Average Acceleration (gravity question)

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SUMMARY

The discussion centers on calculating the time it takes for an object to fall to Earth from a height of 2500 km, considering the variation in gravitational acceleration. The formula presented, t = √(2D/g), is applicable for short distances using g = 9.8 m/s²; however, at higher altitudes, g decreases to approximately 5 m/s². The participants explore the feasibility of using an average gravitational acceleration (g_avg) for the calculation, acknowledging that while it may simplify the problem, it could lead to an underestimation of the fall time. The consensus is that for accurate results, especially over large distances, an integral approach is necessary to account for the changing gravitational force.

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clm222
Hello. I was doing some physics and I came across a problem involving gravity.
The problem was finding out how long (in disregaurd to atmosphere) long (time) it would take for a body to reach earth. I'm familiar with distance's derivatives, which give me:
[tex]t=\sqrt{\frac{2D}{g}}[/tex]
where t=time D=distance g=gravitational acceleration.

Although using g=9.8 is fine for short distances, it wouldn't really work at, say 2500km (about g=5) above the surface of the earth.

I'm curious as to how I may find the time it will take the object to fall? I'm not worried about what happens in between the distance. I'm not taking into account other bodies in the universe (its just a hypothetical physics problem). Since the derivative of g will depend on distance, and since D(t) is dependent on g, that doesn't really work. I'm wondering if i can use the average value of g? although the first bit of time will calculate the body being faster, the last bit of time it won't be getting as much speed. Can I do this?

does:
[tex]t=\sqrt{\frac{2D}{g_{avg}}}?[/tex]

can I do this for any varying acceleration? I know how to find averages of functions, please help.
 
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The same question was discussed here yesterday. In the general case, you have to solve an integral.
If the falling distance is not too large (still small compared to the diameter of earth), you might get away with an average g, but this will underestimate the falling time.
 

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