Recently, Mt. Ruapehu in New Zealand erupted (http://www.theaustralian.com.au/news/world/north-island-new-zealand-volcano-mt-tongariro-erupts/story-e6frg6so-1226521155526) and scientists expect another eruption in the near future (http://news.yahoo.com/mount-doom-likely-blow-soon-201446255.html). I would like to examine the basic physics behind this natural phenomenon.
When a volcano erupts, lava flows through a cylindrical tube called a vent. Lava is extremely viscous (up to 100,000 times as viscous as water), and consequently, in order to get it to flow, there must be an increase in pressure. This pressure is usually due to a shift in tectonic plates. The change in pressure needed to make Mt. Ruapehu erupt can be determined using the equation:
P1- P2 = ρg∆h
Mt. Ruapehu is 2291 m high. There are different types of magma with varying densities, but we can assume that the lava flowing from Mt. Ruapehu was basaltic magma which has a density of approximately 2.3 g/cm^3 (2300 kg/m^3). Therefore,
P1-P2 = (2300 kg/m^3)(9.8m/s^2)(2291m) = 51639.14 kg/ms^2
Mt. Ruapehu required a change in pressure of 51639 Pa in order to erupt.
Assuming that the diameter of the vent that the lava erupted through was 1 km, we can calculate the average velocity of the lava as it leaves the vent using,
Vavg. = (R^2 x ∆P)/8ηL
The viscosity of basaltic lava (η) = 50 Pas
L = 2291 m
R = 0.5 km = 500 m
Vavg. = (500^2 m x 51639.14 kg/ms^2)/(8 x 50 Pas x 2291 m) =
1.2909 x 10^10 / 916400 = 14086.64 m/s
Using the velocity, we can determine whether the lava was flowing laminarly or turbulently by calculating RN.
RN = ρvd/η
RN = (2300 kg/m^3)(14086.64 m/s)(1000 m)/50 Pas = 6.479 x 10^8
Because RN> 3000, the lava was flowing turbulently.
Finally, using kinematics, we can determine the height of the eruption.
Vf^2 = Vo^2 + 2a(∆x)
Vf = 0 m/s because at the height of it, it will have no velocity in the x or y direction assuming that it is ejected straight vertically.
Vo = 14086.64 m/s This is the velocity that was calculated for the magma leaving the volcano.
a = -9.8 m/s We can define our coordinate system so that downwards is negative. The only force acting on it that is of relevance for our calculation is gravity (ignore air resistance). The lava will be slowing at a rate of -9.8 m/s^2 as it leaves the volcano.
0 m/s = 14086.64 m/s + 2(-9.8)(∆x)
∆x = 718.706 m
Therefore, a particle of debris launched vertically into the sky from this blast may have reached a height of 718 m above the crest of the volcano.