Calculate Speed of 91.7 kg Person at Equator

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AI Thread Summary
The discussion centers around calculating the speed of a 91.7 kg person at the equator to address misconceptions about Earth's rotation. Participants clarify that the normal and gravitational forces do not equal each other, as the difference provides centripetal acceleration. The correct formula for speed involves the Earth's rotation period, which is 24 hours, not 1 hour. An initial calculation yielded an incorrect speed due to using the wrong time period for Earth's rotation. Ultimately, the focus is on accurately determining the rotational speed using the correct time frame.
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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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