When you drop a ball from above, why doesn't it drill through earth?

[Nugatory]

I wonder how Newton explained this phenomenon since the electromagnetic force wasn't discovered then yet. Didn't this question occur to him that without some kind of very very very strong repulsive force, things would be piercing the whole planet?

Well, he's not around to tell us what he was thinking... But it is safe to say that in the 17th century people mostly took the solidity of objects for granted, as a self-evident proposition needing no explanation; some materials, such as water, are just more penetrable than others, like steel. I expect that Newton and his contemporaries were no more perplexed by the fact that a brick would sit on the surface of the Earth instead of piercing through it, than they were by the fact that a heavy sledgehammer blow would bounce off of it. (And it's worth noting that the forces involved in a sledgehammer blow are much greater than that of gravity acting on a brick).

 

[Nugatory]

I wonder how whether electromagnetic force really existed when the planet started to form. Because the planet must have started with two atoms or something attracting each other.

Didn't the outer electrons of those first two atoms repel each other, since the gravity was way weaker than the electromagnetic force?

Although gravity is a very weak force, it has one advantage over electromagnetism: All particles attract gravitationally, while with electromagnetism there are positive and negative charges, and like charges repel.

A hydrogen atom (for example) is made up of a positive-charged proton and a negative-charged electron. If you get any significant distance away from the atom, it is electrically neutral; the electrical force exerted by the proton will almost completely cancel out the electrical force exerted by the electron. But the gravitational force is still there. Thus, if you have a large number of hydrogen atoms floating around in empty space, their gravitational attraction will tend to pull them together no matter how spread out they are; the electromagnetic forces won't be significant until the atoms are so close that they are starting to bump into one another.

 

[Nugatory]

Does this gravity affect protons and electrons too?

Yes. Look at DaleSpam's post #16 in this thread.

 

[tarekatpf]

Although gravity is a very weak force, it has one advantage over electromagnetism: All particles attract gravitationally, while with electromagnetism there are positive and negative charges, and like charges repel.

A hydrogen atom (for example) is made up of a positive-charged proton and a negative-charged electron. If you get any significant distance away from the atom, it is electrically neutral; the electrical force exerted by the proton will almost completely cancel out the electrical force exerted by the electron. But the gravitational force is still there. Thus, if you have a large number of hydrogen atoms floating around in empty space, their gravitational attraction will tend to pull them together no matter how spread out they are; the electromagnetic forces won't be significant until the atoms are so close that they are starting to bump into one another.

Thank you very much again for your answer.

I have this question now.

Wouldn't the repulsive force exerted by the outer electrons of the red ball ( which represents surface of the Earth ) neutralized by the pull from the protons of the red ball? Then how come the red ball's outer electrons still repel the outer electrons of the green ball ( the repulsive force of the outer electrons would be neutralized by the protons of the green ball too, I think. )

 

[siddharth23]

The answer is reactive force.

 

[tarekatpf]

The answer is reactive force.

I am sorry I did not get which question you're referring to, since I have made a lot of questions.

 

[Drakkith]

Thank you very much again for your answer.

I have this question now.

Wouldn't the repulsive force exerted by the outer electrons of the red ball ( which represents surface of the Earth ) neutralized by the pull from the protons of the red ball? Then how come the red ball's outer electrons still repel the outer electrons of the green ball ( the repulsive force of the outer electrons would be neutralized by the protons of the green ball too, I think. )

(Very very simplified explanation)
No, the electrons of each atom are separated from the nucleus and "touch" each other first. Since the electrons are closer to each other they experience a repulsion. There is also charge screening going on, which is when the charge from the nucleus is screened from being felt by anything outside the atom.

 

[tarekatpf]

(Very very simplified explanation)
No, the electrons of each atom are separated from the nucleus and "touch" each other first. Since the electrons are closer to each other they experience a repulsion. There is also charge screening going on, which is when the charge from the nucleus is screened from being felt by anything outside the atom.

So, suppose I have two atoms. Atom A and Atom B. Centrifugal forces of all the electrons of Atom A are canceled out by centripetal force of all the protons of the Atom A.

Now if I bring atom B, why would the outer electrons be repelled by the outer electrons of atom A? Has any ''new force'' generated spontaneously?

 

[Nugatory]

I have this question now.

Wouldn't the repulsive force exerted by the outer electrons of the red ball ( which represents surface of the Earth ) neutralized by the pull from the protons of the red ball? Then how come the red ball's outer electrons still repel the outer electrons of the green ball ( the repulsive force of the outer electrons would be neutralized by the protons of the green ball too, I think.)

That is indeed the natural next question... and we're getting to where we actually have to play with the math to see how it supports intuition. Here's a fairly simple model that illustrates the underlying physics pretty well.

Imagine that a hydrogen atom consists of one electron and one proton, connected by a rod of length 10^-12 meters. The total electrical force on the the atom from any nearby charged particles can be calculated just by adding the force on the proton and the force on the neutron. We'll make life a bit easier by choosing units in which the charge of the electron is -1 and the charge of the proton is +1 so the force between two particles is $\frac{1}{r^2}$, attractive or repulsive according to the signs of the charges.

Now consider two such atoms lined up on the x-axis with their electrons facing each other, and 1 meter apart measuring proton to proton. What's the total force on the left-hand atom from the electrical particles in the right-hand atom?
1) The protons repel each either with a force of $\frac{1}{r^2} = \frac{1}{1} = 1$
2) The electrons repel each either with force of $\frac{1}{r^2} = \frac{1}{1-(2\times10^{-12})^2} \approx 1-2\times10^{-24}$
3) The left-hand electron and the right-hand proton attract with a force of $\frac{1}{r^2} = \frac{1}{1-(10^{-12})^2} \approx 1-10^{-24}$
4) The left-hand proton and the right-hand electron attract with a force of $\frac{1}{r^2} = \frac{1}{1-(10^{-12})^2} \approx 1-10^{-24}$

Add the two attractive and the two repulsive force and you get... zero, give or take some corrections that are small compared to $10^-24$ from the approximations in #2, #3, #4 above. There's just no detectable attraction or repulsion.

Now consider what it looks like if we try to push the two atoms together so that the electrons are touching and the protons are separated by just $2\times{10^{-12}}$ meters. The repulsive force between the protons is now stronger, as is the attractive force between the proton-electron pairs. But none of these matter, because the the repulsive force between the electrons is $\frac{1}{0^2}$ which is infinite! This tells us that we can't actually push the two atoms that close together - if we try the total force will become ever more strongly repulsive as we push them closer.

 

[tarekatpf]

That is indeed the natural next question... and we're getting to where we actually have to play with the math to see how it supports intuition. Here's a fairly simple model that illustrates the underlying physics pretty well.

Imagine that a hydrogen atom consists of one electron and one proton, connected by a rod of length 10^-12 meters. The total electrical force on the the atom from any nearby charged particles can be calculated just by adding the force on the proton and the force on the neutron. We'll make life a bit easier by choosing units in which the charge of the electron is -1 and the charge of the proton is +1 so the force between two particles is $\frac{1}{r^2}$, attractive or repulsive according to the signs of the charges.

Now consider two such atoms lined up on the x-axis with their electrons facing each other, and 1 meter apart measuring proton to proton. What's the total force on the left-hand atom from the electrical particles in the right-hand atom?
1) The protons repel each either with a force of $\frac{1}{r^2} = \frac{1}{1} = 1$
2) The electrons repel each either with force of $\frac{1}{r^2} = \frac{1}{1-(2\times10^{-12})^2} \approx 1-2\times10^{-24}$
3) The left-hand electron and the right-hand proton attract with a force of $\frac{1}{r^2} = \frac{1}{1-(10^{-12})^2} \approx 1-10^{-24}$
4) The left-hand proton and the right-hand electron attract with a force of $\frac{1}{r^2} = \frac{1}{1-(10^{-12})^2} \approx 1-10^{-24}$

Add the two attractive and the two repulsive force and you get... zero, give or take some corrections that are small compared to $10^-24$ from the approximations in #2, #3, #4 above. There's just no detectable attraction or repulsion.

Now consider what it looks like if we try to push the two atoms together so that the electrons are touching and the protons are separated by just $2\times{10^{-12}}$ meters. The repulsive force between the protons is now stronger, as is the attractive force between the proton-electron pairs. But none of these matter, because the the repulsive force between the electrons is $\frac{1}{0^2}$ which is infinite! This tells us that we can't actually push the two atoms that close together - if we try the total force will become ever more strongly repulsive as we push them closer.

I didn't understand some of your maths.

Shouldn't the force exerted on the electrons of the left hand atom by the right hand atom's electrons be [1 / { 1- ( 2 x 10^-2 ) }^ 2 }] ? Since the distance between two electrons should be { 1- ( 2 x 10^-2 ) }, and the force should be 1/ (distance)^2?

( I don't know how you put those mathematical symbols there? Could you please help me? )

Similarly, shouldn't the attractive force between the electrons and protons of opposite side be ( 1- ( 10^-2) )^2 ?

Please correct me if I am wrong.

 

[Nugatory]

( I don't know how you put those mathematical symbols there? Could you please help me? )

It's documented here: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

 

[Nugatory]

Shouldn't the force exerted on the electrons of the left hand atom by the right hand atom's electrons be [1 / { 1- ( 2 x 10^-2 ) }^ 2 }] ? Since the distance between two electrons should be { 1- ( 2 x 10^-2 ) }, and the force should be 1/ (distance)^2?

Look carefully - I set the separation to 10^-12 not 10^-2.

BTW, I did get a sign wrong. It should be
$\frac{1}{r^2} = \frac{1}{1-(2\times10^{-12})^2} \approx 1+2\times10^{-24}$
(note the $+$ sign in the rightmost term) but that doesn't change the result: the net force is negligible when the atoms are separated by one meter, but very strong when they are s close to one another that the electrons are almost touching.

 

[adi19956]

First of all, I want to say that I am totally ignorant about physics. And I do not understand the general concepts of physics at all.

Now coming to the question.

Suppose, you drop a green tennis ball from above.

Now It usually lands on Earth and stop there ( ignore the bounces for now. ) Suppose, the point at which the green ball lands is a red ball. Since the Earth is made of ''a mass of balls'' ( to get the concept, I am using such metaphors ), below the red ball should be another ball, and below that another ball, and so on, till you reach the opposite point on the earth, which is, suppose, a pink ball.

Now since all the balls from the red ball upto the pink ball is drawing the green ball toward them, how can this one little green ball stop just over the red ball ignoring the pull of all the balls between the red and the pink one?

Shouldn't it just ''break the red ball'' and keep reaching toward the next ball and break that too? I mean, drilling through earth?

I know the Earth is more or less a sphere. So are the pulls from balls around the red ball preventing the green ball from drilling through earth?

Did you post this on The Student Room?

 

[tarekatpf]

Look carefully - I set the separation to 10^-12 not 10^-2.

BTW, I did get a sign wrong. It should be
$\frac{1}{r^2} = \frac{1}{1-(2\times10^{-12})^2} \approx 1+2\times10^{-24}$
(note the $+$ sign in the rightmost term) but that doesn't change the result: the net force is negligible when the atoms are separated by one meter, but very strong when they are s close to one another that the electrons are almost touching.

Sorry, that should indeed be -12, not -2. But let me show in a diagram what I thought:

I didn't understand why you think the force with which the electrons repel each other is [1 / { 1- ( 2 x 10^-2 )^2 }] instead of [1 / { 1- ( 2 x 10^-2 ) }^ 2 }]

 

[tarekatpf]

Did you post this on The Student Room?

Yes, why?

 

[adi19956]

Yes, why?

I saw it there too, weird

 

[Nugatory]

instead of [1 / { 1- ( 2 x 10^-2 ) }^ 2 }]

Because I still can't type .

It is, of course and as your picture shows, $\frac{1}{(1-2\times{10^{-12})^2}}$ for the electron/electron repulsion, $\frac{1}{(1-{10^{-12})^2}}$ for the electron-proton attractions, and $1$ for the proton-proton repulsion.

The result of the arithmetic comes out the same though: when the atoms are separated by one meter the net force between them is negligible because all the forces are equal to one, give or take some tiny correction; but as you push them closer together the electron-electron repulsion dominates.

 

[brocks]

Thank you very much. I wonder why electromagnetic force is not as ''famous'' as or more so than the gravitational force.

Because most people are not physicists. EM forces and fields are a large part of a physics curriculum.

But you have seen how much stronger electromagnetic forces are than gravity, you just didn't realize what you were seeing. Most kids have seen that if you get a small static charge on a comb or balloon or something by rubbing it on their hair, you can pick up bits of paper with it. The paper is being drawn down by gravitational force of the entire earth, and it's being lifted by the measly little static charge on your comb. And the comb wins.

 

[tarekatpf]

Because I still can't type .

It is, of course and as your picture shows, $\frac{1}{(1-2\times{10^{-12})^2}}$ for the electron/electron repulsion, $\frac{1}{(1-{10^{-12})^2}}$ for the electron-proton attractions, and $1$ for the proton-proton repulsion.

The result of the arithmetic comes out the same though: when the atoms are separated by one meter the net force between them is negligible because all the forces are equal to one, give or take some tiny correction; but as you push them closer together the electron-electron repulsion dominates.

Thanks for your great and kind efforts. Your mathematical explanation really helped me a lot.

Thanks, again.

 

[tarekatpf]

Because most people are not physicists. EM forces and fields are a large part of a physics curriculum.

But you have seen how much stronger electromagnetic forces are than gravity, you just didn't realize what you were seeing. Most kids have seen that if you get a small static charge on a comb or balloon or something by rubbing it on their hair, you can pick up bits of paper with it. The paper is being drawn down by gravitational force of the entire earth, and it's being lifted by the measly little static charge on your comb. And the comb wins.

Yep. A classic example. Thanks.