In middle or high school, the free fall experiment is simply described by this formula: s = \frac{1}{2} g t^2 where g = 9.8 \text{ m/s}^2, which is equal to the gravitational acceleration experienced by a 1 kg object on the Earth’s surface.
However, there is an obvious problem with this formula: in reality, on Earth, g is not constant; it varies with distance (i.e., altitude). The above formula can describe free fall motion within a certain range, but when the distance is very large, the formula becomes invalid. For example, consider the following problem:
A Greek myth mentions that a piece of iron dropped by a god took exactly nine days to reach the ground. Considering only the factor of Earth’s gravity, calculate the “height of heaven.”
This is a problem from an international astronomy competition. Clearly, it cannot be calculated using s = \frac{1}{2} g t^2. After searching through the literature, I discovered this formula: t = \sqrt{\frac{r_0}{2GM}} \left( r_0 \cdot \arctan \sqrt{\frac{r_0 - r}{r}} + \sqrt{r(r_0 - r)} \right)
It represents the time required to fall from a distance r_0 from the center of the field to a distance r. It can be proven that when r_0 is very close to r, this formula approximates s = \frac{1}{2} g t^2.
Now we can use the new formula to solve that problem from the astronomy competition.
First, we must adopt a certain mindset: In fact, astronomical theories are quite complex; in practical applications, we should strive to approximate and simplify them. The following calculations will reflect this approach.
We already know t = 9 \times 86400 \text{ s} and r = 6.371 \times 10^6 \text{ m}, and we need to find r_0. We can roughly estimate that r_0 will definitely be much larger than r, so \sqrt{\frac{r_0 - r}{r}} must be very large. Under these circumstances, we can assume \arctan \sqrt{\frac{r_0 - r}{r}} = \frac{\pi}{2}.
Similarly, we can approximate \sqrt{r(r_0 - r)} as \sqrt{r r_0}. Thus: t = \sqrt{\frac{r_0}{2GM}} \left( \frac{\pi r_0}{2} + \sqrt{r r_0} \right)
And we have GM = R^2 g. Let \sqrt{r_0} = x, then: 2 r t \sqrt{2g} = \pi x^3 + 2 \sqrt{r} x^2
If we neglect 2 \sqrt{r} x^2, we can solve for x = 24525, r_0 = 6 \times 10^5 \text{ km}. If we set 2 \sqrt{r} x^2 = x^3, we get x = 22367, r_0 = 5 \times 10^5 \text{ km}. The official answer provided is 510,000 km.
Of course, this is not the official solution, and there is no need to specifically master this method. This is just provided as a line of thought and a method for friends researching astrophysics!
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