Just a drop in the bucket

If approximately 70 % 70 \% of the Earth's surface is covered with water, what is the order of magnitude for the number of raindrops in the world's oceans?

Hint: The order of magnitude of 2478 = 2.478 × 1 0 3 2478 = 2.478 \times 10^{3} is 3 3 .

Details and assumptions

  • The earth's radius is 6380 km 6380~\mbox{km} .
  • The radius of a raindrop is 1 mm 1~\mbox{mm} ; raindrops can be approximated to be spherical.
  • The average ocean depth is 2 km 2~\mbox{km} .


The answer is 26.

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1 solution

Discussions for this problem are now closed

David Mattingly Staff
May 13, 2014

We should first calculate the volume of the ocean: V = 0.7 × 4 π R 2 d V = 0.7 \times 4 \pi R^2 d (a more correct formula would be 0.7 × 4 3 π ( R 3 ( R d ) 3 ) 0.7 \times \frac{4}{3} \pi(R^3 - (R-d)^3) but we do not need this kind of precision since we are trying to find the order of magnitude). On the other hand, that volume should be made out of raindrops: N 4 3 r 3 π N \frac{4}{3} r^3 \pi , where N N is the number of raindrops. Expressing N N from these equations we get N = 1.7096 × 1 0 26 N = 1.7096 \times 10^{26} , which has the order of magnitude 26 26 .

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