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Change in Gravity Affecting Free Fall 
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#1
Mar913, 04:41 PM

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In high school physics, which I am in, we learn that to find the distance that and object falls can be found with the equation, x=v_{0}t+1/2gt^{2}. We also learn that the force of gravity between two objects can be found by F_{g}=Gm_{1}m_{2}/r^{2} and thus acceleration due to gravity can be derived to be g=Gm/r^{2}
This is all well and good on a small scale, but on a large scale, such as something falling from space, the difference in gravity due to change in distance between the objects is too large to be negligible. As an object falls, the force of gravity, and by extension, its acceleration, increases exponentially. I have tried to derive an equation that gives the distance that an object will fall (to Earth) as a function of time and initial height, ignoring air resistance, using the equations above, but my knowledge of calculus is only so great, and I keep getting stuck not knowing which variable to solve for or use. I am sure that there is an e in there somewhere as continuous compounding would be needed, but I'm not sure. Could someone please give me an equation that meets these specifications, and if possible, a stepbystep derivation of said equation? 


#2
Mar913, 05:04 PM

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hi kaikalii! welcome to pf!
x=v_{0}t+1/2gt^{2} comes from x'' = g, which we integrate once to get x' = gt + constant, and again to get x = gt^{2}/2 + (constant)t + constant if instead we use r'' = Gm/r^{2}, we multiply both sides by r' to get r'r'' = Gmr'/r^{2}, integrate that to get 1/2(r')^{2} = Gm/r + constant, or r'/√(Gm/r + constant) = √2 … i don't think that has an integral in terms of ordinary functions 


#3
Mar913, 05:20 PM

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#4
Mar1013, 03:08 AM

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Change in Gravity Affecting Free Fall
hi kaikalii!
(just got up ) ' means derivative, and '' means derivative of derivative (and so on) r' and r'' are a lot easier to write, and "rdash" and "rdoubledash" are a lot easier to say (than dr/dt and …) 


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