Sphere Problem - Please correct

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In summary, if 10 000 cubic km of water is spread over a sphere with radius R = 6370 km, the depth of the water layer would be approximately 2 cm. This can be calculated by dividing the volume of the water by the surface area of the sphere, which is equal to 4/3*π*r^3 and 4*π*r^2, respectively. By converting the units to centimeters, we can determine that the depth of the water layer is 1.96 cm.
  • #1
Physicsrapper
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10 000 cubic km of water can cover all the continents with a layer 7 cm deep.

Find the depth of the water layer if 10 000 cubic km of water is spread over a sphere with radius R = 6370 km.
________________________________________________

My calculations:

Volume of the sphere: 4/3*π*r^3

R* = R + d (d = depht of the water=> searched)
d = R* - R
V* = V + v (V = 4/3*π*6370^3 ; is the little v equal to the 10 000 km^3?)


Would that be correct:

V = 4/3*π*6370^3 = 1.0826969e+12 km^3 (?)
v = 10 000 km^3 (?)

Total V*: 1.0826969e+12

Next we search R*:
Volume of the sphere: V* = 4/3*π*R*^3
=> R*^3 = V*/π*3/4
= (1.0826969e+12/π)*3/4 => take the third root = 1.73205080757

So d ~ 1.73 cm ?

______________________________________

Please correct this! I'm really not sure if it's correct.

Can someone please help and explain what I did wrong?
 
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  • #2
How did you get that number (it is wrong)? I don't understand your calculations. Where do you compare R* with R and where do the "cm" for the final answer come from?

Your approach will get a problem with the accuracy, as you subtract numbers extremely close together. The depth of the water will be tiny compared to the radius of the sphere, it is easier to divide the volume by the surface area of the sphere.
 
  • #3
mfb said:
How did you get that number (it is wrong)? I don't understand your calculations. Where do you compare R* with R and where do the "cm" for the final answer come from?

Your approach will get a problem with the accuracy, as you subtract numbers extremely close together. The depth of the water will be tiny compared to the radius of the sphere, it is easier to divide the volume by the surface area of the sphere.


Volume of the sphere: 4/3*π*r^3 ---> is 10 000 km^3 the volume?
Surface of the sphere: 4*π*r^2 ---> it's 4/3*π*6370^3 = 509.9 * 10^6 , isn't it?

So I wanted to divide 10 000 km^3 by 509.9 * 10^6, but it doesn't funcion.

What did I wrong?
 
  • #4
Physicsrapper said:
Volume of the sphere: 4/3*π*r^3 ---> is 10 000 km^3 the volume?
Surface of the sphere: 4*π*r^2 ---> it's 4/3*π*6370^3 = 509.9 * 10^6 , isn't it?

So I wanted to divide 10 000 km^3 by 509.9 * 10^6, but it doesn't funcion.

What did I wrong?

You mention in the OP that the water can cover the continents; not the surface of the Earth itself. So maybe you should divide by the surface area of the continents --around 150 million square kilometers -- and not by the surface area of the whole earth.
 
  • #5
WWGD said:
You mention in the OP that the water can cover the continents; not the surface of the Earth itself. So maybe you should divide by the surface area of the continents --around 150 million square kilometers -- and not by the surface area of the whole earth.

You mean, 10 000 km^3 / 150 mio km^2
But I can't divide something smaller by something bigger, or can I? I'm confused...
 
  • #6
Physicsrapper said:
You mean, 10 000 km^3 / 150 mio km^2
But I can't divide something smaller by something bigger, or can I? I'm confused...

Why not? Then transform the result into cm. units.
 
  • #7
WWGD said:
Why not? Then transform the result into cm. units.

That gives me 66.6666666667 cm.

But the solution should be around 2 cm. So something must be wrong with that equation, but what... ?
 
  • #8
Sorry I was thinking of the first problem; about the earth, not the second one.
 
  • #9
Surface area sphere = 4πr2
in this case 4 X π X 6370 X6370
Divide 100000 by the answer to get the depth as a fraction of a kilometre. Multiply by 100000 to convert to centimetres - I get 1.96...
 
  • #10
Physicsrapper said:
You mean, 10 000 km^3 / 150 mio km^2
But I can't divide something smaller by something bigger, or can I? I'm confused...

I thought we covered this in basic arithmetic.

What is 1/2? Isn't 2 bigger than 1? Cause, you know, fractions.
 
  • #11
Physicsrapper said:
That gives me 66.6666666667 cm.

But the solution should be around 2 cm. So something must be wrong with that equation, but what... ?

Compare this with the results you just got.
 
  • #12
Physicsrapper said:
That gives me 66.6666666667 cm.

But the solution should be around 2 cm. So something must be wrong with that equation, but what... ?
Use the 509.9 * 10^6 km^2 you calculated instead of the 150 * 10^6 km^2, and be careful with the conversion factor from kilometers to centimeters, then you should get the right result.
 

1. What is the sphere problem and why is it important?

The sphere problem is a mathematical problem that involves finding the maximum volume that can be enclosed by a sphere of a given surface area. It is important because it has applications in various fields such as physics, engineering, and biology, where maximizing volume in a limited space is crucial.

2. What is the formula for calculating the volume of a sphere?

The formula for calculating the volume of a sphere is V = (4/3)πr3, where V is the volume and r is the radius of the sphere.

3. How do you solve the sphere problem?

The sphere problem can be solved by using the formula for volume of a sphere and manipulating it to find the maximum volume. This can be done by taking the derivative of the volume formula with respect to the radius and setting it equal to 0 to find the critical point, which will give the radius for the maximum volume.

4. Is the solution to the sphere problem unique?

Yes, the solution to the sphere problem is unique. This is because the volume of a sphere is a continuous function, and the derivative of the volume formula with respect to the radius is always positive, indicating that there is only one critical point.

5. What are some real-life applications of the sphere problem?

The sphere problem has various real-life applications, such as designing fuel tanks for rockets, maximizing storage space in shipping containers, and determining the optimal size for drug capsules. It also has applications in nature, such as how cells pack together in biological systems.

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