# Objects Moving Apart Under Gravity?

1. Oct 17, 2008

### gonegahgah

Let's say that you shoot two balls straight up into the air (in vacuum) from the surface of the Earth side by side with one at a higher velocity and the other at a lower velocity.

The higher velocity ball will go higher and land later on - of course - than the lower velocity ball. And of course the distance between them will increase.

Will the distance between those balls accelerate as they move further from the Earth?
Will the distance between those balls continue to accelerate as the lower velocity ball begins to fall back to Earth?
Will the distance between those balls continue to accelerate as the higher velocity ball begins to fall back to Earth?

My overall question is do the balls appear to accelerate away from each other throughout the entire trajectory?

2. Oct 18, 2008

### Naty1

]Will the distance between those balls accelerate as they move further from the Earth?
Will the distance between those balls continue to accelerate as the lower velocity ball begins to fall back to Earth?
Will the distance between those balls continue to accelerate as the higher velocity ball begins to fall back to Earth?

My overall question is do the balls appear to accelerate away from each other throughout the entire trajectory?

You best way to answer these questions is to write equations for each ball's motion and compare them. Gravity will accelerate each ball at a steady 32 ft/sec/sec towards the earth...that steady rate of acceleration slows the balls from their initial velocity, stops each ball at a maximum altitude, then speeds the balls up as the return to earth...

so the general answer is they accelerate steadily due to gravity...regardless of their individual masses...weights!!!

v = v' - gt ; v^2 = v'^2 - 2gy ; y = v't - 1/2gt^2.... etc where v' is the initial vertical velocity.

3. Oct 18, 2008

### gonegahgah

My question does actually relate to Cosmology but fair enough maybe someone in homework will help me.

Thanks Naty1. I have tried to find example plotted trajectories of two objects with different velocities by searching the net but the only examples I keep finding are example plotted trajectories of two objects with the same velocity (at different angles).

I need the first and not the later.

It pertains to a further question I have namely how would we know if the Universe were in the process of crunching when it would still appear to us to be expanding?

4. Oct 18, 2008

### yuiop

Will the distance between those balls accelerate as they move further from the Earth?
No, but the distance increases constantly.

Will the distance between those balls continue to accelerate as the lower velocity ball begins to fall back to Earth?
No, but the distance will continue to increase constantly.

Will the distance between those balls continue to accelerate as the higher velocity ball begins to fall back to Earth?
No, but the distance will continue to increase constantly.

My overall question is do the balls appear to accelerate away from each other throughout the entire trajectory?
No, but the distance will continue to increase constantly.
This can be expressed as ball 1 has a constant velocity relative to ball 2 throughout the entire trajectory.

Let's say the ball 1 has an initial velocity of 60 m/s and ball 2 has an initial velocity of 50 m/s then ball 1 will always have a velocity of 10 m/s relative to ball 2.

You can work this out using the equation given by Naty1 for the height (y) of each ball at any given time t and subtracting the results from each other at any given time t so:

ydiff = y1-y2 = (v1*t -1/2gt^2) -(v2*t-1/2gt^2 ) = (v1-v2)t -1/2gt^2+1/2gt^2 = (v1-v2)t

ydiff is directly proportional to t since (v1-v2) is a constant (initial velocities) and the plot of ydiff versus time is imply a straight line.

5. Oct 18, 2008

### yuiop

This is the plot of the equations I gave in the last post.

The blue curve is the trajectory of ball 1 with an initial velocity of 60 m/s and the red curve is the trajectory of ball 2 with an initial velocity of 50 m/s. The diagonal green line is the plot of the difference in height at any time t.

#### Attached Files:

• ###### trajectories.GIF
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6. Oct 18, 2008

### yuiop

First, you should realise that the case of the universe you would have to plot the gravitational accelertion of ball one to ball 2 because there is a not a large central source of gravity represented in your example by the Earth in cosmology. Second you would need to use the equations of General Relativity and not those of newton in cosmolgy. The closest solution is probably represented by something based on the interior Schwarzschild solution or maybe even better by the FRW metric because that allows for the motion of all the expanding/contracting falling objects that are in turn the source of gravitation too. It is far from a simple question. From a personal point of view (just my gut feeling) you are probably right on a deeper level that an expanding universe will look very much like a contracting universe bu that is no simple matter to prove.

All the best ;)

7. Oct 18, 2008

### gonegahgah

In my head I would expect the object with higher initial velocity to always be higher than the object with lower initial velocity. The lower object would therefore be in higher gravity and would therefore accelerate quicker back to Earth than the higher object.

In this respect the lower initial velocity ball would move back to Earth with greater acceleration throughout its path than would the higher initial velocity ball.

In this respect the lower ball's speed would change a greater amount in the same amount of time than the higher ball's speed.

In this respect they would appear to be accelerating away from each other at all times.

Am I correct to say that your formula is showing g to be the same for each of the balls at the same time when it is not?

Can anyone provide me with a plot to confirm or negate this please?

Oops! Just saw you attached picture Kev. Have to wait for it to be approved to view it...

8. Oct 18, 2008

### yuiop

OK, the equations mentioned by Naty1 and used by me are the regular Newtonian ones that assume g is constant and the height above the ground is much smaller than the radius of the Earth. The calculations for varying G are much more complicated but you are right that in general the acceleration for the slower particle at a lower altitude will be greater than the accelration on the faster higher particle at any given time.

On the other hand, you would have to show there is a large central source of gravity for the universe that acts like the Earth in your example.

I think this is far as we can go with this in the homework section. I think you will have to rephrase your question and resubmit it to the Cosmology forum in a manner that makes it clear that you are referring to cosmology.

9. Oct 18, 2008

### gonegahgah

What are supposed to be the big drivers of a crunch? Is it gravity?

If I wait will the kind moderators move it back to the cosmology section?

10. Oct 18, 2008

### yuiop

Yes, good ol gravity. But you need more gravity than there is dark energy which is a sort of negative gravity accelerating the expansion of the universe. The consensus of the Cosmolgy experts at the moment is that the dark energy greatly exceeds the the mutual gravity of galaxies and that the universe will never collapse but will expand at an ever increasing rate until no galaxies or stars are visible from any given location. That is the "big rip" rather the the "bug crunch". There is a minority that think dark energy is just a misinterpretation of the observations but that is definately not the "accepted mainstream view", (at this point of time). Then again, the point of view that Sun does not orbit around the Earth was not the mainstream view back in Galileo's time ;)

The trouble with dark energy is that it requires an extremely fine balance. If the dark energy was initialy some parts per million greater than it is , then the the big rip will already of happened. Some think the extraordinary coincidence required to explain the observations is "too much".

Last edited: Oct 18, 2008
11. Oct 18, 2008

### gonegahgah

Thanks Kev.

Even without dark matter I would like to think there is something like an escape velocity which once it is initiated then things will never fall together again under gravity ever. It works for planets doesn't it? If that is so I don't see why it wouldn't work for a universe spreading apart. That's speculation on my part of course and I don't know how that even remotely fits in with any of today's positions on expansion.

Also from what I understand of it; the universe acts as its own centre of gravity. Assuming there is a centre somewhere - though non-identifiable from within - then there will always be more matter on your side towards, through and beyond the centre then there is on your opposite side away from the centre. So this would then always act as the centre of gravity towards which you are most strongest attracted (ie. there is more stuff towards the inside than towards the outside).

That is assuming a couple of things: that there is a centre somewhere (even if it is unidentifiable as such from our perspective), that there is an edge somewhere, and that the expansion occurs througout the universe from the edges through the centre in a common way.

12. Oct 19, 2008

### gonegahgah

I can see that plot of the two trajectories now Kev. Thanks for that.

It is good because even with this it is possible to see that for each slot of time the two balls constantly expand apart even during their transition and crunch phases.

So 'if' this represented the behaviour our universe then we could not know if we were in a phase of expansion or a phase of contraction by observation.

13. Oct 19, 2008

### Idjot

I would like to try and answer this question by simply visualizing the experiment using 'common sense' if you will:

Ball 1 is moving faster going up = the distance between them increases.

The slower ball 2 stops going up while the faster ball 1 continues upward a little longer = the distance between them increases further.

Ball 1 stops going upward while ball 2 has already started accelerating downward = the distance between them increases further.

Ball 1 starts accelerating downward while Ball 2 continues to accelerate downward =
the distance between them increases further.

Ball 2 hits the ground while Ball 1 continues accelerating downward = the distance between them decreases until Ball 1 hits the ground, finishing the experiment.

Conclusion: The distance between them increases until the "slower" ball hits the ground.

I can see no reason for the distance between them to "accelerate" or, in other words, grow exponentially, throughout the entirety of the "growing" stages. In fact I think that's impossible in this experiment although I'm not sure how to articulate why.

Last edited: Oct 19, 2008
14. Oct 19, 2008

### Idjot

I really would like to understand the calculations involved here and I hope you all will indulge me before closing this discussion. And I am NOT a student, just an inquisitive Idjot:

Suppose each ball weighs 10 lbs.
Suppose Ball 1 is hit upward with 50 lbs of force and Ball 2 is hit upward with 40 lbs of force.

Here are my questions:
1: What will be the max speed of each ball on the way up?
2: What will be the max height of each ball?
3: At what height will each ball reach it's maximum speed?
4: At what rate will each ball slow down?

5: I can understand the balls appearing to accelerate away from one another on the way down, but not on the way up. It seems to me that they would appear to be moving away from each other steadily on the way up, or at least for part of it. My so called 'common sense' tells me that the relationship of the 2 balls couldn't possibly be exactly the same fighting against gravity as submitting to it, and also during separate transitions in states.
Being that G is a set number, shouldn't that mean that the difference in the balls initial velocity could alter the results of said acceleration in distance. In other words, even if they really do appear to accelerate away from one another the whole time, wouldn't that rate have to be different for each stage of the experiment due to the initial difference in velocity against G?

6: Do I sound insane?

15. Oct 20, 2008

### gonegahgah

Hi Idjot

As soon as the balls leave the 'ground' they begin to decelerate. So their fastest speed is at take off and corresponds to their speed at landing.

In the plot provided by Kev the deceleration (g) remains constant throughout the path of the balls (whereas normally g decreases with height).
This means that the speed (not the distance travelled) of each ball will be altered by the same amount during each time period. This is because acceleration acts on speed.

Here is an example showing the height of the balls after time 1, 2, 3, 4, 5. Ball 1 is shown as a) and its speed is shown as s1 and its height is shown as y1. Ball 2 is shown as b) and its speed is shown as s2 and its height is shown as y2. The height is shown as previous height plus current change (note that current change is half the speed):
a) s1=6,y1=0+3; s1=4,y1=3+2; s1=2,y1=5+1; s1=-0,y1=6+0; s1=-2,y1=6-1.
b) s2=4,y2=0+2; s1=2,y2=2+1; s1=0,y2=3+0; s2=-2,y2=3-1; s2=-4,y2=2-2.

So for each time slot you get y1,y2 as:
t1=3,2; t2=5,3; t3=6,3; t4=6,2; t5=5,0

The distance between the two for each time interval goes as:
1, 2, 3, 4, 5

So you can see through this example, and also by the plot provided by Kev, that the change grows by the same amount whether both balls are still climbing, the lower only is dropping or when both are now dropping.

I hope this clarifies it for you idjot. If it doesn't let us know and I, Kev, or someone will try to explain it better.

Last edited: Oct 20, 2008
16. Oct 20, 2008

### Idjot

First let me say thank you for your willingness to help me understand this in it's simplest form.

Based on what you've layed out here let's assume that the time interval between observations is 1 second and the distance measurements are in feet.

That would mean that the balls do not appear to be accelerating away from one another at all but rather they appear to be moving away from one another inertially at 1/2 foot per second each (or one of them moving away at 1 foot per second while the other sits).

No acceleration, right? Just a steady increase in the distance between them, right?

What I'd like to know now is why they would both decelerate at the same rate when one has more momentum.

Last edited: Oct 20, 2008
17. Oct 20, 2008

### gonegahgah

You are correct. Our example has the deceleration force as constant so the two balls would move away from each other at constant speed; as shown.

This is not a real world model though. Under real world conditions the higher ball will feel increasingly less gravity at all times than the lower ball as it gets further apart. So it will experience increasingly less deceleration at all times than the lower ball; except at launch where they briefly both feel the same deceleration. Under these conditions the balls would accelerate apart slightly throughtout their entire journey. But if g were to remain constant their separation would be constant.
Momentum only becomes a consideration where the force is not gravity. Even if the force is not gravity then momentum only becomes a consideration when the objects have different masses. If they are the same mass then the same force will change their speeds by the same amount.
We are dealing with gravity so momentum has no differential effect between the two balls in our examples. They could be two different masses and it wouldn't matter.

Gravity force acts directly upon the speed of objects the same amount no matter what their mass is. Only the amount of gravity is important. And it acts on the speeds - even if the speeds are different - by the same amount. So deceleration of the speed under a constant gravity force will be by a constant amount.

If you have a frictionless car travelling at 100 km/hr and it takes 1 unit of force to slow it down to 90 km/hr then it will also only take another 1 unit of force to slow it down further to 80 km/hr, and so on. The momentum at different speeds for the same mass makes no difference to this. The same amount of force (over any amount of time) changes the speed by the same amount.

Remember that force acts directly upon the speed something is going at - the change in speed being dependent upon its mass; except under gravity. Force does not act directly on the distance travelled. ie.

speed = distance / time ie. (L/T)
acceleration = force / mass ie. ((ML/T^2) / M = L/T^2)
speed2 = speed1 + acceleration * time ie. ((L/T) + (L/T^2) * T) = (L/T) + (L/T) = (L/T)

where L is a unit of length, T is a unit of time and M is a unit of mass,
where force is ML/T^2 and acceleration is L/T^2.

As I said under gravity the mass makes no difference so the above middle formula for acceleration would be different. Have to go but I will think about that; unless someone wants to quickly enlighten me too.

Last edited: Oct 20, 2008
18. Oct 20, 2008

### Idjot

I think I feel the spark of understanding in my brain, but there is still something fuzzy to me here considering some of your statements and how they would relate to the real world model:

Isn't it a difference in momentum that causes the difference in speed?

Isn't it the difference in speed that causes the difference in altitude?

Isn't it the difference in altitude that causes the difference in gravity?

Doesn't that mean that a difference in momentum can cause a difference in gravity, albeit indirectly?

Isn't it the difference in Gravity that causes the difference in deceleration?

Doesn't that mean that a difference in momentum can cause a difference in deceleration, albeit indirectly?

19. Oct 20, 2008

### Idjot

(bold text): This also makes me curious. I would have thought that while looking at the experiment in reverse that the difference in deceleration would become infinitesimally smaller. If there is a brief period where it's identical, how does one determine the threshold of change?

20. Oct 21, 2008

### gonegahgah

Momentum doesn't cause speed; momentum is a property related to speed.
ie. momentum = mass * velocity

Heavier objects are usually harder to slow down then lighter objects because they have greater mass; except when gravity is doing the slowing down.

It is the difference in mass that will make objects harder or easier to slow down. A heavier car will take more force to slow down and speed up than a lighter car. The same force will slow down a heavier car less than it slows down a lighter car.

They are not harder to slow down because they have greater velocity. If they are going faster then it just takes longer to slow them down but the rate of change to the velocity remains the same for the same force over the same time.

There are two parts to momentum: mass & velocity. Only one of these can make it harder or easier to slow down a car; not both (ie. the mass). The other part then follows how long it takes to slow down the car for that mass.

I'm showing two examples following: the first where we have different initial speeds, and the second where we have different massed cars.

I'm using the generalised units L for length, T for time and M for mass. You could consider L to be kms, T to be hrs and M to be 1000 kg if you like.

The formula to work out the new speed is: velocity2 = velocity + (-force / mass) * time.

Example 1:

Let's say a car travelling at 1L/T has a mass of 2M and you apply a reverse force of 1ML/T^2 and you apply this force for 1T.
This will give: 1L/T - 1ML/T^2 / 2M * 1T = 1L/T - .5L/T = .5L/T
Let's say the same car is travelling at 2L/T and the force and time are the same.
This will give: 2L/T - 1ML/T^2 / 2M * 1T = 2L/T - .5L/T = 1.5L/T
Let's say the same car is travelling at 3L/T and the force and time are the same.
This will give: 3L/T - 1ML/T^2 / 2M * 1T = 3L/T - .5L/T = 2.5L/T
Let's say the same car is stopped and the force and time are the same.
This will give: 0L/T - 1ML/T^2 / 2M * 1T = 0L/T - .5L/T = -0.5L/T

So you get the following for the same mass, force & time but different initial speed:
(initial speed, mass, force, time, final speed):
s1=0, m=1, f=1, t=1, s2=-0.5
s1=1, m=1, f=1, t=1, s2=0.5
s1=2, m=1, f=1, t=1, s2=1.5
s1=3, m=1, f=1, t=1, s2=2.5

Example 2:

Let's say a car travelling at 1L/T has a mass of 1M and you apply a reverse force of 1ML/T^2 and you apply this force for 1T.
This will give: 1L/T - 1ML/T^2 / 1M * 1T = 1L/T - 1L/T = 0L/T
Let's say a car twice as heavy starts at the same speed and the force and time are the same.
This will give: 1L/T - 1ML/T^2 / 2M * 1T = 1L/T - .5L/T = 0.5L/T
Let's say a car four times as heavy starts at the same speed and the force and time are the same.
This will give: 1L/T - 1ML/T^2 / 4M * 1T = 1L/T - .25L/T = .75L/T

So you get the following for the same initial speed, force & time but different mass:
(initial speed, mass, force, time, final speed):
s1=1, m=1, f=1, t=1, s2=0
s1=1, m=2, f=1, t=1, s2=0.5
s1=1, m=4, f=1, t=1, s2=0.75

Only mass changes how much acceleration (or deceleration) is achieved for the same force (except under gravity where mass of the ball does not matter). The speed does not change the acceleration. Speed is changed by the acceleration amount. Acceleration is simply the amount the speed will be changed. You would have a catch 22 situation if speed affected acceleration and acceleration affected speed. Which is doing the affecting?

In the above examples you can see that the speed of each car with different initial speeds changed by the same amount even when the car started stopped. This is because acceleration acts directly upon the speed. But in the second example, where the mass changes, the amount of acceleration also changes (except under gravity).

Hopefully that is a bit clearer but keep at me if it isn't.
It is movement that causes the change in altitude. Speed is how quickly that movement causes that change. So yes faster relative speed does mean more distance will be covered if that is what you mean.
Yes.
Very indirectly.

Yes.

Very indirectly.

I say 'very indirectly' because the momentum is nothing without considering where the balls are now which is what really determines the amount of deceleration acting on them. Momentum will help to get them there but it can not be squeezed into any formula for current deceleration. Only the altitude and mass of the Earth can determine this; not the momentum of the balls. Momentum can only be put into formulas that determine where the object is now after also supplying where it was before, the force applied and the time over; and then only where the force is not gravity. Only speed needs to be considered when acting under gravity; not momentum.

21. Oct 21, 2008

### gonegahgah

The difference in deceleration continues to increase throughout the entire flight. The difference in deceleration never decreases between them.

This is because the gap between the balls always grows which means that lower ball will be more lower each time than the higher ball. This means the gravity acting on the lower ball will always increase at a faster rate than the gravity acting on the higher ball because the lower ball will get closer to the Earth sooner than the higher ball at an ever increasing rate.

Last edited: Oct 21, 2008
22. Oct 21, 2008

### Idjot

Thank you for the momentum explanation.

As to the quote above I think you missed where I said "in reverse"

23. Oct 21, 2008

### gonegahgah

Hi Idjot.

You would agree that the bottom ball plateaus (reaches zero up speed) first whilst the top ball is still travelling up.

This means that at this point in time the top ball is still heading towards lower gravity while the bottom ball is beginning its drop back to higher gravity.

That means the gravity on the top ball will continue to decrease and the gravity on the bottom ball will now begin to increase as it drops.

So the acceleration down on the top ball continues to decrease; but the acceleration on the bottom ball begins to increase as it begins to drop into higher gravity.

Remember that acceleration is how much the speed will change by for a given time.

A decreasing acceleration down on the top ball will mean that it will slow down at a slower rate the higher it continues; whereas the increasing acceleration down on the bottom ball means that it will drop at a faster rate as it continues.

The top ball has lessening gravity happening on it over time; and the bottom ball has increasing gravity happening on it over time.

So the acceleration down on the top ball is decreasing; and the acceleration down on the bottom ball is increasing.

So the top ball will change speed slower and continue to rise; whereas the bottom ball will change speed faster and continue to drop.

So the top ball is changing direction ever slower; and the bottom ball is changing direction and moving down ever faster. If the top ball is changing at an ever slower rate and the bottom ball is changing at an ever faster rate then they can not help but to be accelerating apart as a result.

So during this phase where the top ball is still rising and the bottom ball begins to fall the top ball continues to rise into lessening gravity; and the bottom ball begins its drop back into higher gravity. Less gravity means less deceleration for the top ball. More gravity means more acceleration down for the bottom ball. This creates an accelerating difference.

Okay. Now to the next phase where the top ball now begins to plateau.

The bottom ball continues to drop into higher gravity so the amount of acceleration on its speed continues to increase; whereas for the top ball it is plateauing so the change in gravity briefly stops.

At this time the bottom balls higher gravity is still increasing while the top balls lower gravity is plateauing out. The amount of acceleration down on the bottem ball therefore continues to keep increasing ever quicker; while for the top ball the amount of acceleration is presently changing very minutely through the plateau.

The quicker change in acceleration for the bottom ball means that it will increase its pace of acceleration down faster than the acceleration down is increasing for the top ball.

The top ball therefore now drops into into increasing gravity at a slower increasing rate than does the bottom ball which is dropping into increasing gravity at a faster increasing rate.

So again the bottom ball will have increasing gravity happening on it faster than the top ball does.
Because the bottom ball moves into increasing gravity at a faster rate its acceleration also picks up at a faster rate. It therefore picks up speed down faster and faster and faster. The top ball also picks up speed down faster and faster but it is still travelling slower so it drops slower and experiences greater acceleration at a slower rate than the bottom ball which is moving faster into higher gravity.

As you should see from the above; we are now at a point where both balls are dropping; the top ball at a slower speed; and the bottom ball at a faster speed.

The rate of acceleration - which is based on altitude - on the bottom ball therefore increases at a faster rate than it does for the top ball. The bottom ball will therefore change speed faster than the top ball and will therefore appear to accelerate away from the top ball.

How's that? Clear as mud?

24. Oct 22, 2008

### Katrex

Ughh thats horrible, I read that a nearly puked after 2 lines. 32ft per second... Do you realy do physics in imperial mesuments!

Im asuming this is over a distance that g does not change significantly

As for the question in hand. First disntance doesnt accelerate, we know what you mean but even so.

Does the balls apear to accelerate away from each other. the answer is no. Both balls are falling with a constant acceleration, One starts with a higher velocity if you take your mesurments from when the second one starts falling. Both their speeds increase by 9.8ms-2 so the diference in velocities increases at a constant rate. In otherwords they move apart at a constant speed

In a real world situation the accleration changes but its rate of change is contant so they accelerate apart at a constant(although unoticible speed)

Also dear ijot first mks please! secod if you sit something with a force we cant do a caculation from that because we need to know the time over which the force was aplied to calculate the momentum. saying you hit something with 50lb of force doesnt say anything. Just say you give one enough force to give it initial velocity of x and the other initial velocity of y, then use suvat

25. Oct 22, 2008

### Idjot

Much clearer than mud. Thank you. I'm sorry for confusing you into explaining that all over again.

What I meant by "looking at the experiment in reverse" was aimed at understanding the very BEGINNING of the experiment when the balls are first hit upward.

You mentioned before that there was a period of time "at launch when they briefly both feel the same deceleration." I was just wondering when that brief period ended if it ever really began, so I tried to visualize watching it in "rewind" if you will. I tried to say that to me it seems that the difference in deceleration would become smaller and smaller(while watching it in rewind) all the way to the hit, and that there might not ever really be a measurable period of sameness with regard to deceleration.

In other words, I am suggesting (based on what you've explained to me) that the difference in deceleration of the 2 balls is increasing from the very instant they are hit and that there cannot possibly be a brief moment, even a nanosecond, when they both experience the same deceleration because they are being hit with different amounts of force, thereby causing them to be at different heights from the instant they are hit.

How's MY mud looking now?