Calculate Speed of 91.7 kg Person at Equator

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Homework Help Overview

The discussion revolves around calculating the speed of a 91.7 kg person at the equator, in the context of Earth's rotation. The problem references the misconception that Earth's rotation would cause people to be thrown into space, prompting an exploration of the forces acting on a person at the equator.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the balance of normal and gravitational forces and their relation to centripetal acceleration. There are attempts to calculate speed using the formula v = 2πr/T, with some questioning the validity of their results and the time period used for Earth's rotation.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the correct time period for Earth's rotation, and there is a recognition that the initial calculations may have used an incorrect time frame.

Contextual Notes

Participants are working with specific values for Earth's radius and gravitational acceleration, while also addressing the assumption that the Earth rotates once per day. There is a noted confusion regarding the appropriate time period for the calculation.

Maiia
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Homework Statement


An early major objection to the idea that Earth is spinning on its axis was that Earth
would turn so fast at the equator that people would be thrown into space.
Given : radius of Earth = 6.37 × 106 m,
mass of Earth = 5.98 × 1024 kg ,
radius of moon = 1.74 × 106 m, and
g = 9.8 m/s2 .
Show the error in this logic by calculating
the speed of a 91.7 kg person at the equator.
Answer in units of m/s.

well drawing a free body diagram, we have normal and gravitational forces- which equal each other..so isn't there no unbalanced force? because doesn't there need to be one to have centrifugal acceleration?
 
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Hi Maiia,

Maiia said:

Homework Statement


An early major objection to the idea that Earth is spinning on its axis was that Earth
would turn so fast at the equator that people would be thrown into space.
Given : radius of Earth = 6.37 × 106 m,
mass of Earth = 5.98 × 1024 kg ,
radius of moon = 1.74 × 106 m, and
g = 9.8 m/s2 .
Show the error in this logic by calculating
the speed of a 91.7 kg person at the equator.
Answer in units of m/s.

well drawing a free body diagram, we have normal and gravitational forces- which equal each other

The normal and gravitational forces will not equal each other. The difference between these two forces is what will provide the centripetal acceleration, and so you could find how much they differ.

However, to find the real speed of a person at the equator, you just need the quantities given in the problem, and also use the fact that the Earth rotates once per day. What do you get?
 
hmm i did v= 2pir/T so 2pi(6.37x 10^6)/ 3600sec, adn i got 11117.74m/s- isn't that too big?
 
Maiia said:
hmm i did v= 2pir/T so 2pi(6.37x 10^6)/ 3600sec, adn i got 11117.74m/s- isn't that too big?

Yes, that's too big. You divided by a time period of 3600 seconds, so that would be the speed if the Earth spun around once per hour.
 
but I thought it was supposed to be in seconds? cycle/sec
 
How long does it take the Earth to rotate?

Hint: it is longer than 1 hour.

I thought it was supposed to be in seconds?
Yes. Once you answer the above question, then convert that answer into seconds.
 

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