Does the gravity we feel equal G(M/r^2) - our centripetal acceleration

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SUMMARY

The centripetal acceleration required for objects in circular motion, such as those on Earth's surface, is indeed influenced by gravitational force, calculated using the formula a=G(M/r^2), which yields approximately 9.81 m/s². However, at the equator, individuals experience a slightly reduced effective weight due to the centripetal acceleration counteracting a portion of gravitational force. This means that while gravitational pull remains constant, the normal force exerted by the Earth varies, leading to a perception of reduced weight at the equator compared to the poles. This phenomenon is significant in understanding the nuances of weight perception in different geographical locations.

PREREQUISITES
  • Understanding of centripetal acceleration and its role in circular motion
  • Familiarity with gravitational force calculations using G(M/r²)
  • Knowledge of how Earth's rotation affects weight perception
  • Basic concepts of physics related to forces and motion
NEXT STEPS
  • Explore the effects of Earth's rotation on gravitational force variations
  • Study the implications of centripetal acceleration in different geographical locations
  • Investigate the definition and measurement of weight in physics
  • Learn about the Earth's shape and how it affects gravitational pull
USEFUL FOR

Physics students, educators, and anyone interested in the effects of gravity and motion on weight perception in different locations on Earth.

D.Hayward
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We all know that something traveling in cicular motion at a constant speed has a centripetal acceleration towards the axis of rotation. I suppose this means we (travelling around the axis of rotation of the earth) also require a centripetal acceleration. This can be provided by gravity which is calculated by a=G(M/r^2) to be 9.81 m/s^2. However a portion of this has to be 'used' just to stop us drifting into space, this should be equal to our centripetal acceleration. Therefore shouldn't we feel 9.81 m/s^2 minus our centripetal acceleration in everyday life and in physics practicals. Is this true?
 
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D.Hayward said:
We all know that something traveling in cicular motion at a constant speed has a centripetal acceleration towards the axis of rotation. I suppose this means we (travelling around the axis of rotation of the earth) also require a centripetal acceleration. This can be provided by gravity which is calculated by a=G(M/r^2) to be 9.81 m/s^2. However a portion of this has to be 'used' just to stop us drifting into space, this should be equal to our centripetal acceleration. Therefore shouldn't we feel 9.81 m/s^2 minus our centripetal acceleration in everyday life and in physics practicals. Is this true?

Yes. Basically. You would not "weigh" as much at the equator as you would at the poles. However, you have to be very precise about what the word "weigh" means here. More precisely, the Earth would not have to push up on you with as much normal force in order to support you, and so you would "feel" less heavy (although I'm don't think the difference is discernible). Some people define "weight" as this, in which case yes, you would "weigh" less by that definition. Others define your weight as the force with which Earth's gravity pulls on you, which would not be different between the two locations.* Under this definition, your weight is the same, even if the surface of the Earth doesn't have to support as much of it.

*for a perfectly spherical Earth, that is. In reality, Earth is not perfectly spherical and g varies with location on the Earth's surface.

http://curious.astro.cornell.edu/question.php?number=310
 
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