Previously in this article, we used a formula for free fall in a Newtonian gravitational field: t=\sqrt{\frac{r_0}{2GM}}\left(r_0 \cdot \arctan \sqrt{\frac{r_0 -r}{r}}+\sqrt{r(r_0 -r)}\right)\label{eq:1}
Let us attempt to derive this formula.
Meanwhile, as I have gradually delved deeper into my research, I have found that differential equations are extremely important. Some problems that I previously thought were impossible to solve have been gradually resolved using differential equations. In future articles, we will continue to experience the great magic of differential equations! Therefore, I suggest to all friends who aspire to study physics that you must master differential equations; for more advanced studies, partial differential equations are required!
First, the gravitational force on an object of mass m at a distance r from the center of the Earth is \frac{GMm}{r^2}. According to Newton’s second law F=ma, the acceleration acquired by a naturally falling object is \frac{GM}{r^2}. Suppose an object starts falling freely towards the center of the Earth from a distance r. We seek the displacement s as a function of time t, s=s(t).
According to the definition of acceleration, we have: \frac{d^2 s}{dt^2}=a=\frac{GM}{(r-s)^2}. Thus, the problem essentially consists of solving the ordinary differential equation: s''=\frac{GM}{(r-s)^2}
Next, we let s'=v, then s''=v(\frac{dv}{ds}). Substituting this into the above equation: GM(r-s)^{-2} ds=vdv. Integrating both sides: \begin{gathered} \int vdv = \int GM(r - s)^{ - 2} ds = - \int GM( r - s)^{ - 2}d(r - s) \\ \Downarrow\\ \frac{1}{2} v^2 = GM[(r - s)^{ - 1} + C_1] \end{gathered}
According to the physical situation, when t=0, v=s=0, which leads to C_1=-r^{-1}, namely: \frac{1}{2}v^2 = GM[(r - s)^{ - 1} - r^{ - 1}] \quad\Rightarrow\quad \frac{ds}{dt} = v = \sqrt {\frac{2GM}{r}} \sqrt {\frac{s}{r - s}} Integrating both sides: \int dt = \sqrt {\frac{r}{2GM}} \int \left(\sqrt {\frac{r - s}{s}} \right)ds =2\sqrt {\frac{r}{2GM}} \int \left(\sqrt {r - s} \right)d(s^{0.5})
Let s^{0.5}=P, then 2\int \left(\sqrt {r - p^2} \right)dp According to the integration formula: \int \sqrt{a^2-x^2} dx =\frac{a^2}{2} \arcsin \frac{x}{a} +\frac{x}{2} \sqrt{a^2-x^2}+C We obtain: 2\int \left(\sqrt {r - p^2} \right)dp=r\cdot \arcsin \frac{p}{\sqrt{r}}+p \sqrt{r-p^2}+C Substituting s back, we have: 2\int \left(\sqrt {r - s} \right)d(s^{0.5}) =r\cdot \arcsin \sqrt{\frac{s}{r}}+\sqrt{s(r-s)}+C Then: t=\sqrt{\frac{r}{2GM}}\left(r\cdot \arcsin\sqrt{\frac{s}{r}}+\sqrt{s(r-s)}+C\right) When s=0, t=0, then C=-\pi r / 2, yielding: t=\sqrt{\frac{r}{2GM}}\left(r\cdot \arcsin\sqrt{\frac{s}{r}}+\sqrt{s(r-s)}\right) According to the inverse trigonometric formula \arcsin \frac{a}{b} = \arctan\sqrt{\frac{a^2}{b^2-a^2}}, we get: t=\sqrt{\frac{r}{2GM}}\left(r\cdot\arctan\sqrt{\frac{s}{r-s}}+\sqrt{s(r-s)}\right)\label{eq:2}
It is not difficult to see that equations [eq:1] and [eq:2] are equivalent. The r in [eq:2] is actually the r_0 in [eq:1], and s is equivalent to r_0 - r.
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