Gravitational forces and constant acceleration

AI Thread Summary
The discussion centers on why an object's velocity is not directly proportional to the gravitational force acting on it during free fall. While gravitational force does increase slightly as an object falls closer to Earth, the effect is negligible, leading to a nearly constant acceleration. The net force on the object results in acceleration, which is constant near the Earth's surface, despite the gravitational force changing minutely. Air resistance also plays a significant role in limiting acceleration, preventing velocity from increasing indefinitely. Ultimately, the relationship between force, acceleration, and velocity is complex and not linear due to these factors.
UrbanXrisis
Messages
1,192
Reaction score
1
Why isn't it true that an object tends to move with a velocity that is proportional to the force on it? For example, when an object falls, there must be a gravitational force on it, and this gravataional force in the downward direction gets larger and larger as the object falls and gets closer to the earth. Wouldn't this cause the velocity to get larger and larger as the object falls, so the object undergoes a constant acceleration because the force gets larger at a constant rate as the objecct falls?
 
Physics news on Phys.org
Well it does increase for a time. But don't forget that there is air resistance when free falling which slows down the rate of acceleration.
 
UrbanXrisis
Why isn't it true that an object tends to move with a velocity that is proportional to the force on it?
If you mean a net force then of course the velocity will not be proportional. The net force is an accelleration and especially if the accelleration is constant the velocity increases while the accelleration is the same, thus they are not proportional.

For a simple mathematical example take a ball falling from v=0m/s
The accelleration is G(m1m2)/r^2

say it would fall for one second
it would move g/2=4.41m and the velocity would now be 9.82m/s
the velocity has increased a lot

how has the gravitational force been changed?
if r before was (i'm guessing the distance from the center of the Earth to the ball could be 10km or whatever) it now is 9.999km this means that the gravitational force has been changed with 0.001^2=0.000001 while the the velocity has changed infinite times (0-9.82) so it is not proportional

This is the reason we say that g is constant, it changes so little...

so the object undergoes a constant acceleration because the force gets larger at a constant rate as the objecct falls?
constant accelleration means constant force :/
so truly neither the accelleration nor the force (which is the same F=mg) is constant but increases/decreases so little it can be called constant

excuse any errors, it's still morning in sweden, if you have any questions ask away :P
 
Last edited:
UrbanXrisis said:
Why isn't it true that an object tends to move with a velocity that is proportional to the force on it?
No, it moves with an acceleration that is proportional to the force on it.
You are asking why that is true- one very real reason is that force is defined as "rate of change of momentum" not momentum itself.
For example, when an object falls, there must be a gravitational force on it, and this gravataional force in the downward direction gets larger and larger as the object falls and gets closer to the earth. Wouldn't this cause the velocity to get larger and larger as the object falls, so the object undergoes a constant acceleration because the force gets larger at a constant rate as the objecct falls?

Are you assuming that the distance is so great that (1/r2) comes into play? In that case, there would be an increasing acceleration, not and constant acceleration.

In the case that distance involved is small so that the force stays approximately the same ("surface of the earth" problems), then the acceleration would be approximately the same.
 
Last edited by a moderator:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top