Free Fall with non-constant acceleration due to gravity [duplicate]

Question

In a free fall motion, the height of the falling object is constantly changing and its acceleration due to gravity is also changing.

However, we usually ignore the effects of changes in height on the acceleration due to gravity because the change in height is too small compared to the radius of the earth. Suppose we now take this small change into account, making the acceleration due to gravity ($g$) a non-constant value in the free fall. Given the time the object takes to fall to the ground (where the distance to the center of the earth is exactly the earth's radius) and its mass, how can we find the height that it fell from (displacement)?

I first wanted to start by writing the acceleration ($g$) as a function of height ($r$) and displacement as a function of time ($t$).

$$v(t) = \int_{t_1}^{t_2} a(t) \ \mathrm{d}t, \quad d(t) = \int_{t_1}^{t_2} v(t) \ \mathrm{d}t, \quad g(r) = \frac{GM}{r^2}.$$

But I do not have any clues about the next step. I attempted to find a relationship between $r$ and $t$, but it did not work. I was wondering if this problem is solvable by only using single variable calculus, or more advanced mathematics is required.

Answer 1

So, you can use conservation of energy,

$$GMm(\frac{1}{r}-\frac{1}{h})=\frac{1}{2}mv^2\implies v=\sqrt{\frac{2GM}{h} \frac{h-r}{r}}$$

But v is also $-\frac{dr}{dt}$,so we can integrate to get:

$$\sqrt{\frac{2GM}{h}}t=\int_r^h \sqrt{\frac{r}{h-r}}=\sqrt{hr-r^2}+h\arctan{\sqrt{\frac{h-r}{r}}}$$

So, the time taken and height of dropping we can get r and vice versa. Hope this answers your question and let me know if you need more detail