Calculating the Volume of Spheres with Different Radii

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The formula for calculating the volume of a sphere is V = (4/3)πr^3, where r is the radius. For a sphere with a radius of 5.0 m, substituting into the formula gives a volume of approximately 523.6 m³. To find the volume of air around the Earth, one must calculate the volume of a sphere with a radius of 6500 km (Earth's radius plus atmosphere) and subtract the volume of the Earth alone, which has a radius of 6400 km. This method accurately determines the volume of the atmosphere surrounding the Earth. Understanding and applying this formula is crucial for solving related volume problems.
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What is the formula? I have two problems:

1. What is the volume of a sphere with a radius of 5.0 m?

My initial guess was 5^3 = 125, but apparently the answer is 523.6.

2. The radius of the Earth is 6400 km. If the atmosphere is approximately 10 km high, then what is the volume of air around the earth?

Ok to be honest, I have no idea. I assume you find the volume of the earth, then the volume of a sphere with radius 10 km, and subtract the two... but that goes back to me not knowing the forumla of volume of a sphere :D
 
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Revolver said:
What is the formula? I have two problems:

1. What is the volume of a sphere with a radius of 5.0 m?

My initial guess was 5^3 = 125, but apparently the answer is 523.6.
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v= \frac{4 \pi r^3}{3}
2. The radius of the Earth is 6400 km. If the atmosphere is approximately 10 km high, then what is the volume of air around the earth?

Ok to be honest, I have no idea. I assume you find the volume of the earth, then the volume of a sphere with radius 10 km, and subtract the two... but that goes back to me not knowing the forumla of volume of a sphere :D

You might want to rethink that second one.
 
OK, the general formula for the volume of a sphere is V=(4/3)pi(r^3), where r=radius and pi=3.14 approx.
Question 1 you just have to simply substitute r=5 into the equation. For Question 2 you find the volume of the Earth plus the atmosphere together, having a radius of 6500km, and subtract from this the volume of the Earth alone, where r=6400km. Hopefully that helps.
 
It's two spheres but the larger one is the one that includes the atmosphere it would be (6400 + 10)Km and the smaller one would Earth's 6400 km radius to ground.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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