99.9999% sure the book is wrong about this circumference problem

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Homework Help Overview

The problem involves calculating how many balls, with a given circumference, would fit around the Earth's equator when touching each other. The original poster questions the use of the Earth's radius in the calculation, as they believe the problem specifies the circumference rather than the radius.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations for the ball's diameter and the Earth's circumference, comparing their results with the book's answer. Some question the reasoning behind using the radius of the Earth instead of the circumference, while others suggest that small errors in approximation may affect the results.

Discussion Status

There is an ongoing exploration of the calculations and assumptions made in the problem. Some participants express agreement with the original poster's concerns, while others propose that the book's answer could still be correct under certain conditions. Multiple interpretations of the problem are being examined.

Contextual Notes

Participants note potential errors in calculations and the significance of rounding and approximation in their results. There is a mention of the need for clarity regarding the definitions of radius and circumference in the context of the problem.

ISX
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Homework Statement


The circumference of a ball is 29.6 inches. Given that the radius of Earth is about 6400 km, how many balls would it take to circle around the equator with the balls touching one another?

Homework Equations


(d)(pi) = circumference
1 inch = 0.0254 meters
1 km = 1000 meters

The Attempt at a Solution


Ball diameter = ((29.6)(0.0254))/3.14 = 0.239 m
Earth circumference = (6400)(2)(3.14)(1000) = 40,192,000 m

# of balls circumnavigating the equator = 40,192,000/0.239 = 168,167,364

In the back of the book it says 26,700,000 balls (it only uses 3 significant digits) and I can get that if I divide the radius of the Earth by the ball diameter: ((6400)(1000))/0.239 = 26,700,000

I don't see why they are using the radius when it says around the earth, not halfway through it. Am I right here? Didn't know if there was some sort of magic I was missing.
 
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I agree with you, although you should lose a few significant figures.
 
ISX said:

Homework Statement


The circumference of a ball is 29.6 inches. Given that the radius of Earth is about 6400 km, how many balls would it take to circle around the equator with the balls touching one another?


Homework Equations


(d)(pi) = circumference
1 inch = 0.0254 meters
1 km = 1000 meters

The Attempt at a Solution


Ball diameter = ((29.6)(0.0254))/3.14 = 0.239 m
Earth circumference = (6400)(2)(3.14)(1000) = 40,192,000 m

# of balls circumnavigating the equator = 40,192,000/0.239 = 168,167,364

In the back of the book it says 26,700,000 balls (it only uses 3 significant digits) and I can get that if I divide the radius of the Earth by the ball diameter: ((6400)(1000))/0.239 = 26,700,000

I don't see why they are using the radius when it says around the earth, not halfway through it. Am I right here? Didn't know if there was some sort of magic I was missing.

It does look like the book's answer is wrong. But I believe that your answer is slightly wrong as well. Can you think of a small error term that you did not take into account in your calculation?
 
Yes the ball radius makes the Earth's diameter bigger since the ball isn't flat, I didn't think that would hardly make a difference when they always round it giving it a huge tolerance of error.
 
ISX said:
Yes the ball radius makes the Earth's diameter bigger since the ball isn't flat, I didn't think that would hardly make a difference when they always round it giving it a huge tolerance of error.

I'll agree with that. There's other small approximation errors as well. Like the amount of circumference a ball occupies isn't exactly the same as the diameter of the ball. The 'exact' way to do it would be to find the angle subtended by a ball as viewed from the center of the Earth and divide it into 2*pi. Just for fun I did the number the ISX way and then did it that way and looked at the difference. It's pretty close to pi balls. Which you can account for with that radius correction. So you don't even need a 'huge tolerance of error'. The given answer would be right to within a few balls if a more accurate version of pi were used, assuming all of the other numbers were exact.
 
Last edited:
Using your given numbers I get 26,742,651 so I think your text is correct.
 
LCKurtz said:
Using your given numbers I get 26,742,651 so I think your text is correct.

Can you show your work? ISX is 99.9999% sure. Ibix agreed. Berkeman agreed. I did it two different ways and I agree. I'm really interested.
 
Last edited:
LCKurtz said:
Using your given numbers I get 26,742,651 so I think your text is correct.

Care to explain how?
 
Dick said:
Can you show your work? ISX is 99.9999% sure. Ibix agreed. Berkeman agreed. I did it two different ways and I agree. I'm really interested.

ISX said:
Care to explain how?

Now you guys have me worried -- hold on, I'll check...
 
  • #10
Yup. I must have made the same mistake the book did. I used 29.6 as the radius instead of the circumference. :redface:
 
  • #11
Had me worried there lol.
 
  • #12
LCKurtz said:
Yup. I must have made the same mistake the book did. I used 29.6 as the radius instead of the circumference. :redface:

I was just thinking you had responded to the initial post without reading anything of what the responses were further down the line. That can be kind of annoying.
 
Last edited:

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