# Calculating g with centripetal acceleration

• dmayers94
In summary, the formula a = w^2 * r does not give the value for g, or 9.8 m/s^2. Instead, it calculates the net acceleration required to keep an object moving in a circular path with a given radius. This is due to both the gravitational force and the normal force of the ground acting on objects at the Earth's surface. The formula is not meant to give the value for g, which can be found using the formula g = GM/R^2. What was calculated was the centrifugal force caused by the Earth's rotation at the equator, which is not the same as centripetal force.
dmayers94
Hi, I'm trying to find g, or 9.8 m/s^2, with the formula a = w^2 * r. First, I found the value for the angular velocity (2*pi/24 hours) and i converted this to radians per second finding the value 2*pi/86400. I googled the radius for the Earth and got 6,378,100 meters. Finally, i plugged these values into the formula to obtain a centripetal acceleration of 0.0337 m/s^2 which obviously isn't the answer I was looking for. Can someone find the holes in my math or logic? Thanks.

I haven't checked the numbers, but that formula won't give you g anyway. It will instead give you the net acceleration required to keep an object on the Earth's surface moving in the circular path given by the Earth's radius. At the Earth's surface, both the gravitational force and the normal force of the ground act on objects. What you have found is the difference between the accelerations caused by these two forces.

If gravity were the only force acting on an object at the Earth's surface, it would be pulled down into a much tighter "orbit." One with a radius much smaller than earth's. But this is not possible since contact forces with the ground counteract gravity somewhat, and they balance so that we stay on the surface of the Earth at all times.

This formula is not supposed to give you g.

you get g at a distance R from the Earth's centre using g=GM/R^2

where M is mass of earth
G is the Universal Gravitation constant.
For g at surface we use R as radius of earth.

What you are calculating by a=w^2*R is something entirely else.

You have calculated the centripetal acceleration felt by a particle moving in a circle of radius R at the same angular frequency as that of earth.

That has no reason to be g.

Infact its the centripetal acceleration a person on equator will have wrt to the Centre of Mass of Earth (but we don't feel as we are at rest relative to Earth's surface and because its too small) cause we are rotating with earth.

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Gravity is caused by masses attracting each other and not because they are spinning. Non-spinning objects attract each other just as well as spinning objects.

## What is centripetal acceleration?

Centripetal acceleration is the acceleration directed towards the center of a circular motion. It is responsible for keeping an object moving in a circular path.

## How is centripetal acceleration related to g?

Centripetal acceleration is directly proportional to the acceleration due to gravity, g. This means that the value of g can be calculated by measuring the centripetal acceleration of a body in circular motion.

## What is the formula for calculating g with centripetal acceleration?

The formula for calculating g with centripetal acceleration is g = (v^2)/r, where v is the velocity of the object and r is the radius of the circular motion.

## What units are used for measuring centripetal acceleration?

Centripetal acceleration is typically measured in meters per second squared (m/s^2).

## Can centripetal acceleration be negative?

Yes, centripetal acceleration can be negative if the direction of the acceleration is opposite to the direction of motion. This may occur in cases where the object is slowing down or changing direction in its circular motion.

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