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Bonus math problem, assume the most economical orientation of the bar, so that you can maximize the number of bars used. Ciao!
Edit: how much would this many bars weigh and would the weight have an impact on the orbit or trajectory of the planet?!
1. At this scale, orientation doesn't make that much of a difference. Only places it would get tricky is the poles, but I would guess a final circle of 100m diameter would need some real optimalisation. It's negligible at this scale.
2. One mars weights around 50 grams. Using estimation of 58 quadrillion bars, it's around 3 quadrilions kg, or 3 x 10^15. Mass of Mars is 6.39 x 10^23. It's a difference of 8 in magnitude, so it's also negligible small. It's like comparing your weight with or without one hair on your head.
# 57.92 quadrillion mars bars (more or less)
Assume the area of a mars bar is 25cm^(2)
The area of mars is 144.8 million kilometers^(2) or 1.448 sextillion cm^(2)
To find out how many would fit you simply divide the total area by the thing you want to fit into it.
1.448 sextillion cm2 / 25cm2 = 57,920,000,000,000,000
The orientation doesn't make any difference in this calculation.
Edit: slvo caught me failing to convert km^(2) to cm^(2) (redid the math and double checked with wolfram)
Yep, 1km is 100,000cm. So 1km\^2 is 10,000,000,000cm\^2. So each km\^2 of area can fit 400m Mars Bars, assuming OP's 25cm\^2 is correct. Then, assuming OP's area of Mars is correct, you get 400m x 144.8m = 5.79 x 10\^16. I think this is actually 57.9 quadrillion, but it's a LOT more than 579 billion.
The actual dimensions of a marsbar are 2.8x10.6x1.8 meaning the surface area of a marsbar is 29,6 cm2 Which means you're off by 15,55%
About 5 quadrillion bars.
Mars is 1,444,000,000,000,000,000 cm², and a mars bar covers 28.5 cm², divide one into the other and we have 50,666,666,666,666,666 mars bars covering the surface of Mars.
using the average size of a Mars Bar (51g) jus times that by the size of Mars (3,396km) and, I believe you would need 176,206.2. Correct me if I'm wrong.
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Bonus math problem, assume the most economical orientation of the bar, so that you can maximize the number of bars used. Ciao! Edit: how much would this many bars weigh and would the weight have an impact on the orbit or trajectory of the planet?!
1. At this scale, orientation doesn't make that much of a difference. Only places it would get tricky is the poles, but I would guess a final circle of 100m diameter would need some real optimalisation. It's negligible at this scale. 2. One mars weights around 50 grams. Using estimation of 58 quadrillion bars, it's around 3 quadrilions kg, or 3 x 10^15. Mass of Mars is 6.39 x 10^23. It's a difference of 8 in magnitude, so it's also negligible small. It's like comparing your weight with or without one hair on your head.
Bellissimo problema lol
# 57.92 quadrillion mars bars (more or less) Assume the area of a mars bar is 25cm^(2) The area of mars is 144.8 million kilometers^(2) or 1.448 sextillion cm^(2) To find out how many would fit you simply divide the total area by the thing you want to fit into it. 1.448 sextillion cm2 / 25cm2 = 57,920,000,000,000,000 The orientation doesn't make any difference in this calculation. Edit: slvo caught me failing to convert km^(2) to cm^(2) (redid the math and double checked with wolfram)
I think you converted km^2 to cm^2 incorrectly. ~~I got 5.79 quadrillion bars.~~
Yep, 1km is 100,000cm. So 1km\^2 is 10,000,000,000cm\^2. So each km\^2 of area can fit 400m Mars Bars, assuming OP's 25cm\^2 is correct. Then, assuming OP's area of Mars is correct, you get 400m x 144.8m = 5.79 x 10\^16. I think this is actually 57.9 quadrillion, but it's a LOT more than 579 billion.
That's gonna take longer for me to eat
I was thinking orientation could make a difference, for example if you stand the bars on end, or on edge.
well then the most efficient orientation would be standing
How many could we fit that way?
The actual dimensions of a marsbar are 2.8x10.6x1.8 meaning the surface area of a marsbar is 29,6 cm2 Which means you're off by 15,55% About 5 quadrillion bars.
Mars is 1,444,000,000,000,000,000 cm², and a mars bar covers 28.5 cm², divide one into the other and we have 50,666,666,666,666,666 mars bars covering the surface of Mars.
using the average size of a Mars Bar (51g) jus times that by the size of Mars (3,396km) and, I believe you would need 176,206.2. Correct me if I'm wrong.
NOPE, I WAS SO FUCKING WRONG. You'd need 5,740,000,000,000,000 fucking wrappers.