Solving a Problem with Non-Constant g: Expanding a Maclaurin Series

AI Thread Summary
The discussion revolves around solving for the speed of an object dropped from a height h where gravity, g, is not constant. The proposed solution involves using the Maclaurin series to approximate the speed as v = sqrt(2gh) * (1 - (h/2R)). Participants highlight the challenge of applying conservation of energy due to the variable nature of g and suggest using the gravitational formula mMG/(r+h). Despite attempts to follow this guidance, one participant arrives at a slightly incorrect result, prompting requests for full calculations to identify errors. The conversation emphasizes the importance of correctly applying series expansion in physics problems involving variable gravity.
didinyc
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Homework Statement


Suppose an object is dropped from height h above Earth where h<<R, but is large enough so g, the acceleration due to gravity, is NOT constant! Show that speed with which it hits the ground, neglecting friction, is approximately, v= sqrt2gh *(1-(h/2R))

Hint: you will need to expand an expression in Maclaurin series.

The Attempt at a Solution


I tried to use conservation of energy and find g by using the gauss' law i got a close result but i didnt get the 2 in the last solution! Then I realize I cannot use conservation of energy because the g is not constant so i cannot use mgh! any ideas how could i solve this proble
 
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welcome to pf!

hi didinyc! welcome to pf! :smile:
didinyc said:
… because the g is not constant so i cannot use mgh!

so use mMG/(r+h) :wink:
 


tiny-tim said:
hi didinyc! welcome to pf! :smile:so use mMG/(r+h) :wink:

thank you! I tried what you said too but the answer that i got is wrong.It is really close answer except the 2 in front of the R in the v equation! so this is what i got as my result: v= sqrt2gh *(1-(h/R))!
 
hi didinyc! :smile:

(just got up :zzz: …)

(have a square-root: √ :wink:)

show us your full calculations, and then we'll see what went wrong! :smile:
 
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