A contradiction of the equivalence principle?

[jeff einstein]

TL;DR Summary

is the equivelence principle right?

hi, I am a physics student in grade eleven. I do not have much knowledge of physics but I have a doubt:

apart from the maths and the experiments, shouldn't larger masses be attracted to Earth's gravity rather than be attracted the same regardless of mass? if we had a scenario where we have Mercury was (of course this is not true: just imagine) distanced 10000 km from Earth and another scenario where the moon is distanced 10000km (imagine) from the Earth, it's obvious that Mercury will fall into Earth faster than moon as mars has a larger mass relative to the moon. isn't this also supposed to happen when two different masses (one more mass and the other less mass) are held at the same distance and should be attracted to Earth at different rates (maybe even in a really minute level)? so isn't this contradicting the equivalence principle?
please answer.

please don't bring up anything about orbits and stuff because when I mentioned mercury and the moon I just wanted to address two different masses on a larger scale.

 

[berkeman]

Welcome to PF.

hi, I am a physics student in grade eleven. I do not have much knowledge of physics but I have a doubt:

apart from the maths and the experiments, shouldn't larger masses be attracted to Earth's gravity rather than be attracted the same regardless of mass?

Sorry, what does that mean? Are you familiar yet with the equation for the gravitational force between two masses?
$$F = \frac{Gm_1m_2}{r^2}$$

http://hyperphysics.phy-astr.gsu.edu/hbase/grav.html

 

[Ibix]

apart from the maths and the experiments,

With this start, I don't know what answer you expect. We can only answer by reference to a mathematical model (acceleration doesn't depend on the mass) or to experimental demonstrations of the same.

When the two objects are comparable in mass and you can't treat one of them as stationary then the closing rate will depend on both masses, yes. This has nothing to do with the equivalence principle, though, which is only relevant on scales where curvature is negligible (the full formalised version applies only at an event).

 

[Dale]

TL;DR Summary: is the equivelence principle right?

so isn't this contradicting the equivalence principle?
please answer.

What you described has nothing to do with the equivalence principle as far as I can tell.

The equivalence principle is a local principle, where "local" has the specific meaning that it is a small enough region of space and time that tidal effects are negligible. If you are dealing at the scale of an entire planet or the entire moon then that scale is too big. The equivalence principle is neither right nor wrong in such a scenario, it simply doesn't apply.

 

[hmmm27]

You can plug the aforementioned gravity equation $F=\frac{Gm_1m_2}{r^2}$ into the standard $F=ma$ (where $m$ is the sum of the masses) and end up with : $a=\frac{Gm_1m_2}{(m_1+m_2)r^2}$

[EDIT: except that's wrong ; Fast-forward to PeterDonis'post #22 to continue the math. Remainder of original post - long, florid LaTex - deleted to avoid embarassment]

 

[hmmm27]

Also, what do you think the "equivalence principle" is ?

 

[Cutter Ketch]

TL;DR Summary: is the equivelence principle right?

hi, I am a physics student in grade eleven. I do not have much knowledge of physics but I have a doubt:

apart from the maths and the experiments, shouldn't larger masses be attracted to Earth's gravity rather than be attracted the same regardless of mass? if we had a scenario where we have Mercury was (of course this is not true: just imagine) distanced 10000 km from Earth and another scenario where the moon is distanced 10000km (imagine) from the Earth, it's obvious that Mercury will fall into Earth faster than moon as mars has a larger mass relative to the moon. isn't this also supposed to happen when two different masses (one more mass and the other less mass) are held at the same distance and should be attracted to Earth at different rates (maybe even in a really minute level)? so isn't this contradicting the equivalence principle?
please answer.

please don't bring up anything about orbits and stuff because when I mentioned mercury and the moon I just wanted to address two different masses on a larger scale.

Ignoring the issues with your question others have pointed out, I think I understand what you are asking. I believe you have a misconception.

A bigger mass DOES experience a bigger force of gravity. In fact the force is directly proportional to mass. However, bigger masses also have more inertia. A larger force is needed to move them. Inertia, too, is directly proportional to mass (kind of definitional, really). If you think in terms of Newton’s equation: F = ma, the inertia on the right side is explicitly proportional to mass. If the force on the left side is also proportional to mass (as it is for gravity) then the mass appearing on both sides of the equation cancels, and the acceleration is independent of mass.

So, in short, you have misremembered. The FORCE due to gravity is NOT independent of the object’s mass. It is the ACCELERATION due to gravity that is independent of mass. Thus, near the surface of the earth we have the constant “g” which is the acceleration due to gravity $$ g = 9.8 m/s^2 $$ Note that the units are acceleration, not force.

 

[Vanadium 50]

TL;DR Summary: is the equivelence principle right?

apart from the maths and the experiments

With this start, I don't know what answer you expect.

I guess when you name yourself after Einstein you can't be bothered with such petty things.

I think the right question is from @hmmm27 and builds on a similar question from @Dale - what do you think the equivalence principle is. I mean, we can guess at what you misunderstand and try and clarify things, but it will be smoother and likely faster if you were to tell us what you think your example is disproving.

 

[jeff einstein]

OK, maybe I am wrong about the theory of equivalence. what I actually meant was the idea that all masses fall to the earth at the same velocity. and I have been asked about if know the equation relating to this, Yes I do but I clearly mentioned that I do not want equations used for this doubt clearance. after all, I want to clarify one thing, Mercury (larger mass relative to the moon) "would fall to the earth" faster than the moon (smaller mass) both positioned at an equal distance, am I right?

 

[berkeman]

I clearly mentioned that I do not want equations used for this doubt clearance

A word of advice: that's a bad way to approach learning about science.

 

[jeff einstein]

why?

 

[jeff einstein]

hey doesn't mean I am bad at maths or equations just asking how the theory part of this actually works.

 

[jeff einstein]

A word of advice, that's a bad way to approach learning about science.

so is this theory just true in maths??

 

[berkeman]

so is this theory just true in maths??

Specifically which theory? Specifically which maths?

 

[PeterDonis]

the idea that all masses fall to the earth at the same velocity

No, all masses fall to the earth with the same acceleration. That is because the $m$ that appears in $F = ma$ (mass times acceleration) is the same as the $m$ that appears in $F = G m M / r^2$. So the $m$ cancels out of the equation and you have $a = G M / r^2$. The equality of the $m$ in both equations is (the Newtonian version of) the equivalence principle.

Mercury (larger mass relative to the moon) "would fall to the earth" faster than the moon (smaller mass) both positioned at an equal distance, am I right?

No. Both would fall towards the earth with the same acecleration.

 

[topsquark]

so is this theory just true in maths??

Okay, theories can only be proven by taking data in an experiment and seeing if the predictions by theory match the data. How are you possibly going to do this without numbers, what you are calling "maths?"

If you only want to talk conceptually there seems to be some wiggle room in the argument, but I'll give it a quick try.

Say we have Mercury and Earth. According to Newton, the force between them is proportional to the product of their masses. ala Newton's 2nd Law, we may then say that the acceleration of Mercury is independent of the mass of Mercury, so the mass of Mercury (or whatever is considered to be falling does not determine the acceleration.

It's easier if you use the Math:
$F_M = \dfrac{G M_M M_E}{r^2}$

$F_M = M_M a = \dfrac{G M_M M_E}{r^2}$

Solving for a:
$a = \dfrac{G M_E}{r^2}$

-Dan

 

[jeff einstein]

thank you for properly referring to my questions #peterDonis and #topquark !! but how would acceleration be the same when mercury and the moon have their own mass?

 

[jeff einstein]

and cause of their own mass they have their own gravitational force as well

 

[topsquark]

thank you for properly referring to my questions #peterDonis and #topquark !! but how would acceleration be the same when mercury and the moon have their own mass?

As I pointed out, the mass of Mercury and the mass of the moon cancel out when you use Newton's 2nd Law. It's tough to see that only conceptually.

-Dan

 

[jeff einstein]

now i am getting a clearer view. but why exactly do MErcury and moons gravity have no affect here

 

[PeterDonis]

how would acceleration be the same when mercury and the moon have their own mass?

Because of the math that you say you don't want to look at, which I wrote down: the fact that the $m$ in $F = ma$ is the same as the $m$ in $F = G m M / r^2$. This is often described as inertial mass (the $m$ in $F = ma$) being equal to gravitational mass (the $m$ in $F = G m M / r^2$). And that equality, as I said, is the equivalence principle.

why exactly do MErcury and moons gravity have no affect here

Because you're picking the Earth as your reference body and asking about the motion of Mercury and the moon relative to the Earth. But note that in the equation for gravitational force, $F = G m M / r^2$, both masses appear. The force increases with the mass, but the acceleration does not. And the acceleration is what determines the motion.

If you wanted to pick Mercury or the moon as your reference body, you could do that just as easily. Then you would find that the acceleration of the Earth and the moon towards Mercury if both were at the same distance, or the acceleration of Mercury and Earth towards the moon if both were at the same distance, would be the same.

In the general case when you have to solve for the motion of many bodies in the same system, the usual convention is to use the barycenter of the system. For example, astronomers, when calculating the motion of all of the bodies in the solar system, use the solar system barycenter, which is much closer to the Sun than to anything else but is not the same as the center of the Sun. See here for more information about that:

https://en.wikipedia.org/wiki/Barycenter_(astronomy)

 

[PeterDonis]

You can plug the aforementioned gravity equation $F=\frac{Gm_1m_2}{r^2}$ into the standard $F=ma$ (where $m$ is the sum of the masses) and end up with : $a=\frac{Gm_1m_2}{(m_1+m_2)r^2}$

No, this is wrong. $F = ma$ applies separately to the two masses, so you have $F = m_1 a_1$ and $F = m_2 a_2$. Then you can solve for $a_1$ and $a_2$.

 

[jeff einstein]

this i way clearer. but doesn't the mercury have its own gravity that might affect this?

 

[PeterDonis]

this i way clearer. but doesn't the mercury have its own gravity that might affect this?

Which post are you responding to? This question has already been answered.

 

[jeff einstein]

so at last moon and mercury falls at same speed?

 

[PeterDonis]

so at last moon and mercury falls at same speed?

No, at the same acceleration. Acceleration is not the same as speed.

 

[jeff einstein]

thank you very much. i finally give in to your great answer now i will try using the equation and see the result my self. thank you very much!! for all the patience.

 

[PeroK]

No, at the same acceleration. Acceleration is not the same as speed.

Are you sure about that? The acceleration as a function of time varies depending on the mass, because the distance between the centres of mass reduces more quickly in the case where the Earth's acceleration cannot be neglected. Only the initial instantaneous acceleration is the same for a small object and a very large object.

 

[PeterDonis]

thank you very much!!

You're welcome!

 

[PeterDonis]

Are you sure about that?

About the specific statement I made, which is based on the equations I wrote down, yes.

The acceleration as a function of time varies depending on the mass, because the distance between the centres of mass reduces more quickly in the case where the Earth's acceleration cannot be neglected.

Yes (although if we are using Earth-centered coordinates, as the OP is, there is no "Earth's acceleration" so describing the behavior is more complicated), but the OP explicitly ruled all this out by asking us to consider two different objects at the same distance from Earth.

Only the initial instantaneous acceleration is the same for a small object and a very large object.

And the instantaneous acceleration at a given distance is what the OP was asking about.

 

[jeff einstein]

Are you sure about that? The acceleration as a function of time varies depending on the mass, because the distance between the centres of mass reduces more quickly in the case where the Earth's acceleration cannot be neglected. Only the initial instantaneous acceleration is the same for a small object and a very large object.

so for two large objects like mercury and the moon, there would be something else happening as both are large objects?

 

[jeff einstein]

And the instantaneous acceleration at a given distance is what the OP was asking about.

just to make things clear i meant that both object reach the ground at the same time when held at a certain height this is what i meant by "same speed". is this the same as what your saying?

 

[PeroK]

so for two large objects like mercury and the moon, there would be something else happening as both are large objects?

Let's be precise.

If you dropped an object of mass 1kg from the same distance as the Moon, then it would take a certain time to fall to half that distance.

If the Moon were released from rest at the same distance from Earth, then:

a) It would move half that distance in less time than the 1kg object.
b) The Earth would also have moved significantly, so the overall distance between the Moon and the Earth would be less than half the original distance at that time.

There's nothing but gravity at work here.

 

[jeff einstein]

so this is exactly what I am saying Objects with larger mass (the moon in your case) move towards the earth faster than the smaller mass when dropped at the same height.

 

[jeff einstein]

am i right about this?