Answer to Question #103767 in Classical Mechanics for DertySol

Question

1. How many heartbeats in a typical human lifetime? You will provide the data from your general knowledge. Write your answer as a number (NOT as a power of ten).

2. Estimate the mass of water on the earth. The density of water is 1000 kg/ m^3 . Here are some approximations you can use:

Compared with the oceans, lakes and rivers are tiny. Clouds have very low density. There is (still) lots of ice in the polar regions, but much less area than in the oceans. So neglect all except the oceans, which cover roughly 2/3 of the earth.
Looking at charts of the oceans, we see that the depth is typically several thousand m (and a serious search tells me the average depth is 4700 m). The radius of the earth is about 6000 km. The surface area of a sphere is 4πR^2. This is an order or magnitude question. Express your answer in kg, as a power of 10. For example, if you think the answer is 10^4, then write 10^4.

Answer 1

1. According to my knowledge, average heart rate is 70 beats per minute. Average human lives for 80 years, which is 80*360*24*60=41 472 000 minutes. Therefore, there are (roughly) 41 472 000*70=2 903 040 000 heartbeats.
2. The volume of our planet assuming it's a sphere:

$V_E=\frac{4}{3}\pi R^3.$

The volume of the sphere that is 4700 m below earth's surface:

$V_B=\frac{4}{3}\pi (R-4700)^3.$

The volume of the shell 4700 m thick:

$V_S=V_E-V_B=\frac{4}{3}\pi[R^3-(R-4700)^3].$

2/3 of this volume is water:

$V=\frac{2}{3}V_S=\frac{8}{9}\pi[R^3-(R-4700)^3].$

The mass of water:

$m=\rho V= \frac{8}{9}\rho\pi[R^3-(R-4700)^3]=1.60\cdot10^{21}.$

The order or magnitude: 10²¹