Question

A meteorite travelling 9200 m/s strikes the ocean. Determine the shock wave angle it produces (a) in the air just before entering the ocean, and (b) in the water just after entering. Assume \(T = 20^\circ {\rm{C}}\).

Short answer

(a) The shock wave angle produced by a meteorite in the air is \(2.13^\circ \).

(b) The shock wave angle produced by a meteorite in water is \(9.76^\circ \).

Step-by-step solution

Determination of shock wave angle

The shock wave angle is determined by the sin inverse of fraction of speed of sound in medium to the speed of the system. i.e.,\({\rm{sin}}\theta = \frac{c}{u}\).

Given information

Given data:

The speed of meteorite is \(u = 9200\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\).

The standard value for the speed of sound in air and sea water at \(T = 20^\circ{\rm{C}}\)are\({v_a}=343\;{{\rm{m}}\mathord{\left/{\vphantom{{\rm{m}}{\rm{s}}}} \right.} {\rm{s}}}\) and \({v_w} = 1560\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\) respectively.

Evaluation of the shock wave angle produced by a meteorite in the air 

Part (a)

The shock wave angle produced by a meteorite in the air can be calculated as:

\(\begin{aligned}{c}{\rm{sin}}{\theta _1} &= \frac{{{v_a}}}{u}\\{\theta _1} &= {\rm{si}}{{\rm{n}}^{ - 1}}\left( {\frac{{{v_a}}}{u}} \right)\\ &= {\rm{si}}{{\rm{n}}^{ - 1}}\left( {\frac{{343\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}}}{{9200\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}}}} \right)\\ &= 2.13^\circ \end{aligned}\)

Hence, the shock wave angle produced by a meteorite in the air is \(2.13^\circ \).

Evaluation of the shock wave angle produced by a meteorite in water

Part (b)

The shock wave angle produced by a meteorite in water can be calculated as:

\(\begin{aligned}{c}{\rm{sin}}{\theta _2} &= \frac{{{v_w}}}{u}\\{\theta _2} &={\rm{si}}{{\rm{n}}^{1}}\left({\frac{{{v_w}}}{u}}\right)\\&={\rm{si}}{{\rm{n}}^{1}}\left({\frac{{1560\;{{\rm{m}}\mathord{\left/{\vphantom{{\rm{m}}{\rm{s}}}}\right.}{\rm{s}}}}}{{9200\;{{\rm{m}} \mathord{\left/{\vphantom{{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}}}} \right)\\ &= 9.76^\circ \end{aligned}\)

Hence, the shock wave angle produced by a meteorite in water is \(9.76^\circ \).