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

[Screech]

ah, well this will depend on your frame of reference. If we are assuming, again, no drag: if your frame of reference is from the center of the earth (ie, we treat the center of the earth as not being moving, ever), then the bowling ball will get closer to it faster than the lighter object because it will cause the earth to get closer to it at the same time. But that difference is so small it isn't worth discussing 😉 Also, that frame of reference is not really a correct one; we usually want to choose one in which no acceleration is taking place and your question inherently requires that the earth be accelerating (slowly....). So from a frame of reference centered on, say, the sun, the two objects accelerate toward earth at the same rate; one simply gets there first due to the fact that the earth would move towards it slowly. But again, in reality, it would not be possible to ever measure this difference.

Density is irrelevant here; two objects of heavier mass will create a larger force, but a force that is twice as large on an object of twice as much mass will create the same acceleration as a = F/m, so the bowling ball goes the same speed as the lighter object (again, until drag is considered. Which is the bane of high school physics 😉 )

 

[disappoint]

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?

Yes 2 neutron stars would have a greater force of gravity between them than 2 ping pong balls because they have greater mass, all else being equal (distance between them).

^ 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)...

You're right and I corrected my mistake above. I meant to use uppercase G for the Gravitational constant, rather than lowercase g which denotes the acceleration due to gravity on Earth.

 

[moonbogg]

ah, well this will depend on your frame of reference. If we are assuming, again, no drag: if your frame of reference is from the center of the earth (ie, we treat the center of the earth as not being moving, ever), then the bowling ball will get closer to it faster than the lighter object because it will cause the earth to get closer to it at the same time. But that difference is so small it isn't worth discussing 😉 Also, that frame of reference is not really a correct one; we usually want to choose one in which no acceleration is taking place and your question inherently requires that the earth be accelerating (slowly....). So from a frame of reference centered on, say, the sun, the two objects accelerate toward earth at the same rate; one simply gets there first due to the fact that the earth would move towards it slowly. But again, in reality, it would not be possible to ever measure this difference.

Density is irrelevant here; two objects of heavier mass will create a larger force, but a force that is twice as large on an object of twice as much mass will create the same acceleration as a = F/m, so the bowling ball goes the same speed as the lighter object (again, until drag is considered. Which is the bane of high school physics 😉 )

Ah. So you can have stronger gravity, but not faster gravity? If you held a bowling ball 10 feet off the surface of a neutron star and then let it go, would it fall faster than it would fall on earth? Acceleration should be the same, right?

 

[Screech]

probably not -- as both the mass of the large object (earth or neutron star) as well as the distance from the center of the bowling ball to the center of the earth or star, would be different. i would assume the neutron star would have a higher mass and a radius such that the increase in mass overshadows any difference in radius of one vs the other but really don't know the numbers off the top of my head.

TO answer th earlier part; gravity generally describes the force acting on an object, not the speed of the resulting object from acceleration due to gravity. So the gravitational forces on heavier or lighter objects are all stronger or weaker but for the same spot on earth the resulting accelerations (and speeds) are all more or less the same.

....time to get back to work from lunch..... 😉

 

[Hayabusa Rider]

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.

So in my scenario the radii of the objects remains the same and so the acceleration of the two bodies, one being far more massive than the Earth does not cause contact sooner?

 

[moonbogg]

So in my scenario the radii of the objects remains the same and so the acceleration of the two bodies, one being far more massive than the Earth does not cause contact sooner?

Can't we just get the answer by watching people walk on the moon? Stronger gravity = faster acceleration.

Stuff on Jupiter falls at 26m/second squared at the surface. Earth's is 9.81. There you go! Answers my questions at least.

 

[Paul98]

So in my scenario the radii of the objects remains the same and so the acceleration of the two bodies, one being far more massive than the Earth does not cause contact sooner?

Look at the equations F = G*(m1*m2)/r^2

G is gravitational constant
m1 and m2 are masses of two objects
r is distance between the objects

then look at the F = ma formula

if you set them equal you can calculate acceleration. The acceleration of the more massive object is smaller than the less massive. The reason we don't both to calculate the acceleration of the earth when using things like a bowling ball or volleyball is that the earth's acceleration is basically 0.

you end up with
for acceleration of m2
a = G*(m1)/r^2
for acceleration of m1
a = G*(m2)/r^2

It would obviously come into contact sooner if the other object is more massive than the earth. Not only would the object be accelerated towards the earth but the earth would also be accelerating at a good amount towards the object.

Just as if you were to increase the mass of the earth by a lot the acceleration at the earths surface would also increase. Same if you were to shrink the size of the earth while keeping same mass.

Edit this is ignoring the effect of time dilation.

 

[moonbogg]

Look at the equations F = G*(m1*m2)/r^2

G is gravitational constant
m1 and m2 are masses of two objects
r is distance between the objects

then look at the F = ma formula

if you set them equal you can calculate acceleration. The acceleration of the more massive object is smaller than the less massive. The reason we don't both to calculate the acceleration of the earth when using things like a bowling ball or volleyball is that the earth's acceleration is basically 0.

you end up with
for acceleration of m2
a = G*(m1)/r^2
for acceleration of m1
a = G*(m2)/r^2

It would obviously come into contact sooner if the other object is more massive than the earth. Not only would the object be accelerated towards the earth but the earth would also be accelerating at a good amount towards the object.

Just as if you were to increase the mass of the earth by a lot the acceleration at the earths surface would also increase. Same if you were to shrink the size of the earth while keeping same mass.

Edit this is ignoring the effect of time dilation.

So the only way to make something on earth fall faster would be to have an object so dense, that it causes the earth to move toward it, correct? Otherwise, the only thing doing the pulling would be the earth, and it can only pull so fast no matter what its pulling on. Right?
So if you calculate the pull of a bowling ball on the earth, you would get a value. It wouldn't be worth considering, but there would still be a value. That value would be greater than the value produced by a feather, right?

 

[Paul98]

So the only way to make something on earth fall faster would be to have an object so dense, that it causes the earth to move toward it, correct? Otherwise, the only thing doing the pulling would be the earth, and it can only pull so fast no matter what its pulling on. Right?
So if you calculate the pull of a bowling ball on the earth, you would get a value. It wouldn't be worth considering, but there would still be a value. That value would be greater than the value produced by a feather, right?

correct

 

[moonbogg]

correct

So, bowling balls obviously don't fall faster than anything else (in any practical way), but at least now we have touched on a deeper understanding of why that is and that bowling balls DO produce a greater velocity value compared to less dense objects due to the earth moving toward them at a faster rate. And we did it without getting lost in all that painful math!

 

[Moonbeam]

The OP states the experiment will be taking place in earth's atmosphere and that changes everything compared to the moon. Air resistance is the same on the same sized objects but the effects of air resistance are affected by the mass the resistance is applied to.

 

[sm625]

To the naked eye it would appear that they hit the ground at the same time. But the iron ball would actually hit the ground first, because the volleyball is more porous. It allows a very small fraction of air to pass through it which changes its Fd with respect to the iron ball. You would have to use a very precise and fast camera to catch this differential from a drop distance of a few meters. But I believe you would see the iron ball hit first.

Also, if you release an iron ball and a volleyball at the same time, the material of the volleyball would probably affect its release time by a few milliseconds causing its descent be delayed. Static charge may also affect its release time. The magnetic field around the iron ball could affect its release time with respect to the volleyball. All of these and more could affect an experiment where you are trying to measure drop times to a high degree of accuracy.

 

[Moonbeam]

To the naked eye it would appear that they hit the ground at the same time. But the iron ball would actually hit the ground first, because the volleyball is more porous. It allows a very small fraction of air to pass through it which changes its Fd with respect to the iron ball. You would have to use a very precise and fast camera to catch this differential from a drop distance of a few meters. But I believe you would see the iron ball hit first.

Also, if you release an iron ball and a volleyball at the same time, the material of the volleyball would probably affect its release time by a few milliseconds causing its descent be delayed. Static charge may also affect its release time. The magnetic field around the iron ball could affect its release time with respect to the volleyball. All of these and more could affect an experiment where you are trying to measure drop times to a high degree of accuracy.

If the volley ball was porous enough to produce an effect that could be measured it wouldn't be a volley ball because it would deflate. But if you dropped a lead volley ball sized sphere and a volley ball from a plane at high altitude at the same time into a storm with extremely turbulent weather the lead ball would hit the ground first and in a very different place than the volley ball.

 

[Unitary]

I don't think that they will land at the same time as the masses don't cancel out when you include wind resistance.

True, when looking at the standard gravitational motion equation F=ma (without wind resistance) the masses do cancel out, it no longer does when you include the wind resistance term (proportional to velocity or velocity squared) as there is no mass term here.

Another way to think about it is what is the terminal velocity of a falling iron volleyball compared to a volleyball. Certainly, the iron one will be faster. Dropping the two off the tower of pisa doesn't allow enough time for the wind resistance to make a significant measurable change (but there is a change however small) in the position of the two.

 

[Paul98]

Yeah not much a discussion if you just were trying to see of these two objects will hit the ground at the same time in earth's atmosphere.

Seems like this was more of the question he really wanted to talk about

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?

The question being why would acceleration due to gravity be the same for the heavier ball...

 

[DrPizza]

I've seen a lot of incorrect answers above, but didn't read them all.

The OP said NOT to ignore air resistance. If you don't, then the bowling ball hits the ground first. This would be a lot easier to see if you had a smooth styrofoam ball rather than volleyball of identical size to compare it to - and simply thinking about a mental experiment from the top of a very tall building, using the identical sized bowling ball vs. smooth styrofoam ball should be sufficient to see that the bowling ball is going to win. But, if you only drop them a few inches, it would be incredibly difficult to discern the difference in time between the two.

Reason: drag force. They would experience the same drag force - (if/when they're moving at the same speed). Since acceleration and forces are vectors, we can break them up into components. For both objects, the force due to gravity, divided by the objects mass gives the exact same acceleration, roughly 9.81 meters per second squared. However, let's look at another component of acceleration - that due to the drag (air resistance) - while each object (for the same speed) is subjected to the same amount of drag, because F=ma, the object with the greater mass will have a lower component of acceleration (directed away from the ground), and the lighter object will have a greater component of acceleration. So, the net acceleration is the acceleration due to gravity, minus the acceleration due to drag force. The bowling ball wins. And, as the drag force increases (roughly) with the square of the velocity, it turns out that the bowling ball will have a higher terminal velocity. (Drop a balloon out a window.) This is why over a short distance, the velocity is not that great, thus the drag force is very small - and the difference in net accelerations is very close to zero. But, over greater distances, as the velocities increase, the drag forces increase, causing the lighter object to have a much lower net acceleration than the heavier object of the same size and shape.


Bowling ball wins.

 

[Paul98]

Yeah with air resistance it's just as simple if using F=ma, force of atmosphere is similar( not the same since it increases with velocity.), masses are different. Thus acceleration is different.

 

[CLite]

As the two previous posters point out the object's mass makes a difference even if all other parameters are considered identical.

OP could semi-accurately calculate time of arrival of different objects. The only annoying thing is solving for acceleration with the quadratic equation and then trying to integrate it twice. You might need to do numerical integration, the left-hand rule is painless in excel.

F = - Fdrag + Fgravity
Fgravity = G*mM/r^2
M = earth mass, m = object mass
Fdrag = Cd*rho*v^2*A/2
Cd = drag coefficient, rho = fluid (air) density, A = object cross-sectional area
a = F/m
a = - Cd*rho*v^2*A/(2m) + G*M/r^2
a = - Cd*rho*(at)^2*A/(2m) + G*M/r^2
Cd*rho*t^2*A/(2m) * a^2 + a – 32.2 ft/s^2 = 0
Integrating “a” at this point is going to be bit messy, but suffice it to say it’s a function of (m,t) – assuming Cd/rho/M/r/A are all constants.
Integrate the “v” and you get d = FUNC(m,t)
Reverse the equation and t = FUNC(m,d).

 

[KMc]

I feel like we're overcomplicating the explanation. The only needed equation here is F=ma. More specifically, rewritten as a=F/m.

While it's true the iron ball weighs more and therefore has a larger gravitational force (F), it accelerates more slowly because of it's larger mass (m). So, the direct affect of the greater force on the iron ball is offset by the inverse affect of its greater mass; and so both objects will accelerate at the same rate ~ 9.8 m/s^2.

Said another way, the ratio of any two free-falling objects' force and mass are constant.

 

[Mark R]

This.
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.

No.

The human is lifting the item several times further than the ant, and at a velocity several times greater (because the distance is greater).

I believe that I saw this as an interview question that was asked for programmer interviews at a technology company:

Imagine that you have been shrunk to 1/10 of your size in each dimension, and that your body density remains the same. You are standing in a blender, with a timer counting down before the blades start. What do you do?

The answer is you jump. The reason that is the case is as follows:
Let's assume that muscle can release x Joules of energy per gram of muscle mass (n).
Jumping converts muscle kinetic energy into gravitational potential energy, giving a peak height h = E/mg (where is m is mass, and g is gravitational acceleration)
You can work out the muscle constant x: x = mgh/n = rgh where r is lean body percentage.

You can rearrange this and see that h (jump height) = x/rg. In other words, jump height doesn't depend on body mass. So, the shrink process doesn't affect how high you can jump. Even though a shrunk body is 10x smaller, the jump height is unaffected.

 

[CLite]

I feel like we're overcomplicating the explanation. The only needed equation here is F=ma. More specifically, rewritten as a=F/m.

While it's true the iron ball weighs more and therefore has a larger gravitational force (F), it accelerates more slowly because of it's larger mass (m). So, the direct affect of the greater force on the iron ball is offset by the inverse affect of its greater mass; and so both objects will accelerate at the same rate ~ 9.8 m/s^2.

Said another way, the ratio of any two free-falling objects' force and mass are constant.

You did not read the thread very well. The smaller mass is not canceled out in drag force, as pointed out by all three posts preceding yours.

 

[Dr. Zaus]

The gravity from the iron ball will attract the volley ball rubber-banding the volley ball past the iron ball.

Like the Klingon Bird-of-Prey and the sun in Startrek IV

 

[KMc]

You did not read the thread very well. The smaller mass is not canceled out in drag force, as pointed out by all three posts preceding yours.

Ok, sorry, I read the original post too quickly. Take it easy.

So, we can't ignore aerodynamic drag. Alright, is it ok if for the purpose of demonstration that we just initially assume both objects have the same shape, volume, diameter and surface roughness? In this case, then we could assume that they both have the same aerodynamic drag coefficient, CD.

Then, CD=D/q where D is the aerodynamic drag of the object, q=1/2*rho*v^2 and rho is the density of the air.

Then, D= CD*1/2*rho*V^2

So, each object's net vertical force is the difference between its gravitational force and its aerodynamic force. Therefore, a=(F-D)/m. Tricky part here is that D is dependent on the current velocity of the object. so, at time 0, velocity is 0 and both objects begin accelerating downward at 9.8 m/s^2. But as the velocity of the 2 objects increases, aerodynamic drag begins to come into effect and slow the acceleration of each object.

So, the question becomes which object slows more? Clearly, if they both have the same drag coefficient, the ratio of F/D is larger at every point in time for the heavier object. Therefore, the heavier object's acceleration is less impacted by the aerodynamic drag and will hit the ground first.

Now, if they don't have the same drag coefficient, then the problem is the same, but you'd need some quantifiable values to make a determination.