
#1
May513, 10:44 AM

P: 2

It has occurred to me that gravitational force on an object will change as the object falls toward the Earth. I modified Newton's equation F=GMm/(r^2) to reflect the change in force over time:
F = GMm/[r.5a(t^2)]^2 Now that I have the equation I'm trying to figure out the instantaneous change in force. I assume it would be the derivative of the above equation, but I'm not the best at derivatives. Can anyone help me out? 



#2
May513, 11:08 AM

P: 417

First of all, this will only work over a small enough region for the acceleration to be considered constant. The term ##r_0\frac12at^2##, I assume you are taking from the constant acceleration kinematics equations and assuming ##v_0=0##. If that is the case, your construction of the function requires the ##F'\approx0## just as happens near the surface of the earth (where g is approximately constant). I did a quick check using a variable acceleration and it doesn't seem to be very illuminating (unless I'm missing a simplification).




#3
May513, 11:28 AM

P: 2

There seems to have been a [Math Processing Error] in your reply which looks to be important.
To clear things up. In the equation all variables are considered to be constant except "F" and "t." Initial velocity of "m" is to be considered 0. "r" is to be considered the initial distance between "m" and "M." 



#4
May513, 11:42 AM

P: 31

Instantaneous Change in Gravitational Force
why don't you do
dF/dt=(dF/dr)(dr/dt) F=GMm/r^{2} dF/dr=(2GMm/r^{3})(dr/dt) Now take the initial distance between the 2 bodies as R the conserve energy (calculate velocity form COM frame {for simplicity}) you should ger a dr/dt plug it in and you're done. 



#5
May513, 02:50 PM

Mentor
P: 15,568

Popper's answer is unhelpful. There's nothing wrong with using Newtonian gravity in this limit, and this is a discussion of a moving body through a static field, so the speed of gravitational propagation doesn't enter into this problem.




#6
May513, 03:23 PM

P: 102

To find the force as a function of time first calculate the acceleration of the falling body and then multiply the result by the body's mass. 



#7
May513, 07:17 PM

P: 417

You can't consider the acceleration to be constant. Differentiating the equation you gave will not give the answer you are looking for. Arkavo's explanation works (but the last line should start with dF/dt), but to compute ##\frac{dr}{dt}##, you either have to integrate the force or make use of conservation of energy. I think the latter works better, but again, as a function of the distance, it will be approximately 0. If you want to try it, just set ##E_{total}=\frac{GMm}{R}## where ##R## is the starting distance and solve for ##v##. Edit: clearly my first sentence is silly. The acceleration IS (almost) constant. What I meant was the acceleration isn't constant over the entirety of the fall. 


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