Calculate Speed of 91.7 kg Person at Equator

In summary, the initial objection that the Earth's rotation would cause people at the equator to be thrown into space is incorrect. Using the given values of the Earth's radius, mass, and the force of gravity, we can calculate the speed of a 91.7 kg person at the equator to be approximately 11117.74 m/s. However, this is incorrect as it is based on a time period of 1 hour, whereas the Earth actually rotates once per day. When converted to seconds, the correct speed is significantly lower.
  • #1
Maiia
79
0

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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  • #2
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?
 
  • #3
hmm i did v= 2pir/T so 2pi(6.37x 10^6)/ 3600sec, adn i got 11117.74m/s- isn't that too big?
 
  • #4
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.
 
  • #5
but I thought it was supposed to be in seconds? cycle/sec
 
  • #6
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.
 

Related to Calculate Speed of 91.7 kg Person at Equator

What is the formula for calculating speed?

The formula for calculating speed is distance divided by time. In this case, we will use the distance traveled around the Earth's equator, which is approximately 40,075 km, and the time it takes for one full rotation, which is 24 hours.

What is the weight of a 91.7 kg person?

The weight of a 91.7 kg person is approximately 202 pounds.

How do you convert kilograms to pounds?

To convert kilograms to pounds, you can use the formula: 1 kg = 2.20462 lbs. Therefore, a 91.7 kg person would weigh approximately 202 pounds.

What is the circumference of the Earth at the equator?

The circumference of the Earth at the equator is approximately 40,075 km or 24,901 miles.

How fast does a person at the equator travel in 24 hours?

A person at the equator travels at a speed of approximately 1,674.34 km/h or 1,040.40 mph in 24 hours.

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