Why does Newton's Gravitation Law work for objects nearby?

[Dirkg]

I am having a problem with visualizing why Newton's Law of Universal Gravitation (NUGR) holds true when objects get close together. F=Gmm/r^2 makes sense for planets that are far away because each object can be treated as a point with gravity acting at the center of each object. For a person standing on the surface however, there is a significant portion of the mass of the Earth that is not directly below the person. Of course the mass is evenly distributed on each side, so the direction of the force is straight down, but only the vertical components of each part of the mass of the sphere is acting downwards, the horizontal components will all cancel out. If NUGR holds true for each portion of the planetary mass then the resultant force vector will calculate to a lower magnitude than the whole Earth because of the horizontal forces cancelling out. I know that I must be visualizing something wrong, but I don't see it. Before I started thinking about this, I just accepted that gravity acts from the center of the planet because the sum of the gravitational force from every point within the sphere adds up to the force we feel as if it is actually coming from the center. If that is the case then shouldn't NUGR take into account how much of the mass is not directly below?

 

[A.T.]

I am having a problem with visualizing why Newton's Law of Universal Gravitation (NUGR) holds true when objects get close together.

https://en.wikipedia.org/wiki/Shell_theorem

 

[Dirkg]

Thanks, although my calculus is (very) rusty, I understand the concepts well enough to grasp what is happening. It still seems odd to me that it should work out, but numbers don't lie. I will work on understanding the math better. It is intuitive to me that from inside a shell, all the forces cancel out, and the lower shells still act on the person, but I'm going to have to use a spreadsheet and work out different values of:

as the angles vary to be able to conceptualize this.
Thanks again

 

[Dr.D]

It may help to think of the Earth as comprised of a vast number of particles, each of which interacts with the person on the surface according Newton's Law of Gravitation. If you set up the differential force between a typical particle and the man, then integrate over the entire volume of the earth, you should get the usual, expected result.

 

[Dirkg]

That was the part that was bothering me. Starting from a more basic case, if you start with a hemisphere with a person at the rim. The mass is half of the sphere and r is greater than it was for the sphere. Therefore GMM/r^2 is less than it would be for half of the sphere. Also the force is acting at an angle.
Bringing a second hemisphere into complete the sphere and adding their force vectors results in an even smaller value because only the vertical components add.
What am I doing wrong? Is r somehow smaller for a hemisphere?

 

[Dr.D]

Set the whole problem up in spherical coordinates, and put your person on the polar axis for simplicity. Use sine and cosine functions to resolve the components of the force as needed. But do work with a differential mass element in the Earth to start; don't try the hemisphere. Bear in mind that the r value needed is from the point in the Earth to the man, not the radius of the earth.

 

[jbriggs444]

The mass is half of the sphere and r is greater than it was for the sphere.

A hemisphere is not a sphere. It does not gravitate as if its mass were concentrated at its center of mass. Is that not the point of this thread -- to figure out when an object will gravitate as if its mass were concentrated at its center of mass.

If the shell theorem holds then each hemisphere (when split along a vertical plane) must gravitate more strongly than an equivalent mass located at the hemisphere's center of mass. When the second hemisphere is added, the "horizontal" components of the pair cancel and the "vertical" components add.

 

[Herman Trivilino]

For a person standing on the surface however, there is a significant portion of the mass of the Earth that is not directly below the person.

Legend has it that Newton put away his work on gravity because of this very issue. He eventually got back to it after inventing the mathematics he needed to prove what has now become known as the shell theorem. A spherically symmetric distribution of mass acts as though all of it were concentrated at its center.

 

[Dirkg]

A hemisphere is not a sphere. It does not gravitate as if its mass were concentrated at its center of mass. Is that not the point of this thread -- to figure out when an object will gravitate as if its mass were concentrated at its center of mass.

If the shell theorem holds then each hemisphere (when split along a vertical plane) must gravitate more strongly than an equivalent mass located at the hemisphere's center of mass. When the second hemisphere is added, the "horizontal" components of the pair cancel and the "vertical" components add.

YES! thank you, that is what is bothering me. It is not intuitive that the hemisphere would cause a force greater than 1/2 the force from the sphere. But it would have to if two hemispheres added together are going to add up to F=mg

 

[Dirkg]

Legend has it that Newton put away his work on gravity because of this very issue. He eventually got back to it after inventing the mathematics he needed to prove what has now become known as the shell theorem. A spherically symmetric distribution of mass acts as though all of it were concentrated at its center.

Well it is nice to know I'm in good company. It led to me thinking about hypothetical non-spherical planets (i.e. hemisphere) and started to make my brain hurt. I naturally assumed someone had already worked out all the math and I just couldn't find it on the internet. Turns out that it isn't something that can be easily explained, but it looks like the hemisphere would have a gravitational pull greater than 1/2 of the sphere.

 

[Herman Trivilino]

Well it is nice to know I'm in good company. It led to me thinking about hypothetical non-spherical planets (i.e. hemisphere) and started to make my brain hurt. I naturally assumed someone had already worked out all the math and I just couldn't find it on the internet.

I'm sure it's been worked out. It requires integral calculus. It's actually the same math that you'd use to find the electric field due to a distribution of charge, which can be found in calculus-based introductory physics books, or textbooks for the junior-level electromagnetism course. There's an oldie but goodie little book called Div, Grad, Curl and All That. You can find it on Amazon, it's sort of written for the layman although it is rigorous.

 

[jbriggs444]

YES! thank you, that is what is bothering me. It is not intuitive that the hemisphere would cause a force greater than 1/2 the force from the sphere. But it would have to if two hemispheres added together are going to add up to F=mg

Not a proof -- just something to hang an intuition on...

A sphere is wide. A hemisphere is narrow. It is intuitive to expect the gravity from a wide object (where the force comes more horizontally than vertically) to gravitate less strongly per unit mass than a narrow object (where the forces are all well aligned vertically).

 

[Dirkg]

Not a proof -- just something to hang an intuition on...

A sphere is wide. A hemisphere is narrow. It is intuitive to expect the gravity from a wide object (where the force comes more horizontally than vertically) to gravitate less strongly per unit mass than a narrow object (where the forces are all well aligned vertically).

That doesn't really help in understanding my conundrum. If that were the case, the mass of a sphere and a point mass would not have the came gravity. The point mass is perfectly aligned.

 

[jbriggs444]

That doesn't really help in understanding my conundrum. If that were the case, the mass of a sphere and a point mass would not have the came gravity. The point mass is perfectly aligned.

That argument was aimed at distinguishing between hemisphere and sphere.

To distinguish between sphere and point, you have to factor in that the part of the sphere that is twice as close gravitates more than twice as much as the part that is twice as far away. Inverse square and all that.

[Again, nothing formal. Just a token to keep the intuition happy]

 

[Dirkg]

Oh, right. So maybe it does make sense that the hemisphere pulls with more than half the force of the sphere, given the asymmetry of the hemisphere.

 

[PeroK]

That doesn't really help in understanding my conundrum. If that were the case, the mass of a sphere and a point mass would not have the came gravity. The point mass is perfectly aligned.

I'm not sure that there is an intuitive answer why the gravity of a sphere is,equivalent to the gravity of a point mass.

It's something that emerges from the power of the integral calculus.

The only way I know to understand it is to "do the maths".

That the gravity inside a hollow sphere is zero does, by contrast, have a neat visualisation.

 

[Dirkg]

Actually I can visualize the sphere as a point. My sticking point is more with the gravitational pull from a hemisphere being more than half the pull of the sphere.

 

[PeroK]

Actually I can visualize the sphere as a point. My sticking point is more with the gravitational pull from a hemisphere being more than half the pull of the sphere.

If you are talking about "front" and "back" hemispheres then the front one is closer so contributes more than half the pull!

 

[jbriggs444]

If you are talking about "front" and "back" hemispheres then the front one is closer so contributes more than half the pull!

Which one can easily see. A sphere half the size (1/8 the mass) and twice as close (4 times the pull per unit mass) will fit within the "front" hemisphere. That's half the total pull accounted for and we've not counted the entire front hemisphere.

 

[Dirkg]

If you are talking about "front" and "back" hemispheres then the front one is closer so contributes more than half the pull!

No, Standing at the north pole and slice the Earth along a meridian. Since the pull from the two halves together equals the pull from the whole sphere, The pull from each hemisphere must be larger than half, because the direction of the pull from a hemisphere is at angle. The vertical component of the force would be 1/2, but there would be a horizontal component as well, Therefore the total force from an hemisphere must be more than half of the magnitude of the force from a sphere.

 

[PeroK]

No, Standing at the north pole and slice the Earth along a meridian. Since the pull from the two halves together equals the pull from the whole sphere, The pull from each hemisphere must be larger than half, because the direction of the pull from a hemisphere is at angle. The vertical component of the force would be 1/2, but there would be a horizontal component as well, Therefore the total force from an hemisphere must be more than half of the magnitude of the force from a sphere.
View attachment 216704

The pull from the hemisphere has components in two directions - the pull in the third direction cancels out by symmetry.

The "vertical" pull is half the pull of a sphere. If you add another hemisphere, then the vertical pull doubles, but the horizontal pull cancels out.

 

[Dirkg]

Yes, That is what I said. My point is that the magnitude of the force from a hemisphere (by itself) must be more than half of the magnitude of the force of gravity from the whole sphere. It seems counter-intuitive, but the vector addition is pretty straight forward. That was the whole point of this thread.
It would appear from F=GMm/r^2 that a hemisphere would have less gravitational pull because the mass is half ,and (based on my visualization) r is a longer distance than the radius of the sphere. I assume that somehow, summing up the force from each point in the hemisphere, results in F actually being more. What I have been able to grasp so for is that F=GMm/r^2 is for the simplified case of a sphere. And that something as asymmetric as a hemisphere would require either a different equation or r is measured to a point that is closer to the object (for some reason).

 

[PeroK]

Yes, That is what I said. My point is that the magnitude of the force from a hemisphere (by itself) must be more than half of the magnitude of the force of gravity from the whole sphere. It seems counter-intuitive, but the vector addition is pretty straight forward. That was the whole point of this thread.
It would appear from F=GMm/r^2 that a hemisphere would have less gravitational pull because the mass is half ,and (based on my visualization) r is a longer distance than the radius of the sphere. I assume that somehow, summing up the force from each point in the hemisphere, results in F actually being more. What I have been able to grasp so for is that F=GMm/r^2 is for the simplified case of a sphere. And that something as asymmetric as a hemisphere would require either a different equation or r is measured to a point that is closer to the object (for some reason).

You are making this too complicated. Just think of two equal point masses and, say, you are equidistant from them. The magnitude of the total force could be anything from 0 (if you are directly between them) to just less than the sum of the two individual magnitudes ( if you are far away).

If you take a sphere as a set of point particles, then much of the force from each particle (if you are standing on the surface) is canceled through symmetry.

The total magnitude is much less than the sum of the magnitude of each particle.

In fact, in proving the shell theorem the maths is simplified greatly by ignoring the symmetric cancelling forces.

 

[Dirkg]

You are making this too complicated. Just think of two equal point masses and, say, you are equidistant from them. The magnitude of the total force could be anything from 0 (if you are directly between them) to just less than the sum of the two individual magnitudes ( if you are far away).

If you take a sphere as a set of point particles, then much of the force from each particle (if you are standing on the surface) is canceled through symmetry.

The total magnitude is much less than the sum of the magnitude of each particle.

In fact, in proving the shell theorem the maths is simplified greatly by ignoring the symmetric cancelling forces.

Actually, that makes sense to me, and I can now visualize how it works for a sphere because of this thread. Maybe I should start a new thread to bring up the paradox I think I see with respect to a hemisphere. Namely, that the force of gravity from a hemisphere must be more than half the force of gravity from the whole sphere.

 

[PeroK]

Actually, that makes sense to me, and I can now visualize how it works for a sphere because of this thread. Maybe I should start a new thread to bring up the paradox I think I see with respect to a hemisphere. Namely, that the force of gravity from a hemisphere must be more than half the force of gravity from the whole sphere.

It can't be less than half! Because then if you put two hemispheres together...

It can't be equal, because some of it is lateral.

It must be more.

Because gravity reduces as $1/r^2$ you can't appeal to centre of mass arguments, which are based on $r$.

The shell theorem, in that respect, is a mathematical oddity.

 

[Dirkg]

Actually, that makes sense to me, and I can now visualize how it works for a sphere because of this thread. Maybe I should start a new thread to bring up the paradox I think I see with respect to a hemisphere. Namely, that the force of gravity from a hemisphere must be more than half the force of gravity from the whole sphere.

Yes, and in #15, I realized that is a reasonable explanation. It also seems like a paradox though, just because I try to visualize gravity with an equation in the form of GMm/r^2 and the only way for the magnitude of the force from a hemisphere to be more than half is if r gets shorter. That is obviously not what is happening. If I could still do the calculus, I would figure out the gravitation equation for a hemisphere and would be satisfied.

 

[Dirkg]

Thank everyone. I am realizing that if I want to understand this better I need to brush up on the calculus that I forgot 30 years ago. Nobody else has bothered to investigate this situation close enough to show me a nice diagram that clears it all up; probably because it is such a special case (and physically impossible).

 

[jbriggs444]

the only way for the magnitude of the force from a hemisphere to be more than half is if r gets shorter

See #12. The other explanation is that narrow shapes "waste" less gravity on horizontal forces that cancel out.

 

[Herman Trivilino]

It also seems like a paradox though, just because I try to visualize gravity with an equation in the form of GMm/r^2 and the only way for the magnitude of the force from a hemisphere to be more than half is if r gets shorter.

It's not a paradox when you consider that $G\frac{Mm}{r^2}$ is valid only for point masses. Making $r$ smaller therefore means only that you are moving two point masses closer together. Nothing more!

You don't really need calculus to begin getting a feel for this. In fact, skipping over what I'm about to recommend leaves many students stupefied by the calculus treatment.

Start with just two particles of equal mass such that the sum of their masses is $M$ and find the force they exert on a particle of mass $m$. Then add a third so that their sum is $M$ and so on. Do this for a collection of point masses distributed uniformly throughout your hemisphere. A daunting task, I know, but as you go through the process you build an intuition for what the calculus is doing for you. It in effect finds the force as the number of particles $N$ approaches infinity, with each particle having a mass $\frac{M}{N}$.

 

[jerromyjon]

Nobody else has bothered to investigate this situation close enough to show me a nice diagram that clears it all up

I know exactly what you mean and you are correct. The total "gravity" from two individually calculated bisected hemispheres is much more than the entire sphere calculated as a whole. Brilliant. I don't know how to exactly describe the geometric contributions but the shell theorem just provides a tidy bookkeeping system to give the net attraction. You are referring to the gross attraction.