B A contradiction of the equivalence principle?

  • Thread starter Thread starter jeff einstein
  • Start date Start date
  • Tags Tags
    Confused Paradox
Click For Summary
Larger masses do experience a greater gravitational force, but they also have more inertia, leading to the same acceleration for all objects in free fall near Earth's surface, regardless of mass. This principle is rooted in Newton's laws, where the mass cancels out in the equations for gravitational force and acceleration. The equivalence principle, which states that gravitational mass and inertial mass are equivalent, applies in local contexts where tidal effects are negligible. Therefore, in the hypothetical scenario of Mercury and the Moon falling towards Earth from the same distance, both would accelerate at the same rate, not fall at different speeds. Understanding this concept is crucial for grasping the fundamentals of gravitational physics.
  • #91
here:
E3F077FB-678F-49C1-A911-7730621462A6 (3).jpeg
 
Last edited by a moderator:
Physics news on Phys.org
  • #92
DaveC426913 said:
Yes, you are correct.

You can convert F=ma to a=F/m.

For a given F: If m is small, then a is large. If m is large, then a is small.Looked at another way: Newton's first law says the F is equal and opposite, so
m1a1 = m2a2

And again, if m1 is smaller than m2, then a1 will have to be larger than a2
so sir i was right from the start i have tried to prove my point that different masses have different accelerations when dropped at same height
 
  • Sad
Likes russ_watters
  • #93
am i right?
 
  • #94
No. Different masses do not have different accelerations.

We have already covered this
 
  • Like
Likes russ_watters
  • #95
Jeff, are you even considering the messages you are getting? You respond within seconds and we are covering the same ground over and over and over.
 
  • Like
Likes bhobba, russ_watters, PeterDonis and 1 other person
  • #96
DaveC426913 said:
Yes, you are correct.

You can convert F=ma to a=F/m.

For a given F: If m is small, then a is large. If m is large, then a is small.Looked at another way: Newton's first law says the F is equal and opposite, so
m1a1 = m2a2

And again, if m1 is smaller than m2, then a1 will have to be larger than a2
but if this is true....
 
  • #97
disprove this i guess...
 
  • Sad
Likes Dale
  • #98
jeff einstein said:
disprove this i guess...
Sorry, we are now mixing up two things. I may be culpable for that.

Two different masses both being under the influence of the same gravity will have the same acceleration. (1)

But that's not the scenario you have diagrammed, above. The scenario you have diagrammed is about different masses under the influence of different gravity (i.e. each others'). (2)

1694713002286.png
 
Last edited:
  • #99
jeff einstein said:
but if this is true....
jeff einstein said:
disprove this i guess...
You have been instructed by multiple people over multiple pages carefully explaining the situation. Carefully and repeatedly explaining where your original statements were wrong and also where they were almost right. With multiple corrections from multiple people carefully explaining the physics and the source of the confusion.

Then one single person gives one single confused answer after telling you that they were confused because of a mistaken crop, and immediately you just throw out everything you were taught by all of the previous people?

That shows that you are not seeking to learn, you are just seeking validation.

Thread closed. This topic has been done. Do not re-open, just re-read.
 
  • Like
Likes bhobba, vanhees71, weirdoguy and 7 others
  • #100
PeterDonis said:
The acceleration of both objects, however, is not the same, because it is the force divided by the mass of the object.

I know the topic is closed, and the OP has probably not done enough math yet to understand the full answer, but this is the so-called two-body problem:

https://en.wikipedia.org/wiki/Two-body_problem

I post it to stimulate the study of vector calculus and its applications in the future.

Way back when I studied vector calculus, this important application, interestingly, wasn't covered

Strang's calculus textbook is a good place to learn it:
https://ocw.mit.edu/ans7870/resources/Strang/Edited/Calculus/Calculus.pdf

Thanks
Bill
 
Last edited:

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
1K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
10
Views
2K