A shell is fired upward with a velocity v_0. Considering only gravitational factors, at what distance will the shell reach before it begins to fall freely?
For this problem, we first adopt the approach of a high school student. Considering Earth’s gravity, we assume the shell is in uniformly accelerated motion with an acceleration of -g (-9.8\text{m/s}^2). According to the formula v_t^2 - v_0^2 = 2as, we can obtain s = \frac{v_0^2}{2g}. This is the maximum distance the shell can travel.
However, looking at this formula, we find that this "distance" is always finite. In other words, as long as v_0 does not tend toward infinity, s will not be infinite. But we have also heard Newton say: If a shell is fired from Earth at a certain speed (which we now call the Second Cosmic Velocity), it will never return. Is there a contradiction between the two?
Friends who have read my previous article will immediately have a clue. This acceleration a is not constant. Congratulations, you are right! But what is the specific situation? Please read on—
Let the distance of the shell be s, then during the motion: s'' = v' = -\frac{GM}{(r+s)^2}
Let s'' = v \frac{dv}{ds}, and substitute it in (how familiar this process is): v dv = -GM(r+s)^{-2} ds Integrate both sides: \int v dv = \int -GM(r+s)^{-2} ds \frac{1}{2} v^2 = GM(r+s)^{-1} + C The following treatment is somewhat different: When v = v_0, we have s = 0, then: C = \frac{1}{2} v_0^2 - GMr^{-1} This yields: v^2 = v_0^2 + 2GM[(r+s)^{-1} - r^{-1}] The shell reaches its furthest point when v = 0. Thus: 0 = v_0^2 + 2GM[(r+s)^{-1} - r^{-1}] s = \frac{v_0^2 r^2}{2GM - v_0^2 r} \quad \text{--- --- --- --- (A)}
Since GM = r^2 g, where g is the gravity experienced by a 1kg object on the Earth’s surface, substituting this in gives: s = \frac{v_0^2 r}{2rg - v_0^2} = \frac{v_0^2}{2g - \frac{v_0^2}{r}}
If \frac{v_0^2}{r} is very small, it can be ignored, yielding the low-speed approximation of the Galilean formula: s = \frac{v_0^2}{2g}. This is consistent with the result at the beginning of the article.
We find that when v = \sqrt{\frac{2GM}{r}}, the denominator becomes zero, which means s \to \infty. At this point, it is what Newton called "never to return" (simply not returning to Earth). Thus, we have naturally derived the Second Cosmic Velocity! v = \sqrt{\frac{2GM}{r}}
Yeah! Let’s cheer! Science should be like this; even for a small achievement, one should be joyful. But remember, do not be complacent!
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