# How much energy is contained in a 40 meter wave?

Consider a wave that stands 40 meters high in the sea on very deep water. How much energy would approximately be contained in this wave if it was 100 meters wide and had been produced by wind?

Is this question even answerable? I read somewhere that the tsunami that caused massive destruction in South-East Asia a few years ago, was just something like a meter tall in open sea..

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See this site, which estimates that the southeast Asia tsunami had 0.5 terawatts total power – 1 gigawatt per kilometer of coastline. If the tsunami "lasted" a minute, that's about 7 kt. It also says that during the time the tsunami was only a meter high, it also traveled at 220 m/s and had a wavelength (i.e., in the direction of travel) of 300 km. – rdhs Dec 8 '11 at 2:07
Please note that the tsunami wave is a wave from the top to the bottom of the ocean, like a tidal wave, not a surface wave as the wind produced ones. – anna v Dec 8 '11 at 5:33
how fast is the wave travelling? And see hyperphysics.phy-astr.gsu.edu/hbase/watwav.html – EnergyNumbers Dec 8 '11 at 14:11

We have $40*100*1000=4\times10^6$ cubic meters of water per kilometer, which is $4\times 10^9$ kilograms of water per kilometer, average 20 meters above sea level, which leads to $4\times 10^9 \times 20 \times 10 = 8\times 10^{11} J/km$ in gravitational potential energy. Kinetic energy should be about the same as potential so lets say $1.6\times 10^{12} J/km$ or $0.3$ kilotons per kilometer. For reference the Hiroshima A-bomb was about $10$ kilotons yield.

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I am sorry, but what do you mean by "per kilometer"? I am asking about a single wave with a defined width, so I am uncertain of whether there is a misunderstanding. – David Dec 8 '11 at 11:51
I assumed when you said 100m wide you meant 100m wide in the direction of travel. Of course, the wave front will stretch out perpendicular to the direction of travel for potentially thousands of kilometers, so it makes sense to talk of energy per kilometer. – user1631 Dec 8 '11 at 18:17
I meant 100m wide perpendicular to the direction of travel. I do not know whether 100m is a reasonable assumption regarding the length of the wave. – David Dec 8 '11 at 22:26
Also, are you assuming a perfectly square wave traveling at the arbitrary speed of 19.8 m/s? – rdhs Dec 9 '11 at 0:18
It was an order of magnitude estimate, so assuming a square wave should be adequate for estimating gravitational potential energy. I could have made it a Gaussian or something but what's the point? I made no assumption about the propagation speed. It is not needed since the energy of a wave will be split equally between kinetic and potential. – user1631 Dec 9 '11 at 1:01