Relationship between spherical volume and rectangular volume

Question

I wanted to calculate the height of water above a perfectly smooth earth sphere if all the water on earth surrounded the sphere.

I found the volume of all water on line to be 322.5 cu miles. Using a 400mile radius for my earth sphere, I calculated the volume of this sphere (268.082E9 cu miles) and then added the 322.5E6 cu miles of water and solved for the difference in the two radii (with and without surrounding water). Answer 1.6 miles.

All great, but then a friend did the calculation by dividing the surface area of my earth sphere (201.062E6 sq miles into the volume of water (322.5E6 cu miles) and got the same answer--1.6 mile high. He claims you can use the rectangular volume calculation and get the same answer as with the spherical volume calculation because somehow the earth's gravitational system equates the two.

I don't get it and think he is blowing smoke, but I can't figure out why I get the same answer using both methods (spherical and rectangular). Any help?

Answer 1

Your friend didn't get exactly the same answer as you because they are using an approximation to the actual volume.

You have calculated the volume as the difference of two spheres, i.e. $$\frac 43\pi[(R+h)^3-R^3]=\frac 43\pi R^2h[1+\frac hR+\frac{h^2}{3R^2}]$$

Since $h$, the depth of water is so much smaller than $R$, the radius of the earth sphere, the last two terms in the square bracket are very small, so that the volume is approximately $$4\pi R^2h$$ which is the surface area of the earth sphere multiplied by the depth of water.

Answer 2

The computed depth of water is tiny compared to Earths radius. The volume of a big sphere with radius $r$ compared to the volume of a big sphere with radius $r + h$ is $$ \frac{4}{3}\pi (r+h)^3 = \frac{4}{3}\pi (r^3 + 3r^2 h + 3r h^2 + h^3), $$ where the $4\pi r^2 h$ term completely dominates the difference from $4\pi r^3/3$. This is what your friend computes, assuming that Earth is "flat", which gives a volume $V = A h = 4\pi r^2 h$.

Answer 3

Let $R$ be the radius of the outer sphere and $r$ be the radius of the inner sphere.

The volume $V$ of the water is the difference between the two volumes of the two spheres, i.e $$V = \frac{4\pi}{3}(R^3 - r^3)$$

We can factor the difference of two cubes to get $$V= \frac{4\pi}{3}(R-r)(R^2+Rr+r^2)$$

Now we do as your friend did and divide by the volume of the inner sphere to get $$\frac{V}{4\pi r^2}=(R-r)\frac{1}{3}(\frac{R^2}{r^2} + \frac{R}{r} + 1)$$

Notice that if $R^2$ is close to $r^2$ and $R$ is close to $r$ the expression with the fractions is approximately $3$ and so cancels with $\frac{1}{3}$ to give $$\frac{V}{4\pi r^2}\approx (R-r)$$

Answer 4

All three answers are excellent. If I carry out the calculation to many decimal points I do generate a difference (small) between the spherical and the rectangular approaches. Thanks to all who responded CBMc