Distance travelled of an object falling from non-negligible height

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In free-fall scenarios, the acceleration due to gravity is typically considered constant at 9.8 m/s², but this approximation fails at significant heights where gravity varies. For an object falling towards an atmosphereless planet, the gravitational force can be described by the equation F=ma, leading to the relationship GM/r²(t) = d²/dt² r(t). To derive the distance fallen as a function of time and initial height, one must utilize calculus concepts like derivatives and integrals. A velocity-time graph can be constructed, where the area under the curve represents displacement. Understanding these principles requires a foundational knowledge of calculus.
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In a free-fall problem, 9.8 m/s^2 works fine as long as you stay close to the earth. However, if you are falling from very high up, the change in acceleration due to gravity is no longer negligible.

Let's say there's an atmosphereless planet of mass M, and another object of negligible mass is falling toward it after being dropped from a point a distance r from the center of the planet. How could you derive an equation to find the distance fallen d as a function of time t and initial height r? Just so you are aware, I have a very limited knowledge of calculus (how to find derivatives and integrals), so please answer in terms I can understand.
 
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You will need to learn more calculus in order to understand this sort of physics.
If you just do F=ma as usual you get:
$$\frac{GM}{r^2(t)} = \frac{d^2}{dt^2}r(t)$$

Basically you construct a v-t graph - which will be a curve.
The area under it is the displacement.
 
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