Calculate G-Force: How to Measure Gravity's Force

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SUMMARY

The calculation of G-Force involves using the formula V^2/R, where V represents velocity in meters per second and R is the radius in meters. G-Force is equivalent to centripetal acceleration, which can also be derived from linear acceleration. For example, a rider on a merry-go-round with a radius of 15 meters and a velocity of 8 m/s experiences approximately 0.2724 G's, or 27.24% of Earth's gravitational pull. This calculation illustrates the relationship between velocity, radius, and the perceived gravitational force in a circular motion.

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  • Understanding of basic physics concepts such as force and acceleration
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  • Knowledge of Newton's law of gravitation
  • Ability to perform basic mathematical operations involving squares and division
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How do you calculate G-Force? I know that one g is the force gravity exerts on a particular mass (I think). But how exactly do you figure it out, particularly with objects spinning in a circle, like a merry-go-round?
 
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V^2/R

Velocity in m/s with that number squared divided by the radius in meters.
 
Ferraridude said:
V^2/R

Velocity in m/s with that number squared divided by the radius in meters.

So, G-Force is the same as centripetal acceleration?
 
SignSeeker7 said:
So, G-Force is the same as centripetal acceleration?

"G force" is just the ratio of a given force to the force due to gravity on the same object. Think of it as a unit of measure, it's not a special kind of force. Typically, "G forces" are referring to pseudo forces experienced in an accelerating reference frame.

Now, if you wanted to know the gravitational attraction between two objects, then you'd use Newton's law of gravitation.
 
SignSeeker7 said:
So, G-Force is the same as centripetal acceleration?

Yes, if you are going in a circle, or even changing direction. You can also get g force from linear acceleration. If you were in a rocket that could accelerate at 4 g's straight up, or 39.2 m/sec^2 you would be subject to 5 g's. Four from the rocket and one from the Earth.
 
If you were going around a merry go round with a radius of 15 meters at a velocity of 8 m/s, this is how you would calculate the G-forces.

8^2=64
64/15=approximately 2.67.

So now, you have your m/s^2 for the acceleration.

2.67/9.8 m/s^2 (9.8 m/s^2 is the acceleration of the Earth's gravity)=.2724

So, we can now see that the centripetal force on a merry go round rider with that velocity around a circle with that radius, we are feeling .2724 G's of acceleration, or 27.24% of Earth's gravitational pull.
 
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I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question. Ex. 1 - A moving object impacts a static object at a constant velocity. Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact. Assuming the mass of the objects is the same and the velocity at the moment of impact...

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