Distance travelled of an object falling from non-negligible height

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SUMMARY

The discussion focuses on deriving the distance fallen by an object in free fall from a significant height, specifically in the context of an atmosphereless planet. The gravitational acceleration of 9.8 m/s² is applicable only near Earth's surface, while the equation of motion changes at greater distances. The key equation presented is derived from Newton's second law, F=ma, leading to the differential equation $$\frac{GM}{r^2(t)} = \frac{d^2}{dt^2}r(t)$$, which describes the relationship between gravitational force and the object's motion. Understanding this requires a foundational knowledge of calculus, particularly derivatives and integrals.

PREREQUISITES
  • Basic understanding of Newton's laws of motion
  • Familiarity with gravitational force concepts
  • Knowledge of calculus, specifically derivatives and integrals
  • Understanding of motion graphs, particularly velocity-time graphs
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  • Study the derivation of gravitational force equations in classical mechanics
  • Learn about differential equations and their applications in physics
  • Explore the concept of gravitational acceleration variations with distance
  • Practice constructing and interpreting velocity-time graphs in motion analysis
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the complexities of motion under varying gravitational conditions, particularly in advanced mechanics contexts.

kaikalii
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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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