The general definition of free fall is: Considering only the gravitational factors of the attracting and attracted bodies, ignoring other motions and atmospheric friction, the motion of an object starting from rest (relative to the attracting body) as it approaches the attracting body. According to this definition, assume the Earth is a uniform sphere with radius r and mass M, and an object falls freely from a height h above the Earth’s surface. We seek to find the time t it takes to reach the ground, or to find h based on the time t.
Let s be the distance between the falling object and the Earth’s surface at time t. Ignoring the small mass of the object, we can set up the following differential equation: \frac{d^2 s}{dt^2} = -\frac{GM}{(r+s)^2} \tag{1} with the initial conditions t=0, s=h, \dot{s}=v=0.
In practical applications, we do not need to find the exact solution to this differential equation, as the solution is extremely cumbersome and has been discussed previously. We only need to find an approximate solution with sufficient accuracy. According to the Taylor series expansion: f(x) = f(x_0) + f'(x_0)(x-x_0) + f''(x_0)\frac{(x-x_0)^2}{2!} + f'''(x_0)\frac{(x-x_0)^3}{3!} + \dots
For the differential equation (1) above, we already have s(0)=h, s'(0)=0, s''(0)=-\frac{GM}{(r+h)^2}. Since \frac{d\ddot{s}}{dt} = \dot{s}\frac{d\ddot{s}}{ds}, and it is not difficult to prove that \frac{d\ddot{s}}{ds} is finite, it follows that s'''(0)=0. Thus, we can write the approximate solution of the differential equation as: s = h - \frac{GM}{2(r+h)^2}t^2 \tag{2} Its truncation error is O(t^4). If we solve for the time taken to reach the ground, then s=0, which gives: h = \frac{GM}{2(r+h)^2}t^2 \tag{3} Additionally, we have GM = r^2 g, where g is the gravitational acceleration at the Earth’s surface. Thus, (3) can be rewritten as: h = \frac{r^2 g}{2(r+h)^2}t^2 \tag{4} How accurate is the above formula? Let us verify it from two perspectives: when h is very small and when h is very large.
First, for the case where h is much smaller than r (h \ll r), we have \frac{r^2}{(r+h)^2} \approx 1. Thus, (4) reduces to: h = \frac{g}{2}t^2 \tag{5} This is exactly the free-fall formula we encountered in high school!
Secondly, for the case where r is much smaller than h (r \ll h), let us use this formula to solve a previous problem:
An object falls freely and reaches the ground after 9 days. What was the initial height of the object when it started falling?
Since r \ll h, we get: h(r+h)^2 = \frac{r^2 g}{2}t^2 \approx h^3 \tag{6} Substituting r = 6,371,000\,\text{m}, t = 9 \times 86,400\,\text{s}, g = 9.8\,\text{m/s}^2 into (6), we can calculate: h = 515,482,465\,\text{m} = 51.5 \times 10^4\,\text{km}. This is almost exactly equal to the official answer!
It can be seen that the corrected free-fall formula has high accuracy! Therefore, friends participating in the Astronomy Olympiad might want to master this formula; perhaps the graders will even give you extra points! (Creativity points ^_^)
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