Average Acceleration (gravity question)

AI Thread 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 with distance. The initial formula used, t = √(2D/g), applies well for short distances but becomes inaccurate at larger heights due to the decrease in g. Participants suggest that while using an average gravitational acceleration (g_avg) might simplify calculations, it would likely underestimate the total falling time. For more accurate results, especially over significant distances, solving an integral is recommended. The conversation emphasizes the complexity of varying acceleration in gravitational problems.
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:
t=\sqrt{\frac{2D}{g}}
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 dependant 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:
t=\sqrt{\frac{2D}{g_{avg}}}?

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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