Basic physics question

June 10, 2024 10:54 AM

I’ve managed to confuse myself (it’s not hard) about objects supposedly accelerating at the same rate when you drop them. More inside.

First of all, assume a spherical cow. Air resistance, etc. can be omitted.

So, I do three experiments of dropping items from 30 ft up, stationary relative to the Earth’s frame of reference.

First I drop a ball bearing, then a bowling ball, and both take the same time to hit the ground as measured by the most accurate stopwatch I have. Great, acceleration appears to be constant.

So for my next trick I drop a neutron star. (Or maybe just Mars)

Am I going to get the same result, or is the mutual attraction of the masses going to make a noticeable difference in how quickly (from the Earth’s frame of reference) the star hits the ground?

And if it does, then arguing backwards would it not be fair to say that the ball bearing and the bowling ball accelerated at different rates, but I couldn’t measure them?

posted by Tell Me No Lies to Science & Nature (14 answers total)

Am I going to get the same result, or is the mutual attraction of the masses going to make a noticeable difference in how quickly (from the Earth’s frame of reference) the star hits the ground?

It will be different. In the first two examples, the mass of the ball bearing, etc, are negligible when compared to the mass of the earth. I suppose theoretically the bowling ball does fall slightly faster than the ball bearing, because it attracts the Earth back more, but it's only relevant if you're calculating/timing things to 24 decimal places of precision.

The neutron star on the other hand, is about a million times more massive than the Earth, so it's not negligible for the calculations, and the gravitational attraction between the two bodies at 30 ft distance will be noticeably more than 10 m/s², assuming anyone's still alive to notice it. It's probably fine since in this exercise, humans are points with no mass.
posted by aubilenon at 11:17 AM on June 10 [6 favorites]

Different masses on Earth experience (almost) the same acceleration because their mass is negligible w/r/t the Earth.

The sun has a gravitational acceleration of 274.1m/s^2 at the photosphere, so if you dropped the Earth into the Sun, that's approximately how much acceleration it would experience.
posted by BungaDunga at 11:19 AM on June 10

Conveniently, a neutron star is about the same mass as the sun.

I want to add in the neutron star example, the mass of the Earth is the rounding error, and to the first several digits of calculation or measurement, it will fall towards the neutron star the same speed as the ball bearing or bowling ball. There are some practical difficulties with doing this experiment as it is classically depited though, as the increased gravity on the neutron star would make it difficult to construct a Leaning Tower of Pisa that wouldn't just fall over.
posted by aubilenon at 11:24 AM on June 10 [3 favorites]

Another factor in theoretical calculations would be that such experiments are thought to be done in a vacuum
posted by falsedmitri at 11:26 AM on June 10

Best answer: Let's look at the formulas!

F = G (m₁ * m₂) / r²

F is the force exerted on each object. G is a constant. m₁ and m₂ are the two masses. r is the distance between their centers of masses.

So the force is different in all of these examples (ball bearing, bowling ball, neutron star).

But then we have F=ma, which we can turn into a = F/m to get a, the acceleration of any object in this system.

If we find the acceleration of any of the three objects (ball bearing, bowling ball, or neutron star), it all comes out the same:

a_object = F / m_object = G (m_earth * m_object) / (r² * m_object) = G (m_earth) / r²

So the acceleration of the object is the same in all cases!

BUT: what about the acceleration of the Earth?

a_earth = F / m_earth = G (m_earth * m_object) / (r² * m_earth) = G (m_object) / r²

The Earth is being accelerated toward the object, too, and its acceleration will depend on the mass of the object. For the ball bearing and the bowling ball, it's basically zero. They just don't have enough mass to matter compared to how much they are accelerating themselves. But the neutron star mass does matter here, and so the Earth will accelerate a lot, relatively.

So I think your thinking is correct. Technically, the bowling ball pulls the Earth towards it a tiny tiny tiny bit faster than the ball bearing does, and so they don't actually collide at the exact same time. But the neutron star pulls the Earth towards it a lot, and so they reach each other a loooot faster.

[gosh I hope I got that right...]
posted by whatnotever at 11:30 AM on June 10 [12 favorites]

Another factor in theoretical calculations would be that such experiments are thought to be done in a vacuum

And using a spherical cow of uniform density.
posted by Stoneshop at 12:22 PM on June 10 [2 favorites]

Response by poster: Thank you everyone. Last night I had one of those "I've been lied to all my life" realizations and it's nice to see it confirmed.
posted by Tell Me No Lies at 3:41 PM on June 10 [2 favorites]

Just to add on to whatnotever's excellent answer, the universal gravitational constant G specifically relates to the gravity between 2 bodies incororpotating both their individual/combined mass as well as the distance. This relevant bit from Wikipedia article sums it up nicely :

the magnitude of the attractive force (F) between two bodies each with a spherically symmetric density distribution is directly proportional to the product of their masses, m1 and m2,

As a side note it's one of the many places the joke about a spherical cow actual derives.
posted by chasles at 3:55 PM on June 10

To take whatnotever's calculations one step further, here is the amount of acceleration the earth will experience towards the bowling ball vs towards to the tennis ball:

* Bowling ball acceleration (assuming 7.5 kg bowling ball): 1.24*10^-23 m/s^2
* Tennis ball acceleration (assuming 56 g tennis ball): 9.25*10^-26 m/s^2

So that is quite a large difference: 10^-23 vs 10^-26. That's three whole orders of magnitude!

But those numbers are both sooooo tiny. How are are you going to even measure an acceleration that small?

It looks like it will take this many years for the earth to accelerate to the awesome speed of one meter per hour under these two accelerations:

* Bowling ball: 3.7 * 10¹² years
* Tennis ball: 9.5 * 10¹³ years

By way of comparison, the age of the universe is around 14 * 10⁹ years.

It's hard to imagine how you even could measure an acceleration difference so tiny. So it's hard to say that people truly lied to you about this. There is a difference, but it is so amazingly small that it probably can't even be measured at all.

Flip side is, you're definitely thinking like a scientist here: X makes perfect sense. Now what if I change the conditions around X to the absolute uttermost extreme imaginable? Well now, it doesn't seem to make sense? How am I going to reconcile this?

Now you have to figure out what is going on that makes the difference.

That's exactly how interesting discoveries are made. Or in a more mundane setting - leaps of understanding.
posted by flug at 8:05 PM on June 10 [1 favorite]

Response by poster: It's hard to imagine how you even could measure an acceleration difference so tiny.

Dunno. The differences are gigantic compared to Planck lengths.

So it's hard to say that people truly lied to you about this.

The primary culprit is a dual nuclear physics/mechanical engineering PhD who has taken me to task for oversimplification for literally my entire life. He will be chastised appropriately tomorrow evening on our weekly video call. :-)
posted by Tell Me No Lies at 8:22 PM on June 10

Well, if we're concerned about oversimplifying, I think there's at least one more point to consider. The formulas and points I made above assumed that r (the distance) was the same in all cases. However, that is the distance between the centers of the two masses. If you have a neutron star or Mars, its center will have to be farther away from the Earth's surface than 30 ft, unless you also collapse it to be the size of a bowling ball or so...

If, on the other hand, you wanted to put the surface of Mars or the neutron star (what does the surface of a neutron star look like?) at about the same height from the surface of the Earth as the bowling ball or ball bearing, then we have a different r, and things change. Now, the acceleration of Mars (let's just stick with Mars for now) is actually less than that of the other two objects! Does the increased acceleration of Earth make up for that loss in this case? I think so. Am I doing the math? No. No I am not.

On the other hand, it looks like neutron stars are only about ten miles across, so while the star's center will have to be farther away than the ball bearing's, it's not that much farther, all things considered. And as for its surface, apparently we think it's suuuuper smooth. So just a really big ball bearing! Of doom.
posted by whatnotever at 10:11 PM on June 10 [1 favorite]

If you wanted to 'drop' a massive object more comparable in size to a ball bearing than a neutron star is, you could use a black hole with the Earth's mass, which would have a diameter of about 2 cm.
posted by jamjam at 10:34 PM on June 10

However, this black hole you're holding 9m above the surface of the earth would only apply the familiar gravitational acceleration of 10m/s^2 at the centre of the earth and locations at the same distance (60 degrees away along a great circle). At the point antipodal to you people would feel gravity get stronger by a mere 25%, until they got closer.
On the other hand, someone 2m tall watching your experiment from 9m below you would (momentarily) feel be subject to a pull on their head approximately 1.6 times stronger than the pull on their feet, and (forgive me for any errors introduced by ignoring relativity here) about 500,000 times weaker than the pull on your fingertips holding the event horizon at a distance of 1 cm from the singularity.
If you drop the black hole, it'll still have an initial acceleration of 10 m/s^2 towards the earth's centre - but the earth's gravitational pull will get more difficult to calculate the more it... well.

All of which is to say that inverse square laws are not to be taken lightly, and gravity is only neat and linear when the dominant gravitational sources are at an approximately constant distance.
posted by polytope subirb enby-of-piano-dice at 11:42 AM on June 11

Two points, one more about the physics, one more about the philosophy of science, and models in general.

Firstly, the gravitational acceleration of a point mass in a given gravitational that is not experiencing outside force is absolutely independent of mass.

However, you can't physically definite what acceleration means without the concept of reference frames. A reference frame is essentially your coordinate system for measuring position and velocity; deciding exactly where and what you have chosen to be 0 position and 0 velocity when doing your math.

Importantly. the laws of classical physics only apply when you are measuring from an inertial reference frame. So if you want to apply those laws, it is very important to choose an inertial reference frame - one which measured acceleration except by outside forces will be. In this case, we'll use classical mechanics and treat gravity as an outside force.*

For a very obvious example of why a reference frame needs to be inertial, say you are on a hike and were sitting down, and then start walking at one meter per second. If you try to use yourself as reference frame for measurement, the mountain had zero kinetic energy when you were sitting down, and now suddenly has a very large amount of kinetic energy now that you are walking because all that mass is now moving at 1 m/s relative to you. Clearly physics is wrong and energy is not conserved because you've suddenly created all this kinetic energy that wasn't there before!

Well no, the coordinate system centered on you is not an inertial reference frame. All you did was change the coordinate system in a way that has the 0 reference for total momentum and total energy at a different point. No energy actually went anywhere besides to accelerate your body.

So if you aren't using an inertial reference frames, you're going to measure all sorts of crazy changes in the universe that are actually changes in you.

Typically for problems involving gravity, the easiest inertial reference frame to use is the center of mass reference frame. It is basically the average position of the mass in a space. From this reference frame in a system with the Earth and another object, a point mass will be accelerated relative to the center of mass of the system at, well, approximately 9.8 m/s. The Earth is not in fact round but ultimately just Earth shaped, with a non-uniform distribution of mass, the actual gravity near the surface varies quite a bit from place to place so you don't really get much more accuracy than two decimal places there without defining where you are standing. Regardless of what the gravitational acceleration is at a certain point, a mass at that point will be experiencing a force that will accelerate it at that rate.

The fact that Earth will also be accelerated toward the object proportional to the objects mass does not change this. In all inertial reference frames, the acceleration of the object will only depend on the Earth's mass not its own. Conversely the acceleration of the Earth will depend on the mass of the other object.

The trick is, the larger the second object, the more the reference frame of the surface of the Earth is not an inertial reference frame, and the bigger difference there is between the acceleration of an object and the time it takes to fall.

So at some point, it stops being safe to assume "experiences the same acceleration" and "falls in the same amount time" are the same thing.

Which leads the second point. Scientific theories are just models. Models are not real. Or, at any rate, they are only real as far as they give useful predictions. And all of these predictions are necessarily approximate.

The universe is infinitely complicated and chaotic. If you try to model every single detail and every single force and demand infinite precision, then you end up finding that nothing is the same as anything else, most certainly not your model and anything real. So your model isn't valid anywhere.

And mind you the concept of equality itself is only a mental model. So if your definition of equality and sameness does not admit some approximation, it is not a valid model of reality.

I mean you are admitting some measure of approximation. You mention Planck length. But that is just a lower bound to how approximate any model using using the laws of physics as we understand them must be. It is where even accounting for all known physics, even the models themselves tell you they are wrong.

But, that doesn't mean that your particular model is right down to that scale. It almost certainly isn't. The level approximation at which two things are the same under a model depends on all the things you are sweeping under the rug and ignoring. Like air resistance, the local variation of the Earth's magnetic field, the fact that the Earth, solar system, and Milky Way are all rotating, etc. And that's even before getting into measurement errors!

So...

All physics is only a model.
All models are approximate in their definition of equality/difference
The resolution of your model is on the scale of the biggest thing that you "Spherical cow'd" out of your model.

Therefore claiming something is "really" different based on two numbers being different by an amount smaller than the resolution of your model is meaningless.

So why are you insistent that the Earth accelerating towards an object is important, even when that acceleration is meaninglessly small?

And I don't mean that rhetorically, it is a very useful question.

I think some of the friction here is that you actually have two models going on: You have the abstract model of physical theory, which is accurate on its own terms, with inertial reference frames and acceleration and a given gravitational field and perfectly rigid, spherically uniform objects, where the object you are varying the mass of has uniform acceleration.

And then you have the "thought experiment" model, where you are imagining the spherical object as the Earth, the object of varying mass as a ball bearing, a bowling ball, or a neutron star. But instead of an inertial reference frame you have a stopwatch stationary to the surface of the Earth and instead of knowing the acceleration, you are measuring the time to fall.
.
That second model has so much more inherent messiness because of all the things you aren't accounting for. The error bars on your measured fall time prediction are going to be huge. Or at least many orders of magnitude larger than the effect of the gravitational pull of a ball bearing or bowling ball. But not a neutron star.

So for the ball bearing and the bowling ball, within the accuracy of your thought experiment, the surface of the Earth is the same as an inertial reference frame and the measured time to fall is the same as acceleration. But that's not true for something as large as a neutron star. So the ultimate problem is that the thought experiment is only valid for modelling the physical concept of acceleration for very small falling objects, whose gravitational pull on the Earth is much less than the forces of all the other things you are ignoring (the tidal pull of the moon, the pull of the planets, the push of the solar wind, the photon pressure of sunlight, etc)

The universe is big and complicated. Physics is about simplifying it as much as possible. Figuring out what ranges of parameters your simplified results are useful and valid in is the real trick.

*In the framework of general relativity, gravity is regarded as a curvature of spacetime, so a reference frame experiencing gravity is not inertial, but as you might expect, constructing a four dimensional coordinate system that leaves two objects towards each other as not changing in velocity during the process until they collide is a bit complicated and not very useful for this case.
posted by Zalzidrax at 7:39 PM on June 11