99.9999% sure the book is wrong about this circumference problem

AI Thread Summary
The discussion centers on a problem involving the calculation of how many balls, with a circumference of 29.6 inches, would fit around the Earth's equator. The initial calculations yield approximately 168 million balls, while the book states 26.7 million, leading to confusion over the use of Earth's radius instead of its circumference. Participants agree that the book's answer is incorrect, but also acknowledge potential errors in their own calculations, particularly regarding the ball's radius affecting the overall diameter. A more precise method of calculation suggests that the book's answer could be correct with slight adjustments for approximation errors. Ultimately, the conversation highlights the importance of accurate measurements and significant figures in mathematical problems.
ISX
Messages
120
Reaction score
0

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.
 
Physics news on Phys.org
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:
Back
Top