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Turn on, tune in, drop out. Falling objects question.

disappoint

disappoint

Lifer
Speaking of dropping out, let's drop some things.

You have two objects. An iron ball and a volleyball. They are the same diameter and the surface area is the same. This is not taking place in a vacuum but it may as well be because the air resistance will be equal on both.

You drop them from a long height, say the leaning tower of Pisa. So yeah this is that famous drop experiment you've probably heard about.

Question: Do the 2 balls reach the ground at the same time? If so why?

Isn't there more gravitational pull on the heavier ball? It weighs more for sure.

Ok so a little disclosure here. I do know the answer so this is not a homework problem. I'd just like to discuss your answers. I will also answer any questions you have about it.

What I don't know is if you know the answer. That's where you come in. You can post anything that comes to mind and have a fun time discussing it.

I don't believe this belongs in highly technical because the question is quite simple as is the answer. It's basic classical physics and yet some know it and some do not. Also the discussion may branch into other aspects, such as why this isn't common knowledge or even if it should be. Also it won't get many views there as HT is seldom visited but that's another topic.

Ok have fun, stay awesome.:awe:
 
We did this experiment in grade school.

They will reach the ground at the same time. It's very easy to test.
 
AViking said:
We did this experiment in grade school.

They will reach the ground at the same time. It's very easy to test.
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You can do the experiment and see if they hit the ground at the same time or not, but the more important question is why?

Why would they hit the ground at the same time? One is heavier. Does it not have more force from gravity? It would push the scale down farther if you put it on a scale right? It pushes on your hand more when you hold it as compared to the lighter ball. Why would the acceleration due to gravity be the same for a heavier ball?
 
Classical physics tells us why. Weight has nothing to do with it. Mass has nothing to do with it. If you do the math the variable for mass cancels out. I wouldn't expect most people to know that since I don't think most people even take a basic physics class in high school. I would expect everyone to know that two objects land at the same time.

The reason people don't know this is because they either had a poor education or they're the type of person who doesn't care. I divide the world up into two types of people. Those who ask questions and are curious about things and those who aren't. We all know the difference now that we have smartphones with us all the time. The curious ones are the ones who have to look everything up.
 
AViking said:
Classical physics tells us why. Weight has nothing to do with it. Mass has nothing to do with it. If you do the math the variable for mass cancels out. I wouldn't expect most people to know that since I don't think most people even take a basic physics class in high school. I would expect everyone to know that two objects land at the same time.

The reason people don't know this is because they either had a poor education or they're the type of person who doesn't care. I divide the world up into two types of people. Those who ask questions and are curious about things and those who aren't. We all know the difference now that we have smartphones with us all the time. The curious ones are the ones who have to look everything up.
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Not sure which equation you are referring to when you say the mass cancels out. F=GmM/r^2?

I agree with you on curious people vs. people who simply don't care, and that is a good insight I thank you for sharing. Admittedly I hadn't thought about the curiosity aspect until now. Given the easy access of information the first world problem changes from "I had no resource to learn this" to "I don't care" or some other excuse.
 
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That's the force. You're not calculating force though. You are calculating velocity. You know that the height and acceleration of gravity is equal. So the variable is velocity. If one of them goes faster than the other they will reach the ground first. Mass is not constant but when you do the math it becomes irrelevant.
 
AViking said:
That's the force. You're not calculating force though. You are calculating velocity. You know that the height and acceleration of gravity is equal. So the variable is velocity. If one of them goes faster than the other they will reach the ground first. Mass is not constant but when you do the math it becomes irrelevant.
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You're going to make me guess which equation(s) you are referring to? Because this is a drop experiment the initial condition is 0 velocity. If the acceleration is the same as well as distance then the end velocity would turn out the same would it not? You didn't need any equations but I'm curious which you are referring to and how you arrived at your answer. I'm also learning here so thanks for that.
 
AViking said:
That's the force. You're not calculating force though. You are calculating velocity. You know that the height and acceleration of gravity is equal. So the variable is velocity. If one of them goes faster than the other they will reach the ground first. Mass is not constant but when you do the math it becomes irrelevant.
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I don't understand "mass is not constant"
 
Well the mass of the objects is irrelevant. Try it at home to convince yourself but grab a tennis ball and a ping bong ball. No need to break your floors with anything heavier. Drop them and they'll hit the ground at the same time. This should tell you that the mass isn't affecting the outcome.

Next up really is what do you know? You know the height. They are going to be the same for both balls. You know the rate of acceleration. That's gravity. You have the problem setup to ignore friction such as wind resistance. The variable then that you're trying to solve for to prove whether they land at the same time would be velocity. You have different masses so you could plug those in but they will cancel.

Basically the kinetic energy at the bottom will be equal to the potential energy at the top.

So use the kinetic energy equation: E=0.5mv^2
m is mass and v is velocity

Potential Energy: E=mgh
m is mass, g is gravitational constant, and h is height

mgh=0.5mv^2
cancel out m
v=sqrt(2gh)

If the equations had somehow been different and mass had not cancelled out then mass would have been a factor and you would have gotten different answers for different masses.
 
really it's quite simple if you want to use the Force equation, and you are asking why do they both accelerate at the same rate if there is more force on the more massive object. It's because it's more massive thus it takes more force to move. If you were trying to move an iron ball and a volleyball the iron ball has more mass so it takes more force to move.
 
From what I understand, the bowling ball does fall faster, but the difference is not measurable. Also, the bowling ball and a feather dropped together will fall at the same rate in a vacuum, but if dropped separately, the speed is faster for the bowling ball due to its own gravitation pulling on the earth. 1 part in a trillion trillion is the amount the earth will move toward the ball. I know, because I googled.
If dropped together, the space time warpage effects both and is the same for both. Separating them produces their own individual space time bending.
 
Paul98 said:
really it's quite simple if you want to use the Force equation, and you are asking why do they both accelerate at the same rate if there is more force on the more massive object. It's because it's more massive thus it takes more force to move. If you were trying to move an iron ball and a volleyball the iron ball has more mass so it takes more force to move.
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Suppose one takes two spheres of 1000 meters diameter and places them 1000 km from the surface of the Earth. One is made of iron and the other neutronium. In this thought experiment the spheres do not feel the gravitational influence of the other.

Will they contact the surface of the earth at the same time?
 
Hayabusa Rider said:
Suppose one takes two spheres of 1000 meters diameter and places them 1000 km from the surface of the Earth. One is made of iron and the other neutronium. In this thought experiment the spheres do not feel the gravitational influence of the other.

Will they contact the surface of the earth at the same time?
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Something tells me that one of those spheres will pull on the earth so hard that it will rip it apart, contacting it much faster than the other sphere.
 
if we assume drag can be neglected, F=ma = mg and so the more massive object will experience a proportionally more massive force and so an identical acceleration and they will hit the ground at the same time.

If however we are high enough, the terminal velocity of a iron ball is faster (I think....) (v = sqrt((2mg)/(rho*A*C)) )and so even with equal drag coefficients I think the iron ball eventually goes faster. But I've never done the experiment from a plane.
 
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AViking said:
Well the mass of the objects is irrelevant. Try it at home to convince yourself but grab a tennis ball and a ping bong ball. No need to break your floors with anything heavier. Drop them and they'll hit the ground at the same time. This should tell you that the mass isn't affecting the outcome.

Next up really is what do you know? You know the height. They are going to be the same for both balls. You know the rate of acceleration. That's gravity. You have the problem setup to ignore friction such as wind resistance. The variable then that you're trying to solve for to prove whether they land at the same time would be velocity. You have different masses so you could plug those in but they will cancel.

Basically the kinetic energy at the bottom will be equal to the potential energy at the top.

So use the kinetic energy equation: E=0.5mv^2
m is mass and v is velocity

Potential Energy: E=mgh
m is mass, g is gravitational constant, and h is height

mgh=0.5mv^2
cancel out m
v=sqrt(2gh)

If the equations had somehow been different and mass had not cancelled out then mass would have been a factor and you would have gotten different answers for different masses.
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This.

To put it into words: At the same height two objects have given, respective gravitational potential energies (mgh). As they fall, this potential energy turns into kinetic energy((1/2)mv^2).

Now the total amount of energy present is dependent on the mass (which can be calculated using PE = mgh and/or KE = (1/2)mv^2) But the conversion rate of the energy (from PE to KE) for the two objects is the same relative to their masses.

A good (if not entirely accurate) analogy would be an ant lifting a small twig and a human lifting a telephone pole (if this were possible). Both are lifting something ~7x heavier than they are, and they can lift at the same velocity. The human is exerting more energy than the ant, but their velocities remain the same.
 
To go into a tad more detail: at some velocity v1, there will come a time when the gravitational force F=m1g, where m1 is the mass of the volleyball, is equal to the drag force, -.5(pv^2AC). There is also a time when the gravitational force F=m2g, where m2 is the mass of the iron ball, is equal to the drag force -.5(pv^2AC) (we assume the drag force on each for the same velocity is equal, as stated in a slightly nuanced way in the OP). However v2 > v1 as m2 > m1 and so the iron ball hits the ground first if we are not in a vacuum and the height difference is substantial enough for this to come into play.
 
I should probably also add: if you had a small iron ball and a large iron ball I believe they will hit the ground at about the same time from any height as the difference in masses should be approximately offset by the difference in areas for terminal velocity.

edit: er that actually not right, the mass will not scale with area exactly. mass will scale with the volume of the sphere while the area for drag considerations scales with .5 * surface area, and surface area scales with r^2 while volume with r^3, so they don't scale exactly. whoops!
 
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Take an iron feather exactly the same size as a real feather and drop them. Will they hit the ground at the same time, fall an equal distance at the same acceleration? Of course not, because while the air resistance on each will be the same, the mass affected by air resistance is unequal. This is the same reason that all the atmosphere of the earth isn't pooled on the ground. The Brownian motion of the atmosphere imparts a greater force than gravity. Build a light enough feather and it won't fall at all.
 
Ok, if someone could clarify the following...

Lets take air resistance out for starters. I am interested in potential acceleration due to gravitational attraction of two bodies. Stuff falls the same on earth because the earth is doing the pulling. Everything else, compared to the earth, is so small its all basically the same.
In the vacuum of space, would two ping pong balls attract to each other slower than two neutron stars? Do denser objects have the potential to attract faster than less dense objects? I think a bowling ball might fall faster toward a black hole than it does toward the earth, right?
So, do two objects fall the same speed on earth simply due to the relative mass of those objects compared to the earth?
 
^ On earth we assume the acceleration due to gravity to be 9.8 m/s^2, or 'g' in the equation F=mg. Really the actual equation is F = G(m1m2)/r^2 (where capital G here is not the lowercase g above, but rather a different gravitational constant), so the 2 neutron stars would attract a lot faster than the two ping pong balls, even from much further apart; it just so happens that on earth, if m1 is earth, and you are somewhere on the earth's surface (or near it), Gm1/r^2 are all practically constant and in total are equal to 'g'. r is the distance from the center of the earth in this example, not the surface, which is crucial.

You can also see here that the higher you are the less gravitational force will be acting on you. That said, falling from 30 feet will still hurt a lot more than falling from 1 foot 😉

edit: In theory the above equation has no relation to density, ie, 2 planets of the same mass of various densities will have the same gravitational force so long as you are completely outside of their boundary. The black hole however likely has a lot more mass than....pretty much anything and so it will allow for more gravitational force.

edit2: two objects fall at the same speed because velocity v is defined as v = v0 + at, where v0 is initial velocity (let's call it zero for simplicity) and a is acceleration (g in this case) and t is time. As long as the acceleration of two objects is equal they will accelerate to the same speed. Acceleration is defined from F=ma, where F is force, m mass, a acceleration, so a = F/m. We know from the larger gravity equations that gravitational acceleration on earth is ~g, so g = F/m, ie, for an object double the mass of another, it will experience double the force of gravity on it from earth. It should be noted that the earth will also experience double the gravitational force acting on it from that object but that force is so small compared to its mass that F/m is basically zero for all practical intents and purposes. But anyway. If a = g, then v = 0 = at = gt for any object (ignoring drag, of course).

hopefully this rambling is clear 😉
 
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Screech said:
^ On earth we assume the acceleration due to gravity to be 9.8 m/s^2, or 'g' in the equation F=mg. Really the actual equation is F = G(m1m2)/r^2 (where capital G here is not the lowercase g above, but rather a different gravitational constant), so the 2 neutron stars would attract a lot faster than the two ping pong balls, even from much further apart; it just so happens that on earth, if m1 is earth, and you are somewhere on the earth's surface (or near it), Gm1/r^2 are all practically constant and in total are equal to 'g'. r is the distance from the center of the earth in this example, not the surface, which is crucial.

You can also see here that the higher you are the less gravitational force will be acting on you. That said, falling from 30 feet will still hurt a lot more than falling from 1 foot 😉

edit: In theory the above equation has no relation to density, ie, 2 planets of the same mass of various densities will have the same gravitational force so long as you are completely outside of their boundary. The black hole however likely has a lot more mass than....pretty much anything and so it will allow for more gravitational force.

edit2: two objects fall at the same speed because velocity v is defined as v = v0 + at, where v0 is initial velocity (let's call it zero for simplicity) and a is acceleration (g in this case) and t is time. As long as the acceleration of two objects is equal they will accelerate to the same speed. Acceleration is defined from F=ma, where F is force, m mass, a acceleration, so a = F/m. We know from the larger gravity equations that gravitational acceleration on earth is ~g, so g = F/m, ie, for an object double the mass of another, it will experience double the force of gravity on it from earth. It should be noted that the earth will also experience double the gravitational force acting on it from that object but that force is so small compared to its mass that F/m is basically zero for all practical intents and purposes. But anyway. If a = g, then v = 0 = at = gt for any object (ignoring drag, of course).

hopefully this rambling is clear 😉
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So, at least in theory, a bowling ball does fall faster than a lighter object on earth. Its just so small of a difference that it only applies in theory? Perhaps 1 part in a trillion trillion faster due to the bowling ball's own gravitational influence on the earth?
If two dense objects attract faster than two less dense objects, then a bowling ball and the earth are two objects, both being denser than a feather. So the bowling ball should go faster (in theory)?
 
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