Instantaneous Change in Gravitational Force

In summary: 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##.
  • #1
usnkruse
2
0
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?
 
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  • #2
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
Reply to DrewD

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
why don't you do

dF/dt=(dF/dr)(dr/dt)

F=GMm/r2

dF/dr=(-2GMm/r3)(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
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
Vanadium 50 said:
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.

The error in my post was that I misunderstood his question. I thought that he was talking about the change in the gravitational field of a moving body, soI deleted my erroneous response.

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
usnkruse said:
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."


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.
 

1. What is instantaneous change in gravitational force?

Instantaneous change in gravitational force refers to the sudden and immediate alteration in the strength of the gravitational pull between two objects. This change can occur due to a variety of factors, such as a change in the mass or distance between the objects.

2. How is instantaneous change in gravitational force different from regular gravitational force?

Regular gravitational force is the constant pull of gravity between two objects, while instantaneous change in gravitational force is a sudden alteration in this pull. Regular gravitational force can be calculated using the formula F=G(m1m2)/r^2, while instantaneous change in gravitational force may require different calculations depending on the situation.

3. What causes instantaneous change in gravitational force?

Instantaneous change in gravitational force can be caused by a number of factors, including changes in the mass or distance between two objects, the presence of other massive objects nearby, or the effects of general relativity.

4. How can instantaneous change in gravitational force be measured?

Instantaneous change in gravitational force can be measured using devices such as accelerometers or gravimeters. These instruments can detect changes in the strength of gravitational force and record them for analysis.

5. What are the potential implications of instantaneous change in gravitational force?

Instantaneous change in gravitational force can have significant implications for space travel, satellite orbits, and other aspects of celestial mechanics. It can also impact the tides and weather patterns on Earth, as well as the stability of objects in our solar system.

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