mg since r would be 0m
W = m (g – ω^2 r)
So the difference in the acceleration you appear to have due to gravity would just be that ω^2 r, where as g would be 9.8m/s^2
ω = 2π radians/day (speed of the Earth’s rotation)
r = 6378 km (equatorial radius of Earth)
ω^2 r = ~.0337 m/s^2
Earth’s gravity also isn’t uniform. If you look at a gravity map of Earth you can see some regions are higher or lower. James Bay in Canada for example has some.of the lowest gravity in the world.
Latest Answers