Watch the simulation below for better visualization.

Suppose we create an explosion, for example at $2R_{Earth}$ from the center of Earth. Just before the explosion was set off, the explosive was at rest with respect to Earth. For simplicity, let's assume the explosion is spherically symmetric and all the gas particles are initially escaping with exactly the same speed. This implies all particles will have the same kinetic energy at the explosion site and the same total orbital energy during the following flight. Thus the semi-major axis of all the particle trajectories are equal ($E_{total} = \frac{GMm}{2a}$). We conclude, by Kepler's third law ($T_{period}^2 \propto a^3$), all the orbital periods $T_{period}$ are also equal, implying the fast moving gas will again be compressed at the same spot after $T_{peroid}$ elapsed. By energy conservation, the fast-moving gas will heat up to roughly the same temperature as in the initial explosion, creating an "echo" explosion at the same spot.

The assumption of constant velocity might seem absurd, however, if the ball of explosive is set off uniformly, the gas expands initially only from the surface (where there is almost infinite pressure gradient). The gas on the surface will start moving, expand and cool instantaneously, maintaining the sharp edge. This would allow almost all the gas to accelerate to some uniform velocity. The assumption could have been derived also using Bernoulli's law for compressible fluids.

Are these "echo" explosions realistic given that the explosion of a small or large enough bomb (say it's still much smaller than Earth) is "perfect", or are these hindered by some other physical effects? If they are, some rough estimates on the significance of these effects would be appreciated. For example, are the "echoes" hampered significantly by different areas of explosion having different initial potential energies, resulting in a variance of the periods of the gas particles (in case of a very large bomb, the differences in periods increase, for very small ones, the timing has to be extremely precise)?

Edit: The effect is clearly visible in my $F \propto r^{-2}$ force law simulation, which I programmed for this question (sorry for the low quality, I was forced to use gif format): Simulation

  • $\begingroup$ Oh forget everything I said. Deleted my comments. I implicitly assumed you were talking about orbits around the sun. In which case the energies of the particles would all be different and this wouldn't hold. $\endgroup$
    – user12029
    Jan 17, 2015 at 20:35

1 Answer 1


Your analysis would be correct if the explosion truly did impart the exact same speed to every particle, if the Earth had a spherically symmetric mass distribution, and if there were no perturbing influences such as the Moon, the Sun, and the Earth's atmosphere.

Ignoring those perturbing influences, the vis viva equation, $\frac {v^2} {\mu} = \frac 2 r - \frac 1 a$ (where $\mu$ is the Earth's standard gravitational parameter), indicates that the semi-major axis of the orbits will all be the same for all of the particles from an explosion assuming that explosion imparts a uniform speed to each particle involved in the explosion (your key assumption). Given your assumptions, your simulation is correct since the period of a Keplerian orbital is a function of semi-major axis and gravitational parameter, and nothing else.

However, there is no such thing as a nice spherical explosion such as yours, the Earth has a markedly non-spherical mass distribution (the satellites placed in Sun synchronous orbits depend very much on this asymmetry), and there are a number of other objects that perturb orbits away from Keplerian. Those perturbing influences mean that while your "echo" explosion is a very, very nice spherical cow (+1 on the question, BTW), it is not "realistic".

  • $\begingroup$ There is another problem, the question starts out with something that "cools" into having a spherically symmetric velocity distribution (with uniform speed), but then when they converge later this perfect distribution becomes thermal again. It's way way beyond unrealistic to be thermal, then perfectly known and then thermal again. Even a spherical cow should be thermal or not thermal, but never switch back and forth. $\endgroup$
    – Timaeus
    Jan 25, 2015 at 19:08

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