Uniform Circular Motion of a Spherical Earth

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SUMMARY

The discussion centers on the uniform circular motion of objects released from the Earth's surface, specifically comparing the acceleration of objects at the equator versus the North Pole. It is established that objects at the North Pole experience a greater acceleration of 0.0338 m/s² due to the absence of centripetal acceleration, while objects at the equator experience a combination of gravitational acceleration and centripetal acceleration. The confusion arises from misinterpreting the role of centripetal acceleration in this context. The correct understanding is that centripetal acceleration is not an additional force but rather a component of the total acceleration experienced by objects in circular motion.

PREREQUISITES
  • Understanding of gravitational acceleration (g)
  • Knowledge of centripetal acceleration and its formula
  • Familiarity with the concept of uniform circular motion
  • Basic knowledge of Earth's rotation and its effects on objects
NEXT STEPS
  • Study the relationship between gravitational acceleration and centripetal acceleration in rotating systems
  • Explore the implications of Earth's rotation on object motion at different latitudes
  • Learn about the physics of circular motion and its applications in real-world scenarios
  • Investigate the mathematical derivation of centripetal acceleration and its significance
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in understanding the effects of Earth's rotation on object motion.

Gabble1
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Homework Statement


In this question, the Earth is modeled as a uniform sphere of radius 6400km. Objects are released from points just above the Earth's surface at the equator and at the North Pole. Which will fall to the Earth with the greater acceleration and by how much?

Homework Equations


The answer in the book states that objects at the North Pole will fall with a greater acceleration of 0.0338 m/s^2 (3 s.f.) than objects at the equator.

The Attempt at a Solution


I first took a cross section of the Earth at the equator as an example of circular motion.
The centripetal acceleration due to the rotating Earth, a=(omega)^2 * r = (2pi/T)^2 * r = (7.27 * 10^(-5))^2 * 6400000 = 0.0338 m/s^2 <- Correct.

However, at the equator total centripetal acceleration = centripetal acceleration due to rotating Earth + acceleration due to gravity = (0.0338 + g) m/s^2

However, at the North Pole, an object is not rotating with circular motion, thus total centripetal acceleration would simply be acceleration due to gravity, ie. g m/s^2.

Due to this I answered that objects at the equator would fall with a greater acceleration of 0.0338 m/s^2, however this is the opposite answer to the answer given.

Could someone please point out where I am going wrong in my reasoning?

Cheers
 
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Hi Gabble1,

Consider: is centripetal acceleration a real force that pulls an object in, or is it supplied by some other force pulling the object in?
 
Gabble1 said:
at the equator total centripetal acceleration = centripetal acceleration due to rotating Earth + acceleration due to gravity
You seem rather confused. There are not two different centripetal accelerations. There is gravitational acceleration (which is the actual acceleration, since there are no other forces), the centripetal acceleration (which would be the actual acceleration if the object were to stay at the same altitude while it circled the Earth), and the apparent acceleration(i.e. the second derivative of its altitude).
What do you think the relationship is between those three?
 

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