Why do particles in a falling coach get closer together?

AI Thread Summary
Particles in a falling railway coach move closer together due to the nonuniform gravitational field of Earth, which affects them differently based on their positions. As both particles are attracted toward Earth's center, their distance decreases when the coach is in free fall. This phenomenon is not due to gravitational attraction between the particles themselves, which is negligible. Conversely, when the coach rises, the gravitational force weakens according to the Inverse Square Law, preventing any significant change in particle position. The discussion emphasizes the distinction between the effects of gravity in falling versus rising scenarios.
Ashshahril
Messages
4
Reaction score
1
Homework Statement
You are launched upward inside a railway coach in a horizontal position with respect to the surface of Earth, as shown in the figure (attached pdf). After the launch, but while the coach is still rising, you release two ball bearings at opposite ends of the train and at rest with respect to the train.

Riding inside the coach, will you observe the distance between the ball bearings to increase or decrease with time?
Relevant Equations
F ∝ 1/R^2
As widely separated particles within a large enclosed space are differently affected by the nonuniform gravitational field of Earth, to use the Newtonian way of speaking, two particles released side by side are both attracted toward the center of Earth, so they move closer together as measured inside a falling long narrow horizontal railway coach. This has nothing to do with "gravitational attraction" between the particles, which is entirely negligible. (Spacetime Physics, Second Edition by Edwin F. Taylor and John Archibald Wheeler; ISBN:0-7167-2327-1)

So, when the coach fall toward Earth, the particles get attracted by the center of Earth. As a result, their distance decreases. But when the coach rise above, due to the Inverse Square Law, the gravitational force gets weaker and weaker. As a result, the particles mustn't change their position. But, in the case of this question, the distance will decrease according to the answer of “Spacetime Physics” book. But I don't understand why. I know that it is so, so that you cannot distinguish rising from falling (inertial reference frame) but how can the distance between the particles decrease in this particular case?
 

Attachments

Physics news on Phys.org
Ashshahril said:
But when the coach rise above, due to the Inverse Square Law, the gravitational force gets weaker and weaker. As a result, the particles mustn't change their position.
"Weaker and weaker" is not the same thing as non-existent.
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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