Rate of Attraction

[DimaMazin]

I do it via a spreadsheet and vary one parameter at a time and I get the feeling what happens in different situations.

So on the condition that the material has a consistent density you can select r1 and r2 and the spacing (gap) g therefore you can formulate the amount of force, and later divide by the mass based on the chosen radius and gap values to get accelerations.Starting from zero, acceleration determines speed and speed determines distance in time. So at this stage I'm just considering the effects on acceleration.

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

I think we can use it.

 

t={ (gap+r₁+r₂)³ / 2G(M₁+M₂) }^1/2 _* { [(r₁+r₂)_*gap]^1/2/(r₁+r₂+gap) + arccos[(r₁+r₂)/(r₁+r₂+gap)]^1/2 }

r₁ is radius of first body

r₂ is radius of second body

[Robittybob1]

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

I think we can use it.

 

t={ (gap+r₁+r₂)³ / 2G(M₁+M₂) }^1/2 _* { [(r₁+r₂)_*gap]^1/2/(r₁+r₂+gap) + arccos[(r₁+r₂)/(r₁+r₂+gap)]^1/2 }

r₁ is radius of first body

r₂ is radius of second body

Have you tried it? So Y0 = (gap+r1+r2)? and y= r1+r2 and therefore (1-y/Y0) = gap? What is the arccos value going to be?

It looks like you got the algebra right.

 

The syntax for the ACOS function in Microsoft Excel is:

ACOS( number )

The Microsoft Excel ACOS function returns the arccosine (in radians) of a number.

Are we dealing with an angle?

[DimaMazin]

Have you tried it? So Y0 = (gap+r1+r2)? and y= r1+r2 and therefore (1-y/Y0) = gap? What is the arccos value going to be?

It looks like you got the algebra right.

Are we dealing with an angle?

1 - y/y₀= gap/y₀

(y₀^-2)^1/2=y₀^-1

I don't know derivation of the arccos, but I don't see angle here.

 

for A and B:

t ~= {(gap+r)^3/2/(2GM)^1/2} _* {(r_*gap)^1/2/(r+gap) + arccos[r/(r+gap)]^1/2 }

 

for D and C:

t={(gap+2r)^3/2 /(4GM)^1/2} _* {(2r_*gap)^1/2/(2r+gap) + arccos[2r/(2r+gap)]^1/2 }

 

r is radius of big objects

Edited June 13, 2015 by DimaMazin

[Robittybob1]

t={ (gap+r1+r2)3 / 2G(M1+M2) }1/2 * { [(r1+r2)*gap]1/2/(r1+r2+gap) + arccos[(r1+r2)/(r1+r2+gap)]1/2 }

 

slight error in the formula you have to do the fraction, then take the square root then find the arc cosine of that.

t={ (gap+r1+r2)3 / 2G(M1+M2) }1/2 * { [(r1+r2)*gap]1/2/(r1+r2+gap) + arccos[(r1+r2)/(r1+r2+gap)1/2] } might be closer.

 

the result with r1 =10 m and r2 = 1m and gap = 10 m gave result of 64246.08 secs

where as with r1 =10 m and r2 = 10 m and gap = 10 m gave result of 73251.79 secs

So that clearly shows that A-B beats B-C.

 

the result with r1 =10 m and r2 = 1m and gap = 20 m gave result of 110390.14 secs
where as with r1 =10 m and r2 = 10 m and gap = 20 m gave result of 119619.05 secs
So that clearly shows that A-B stills wins.

 

I wonder at what gap they would be equivalent? ( Note I have edited this as I made an error)

It was between a gap of 80 -81 meters that the rates changed.

Edited June 17, 2015 by Robittybob1

[DimaMazin]

Which were masses A and B?

[DimaMazin]

+ arccos[(r1+r2)/(r1+r2+gap)1/2] } might be closer.

 

+arccos[(r₁ + r₂)^1/2/(r₁+r₂+gap)^1/2 ]} then agen.

[Robittybob1]

Which were masses A and B?

I gave the material a density of 1 kg/m^3, which is a bit stupid for it would be a very light material, but both balls have the same density, so their masses were proportional to the cube of their radii via the equation for volume of a sphere. I'll check another time whether density has an effect on the rates.

 

I have checked at the low density:

"I wonder at what gap they would be equivalent? It was between a gap of 80 - 81 meters that the rates changed.

Beyond a gap of 80 - 81 meters B - C would be faster than A - B (according to that formula) using that density.

Edited June 17, 2015 by Robittybob1

[Janus]

Consider 2 separate systems of masses; A and B, C and D. B,C and D are all the same mass. A is some mass a lot smaller than the others. The only gravitational influence is from the masses within in each system. Assuming the edge of each mass is lined up the same as in the image, will the two systems make contact at the same time?

 

4 Spheres.png

 

The "fall time" for two bodies such as in your scenario is found by

 

[math]t = \frac{ \arccos \left( \sqrt{ \frac{x}{r} }\right) + \sqrt {\frac{x}{r} \left ( 1 - \frac{x}{r} \right ) } }{ \sqrt{ 2 \mu } } r^{3/2}[/math]

 

Where r is the starting distance between the centers of the bodies, x is the ending distance and [math]\mu = G(M_1+M_2)[/math].

 

Thus for the left image, x= R_A+R_B and r = d+R_A , and for the right image, x = R_C+R_D and r = d+R_C, where d is the distance from the center of B and the lower edge of A or the distance between the center of D and the edge of C. Since R_B=R_C=R_D, we can reduce the number of variables and say that for the right image, x = 2R_B and r =d+R_b.

 

Also, since M_B=M_C=M_D, M+m can be rewritten as 2M_B. And, assuming an equal density for body A, its mass can be rewritten in terms of M_B as [math] M_B \left ( \frac{R_a}{R_B} \right )^3[/math]

 

Thus it is now just a "simple matter of making the proper substitutions in the first formula to find the fall times for both scenarios and determining which is the shorter.

Edited June 17, 2015 by Janus

[Mordred]

The "fall time" for two bodies such as in your scenario is found by

 

[math]t = \frac{ \arccos \Big( \sqrt{ \frac{x}{r} }\Big) + \sqr{r} \ ( 1 - \frac{x}{r} ) } }{ \sqrt{ 2 \mu } } \, r^{3/2}[/math]

 

 

You had a latex error. I tried finding the error but not sure where it is

 

Edit I see ya found it

Edited June 17, 2015 by Mordred

[Robittybob1]

The "fall time" for two bodies such as in your scenario is found by

 

[math]t = \frac{ \arccos \left( \sqrt{ \frac{x}{r} }\right) + \sqrt {\frac{x}{r} \left ( 1 - \frac{x}{r} \right ) } }{ \sqrt{ 2 \mu } } r^{3/2}[/math]

 

....

Is that the same equation as we are already using as presented in the Wikipedia article?

https://en.wikipedia.org/wiki/Free_fall#Inverse-square_law_gravitational_field

[Robittybob1]

I gave the material a density of 1 kg/m^3, which is a bit stupid for it would be a very light material, but both balls have the same density, so their masses were proportional to the cube of their radii via the equation for volume of a sphere. I'll check another time whether density has an effect on the rates......

In another test of the rates of attraction I increased the density to 10 kg/m^3 the time to impact was reduced by about a 1/3rd.

I wonder if the time to impact is the time for the spheres to touch their outer edges.

 

I should be able to test this if I make the gap = 0 then T should be zero as well.

Strangely enough when the gap is zero the time is exactly 1 so that is a bit odd for it should also be zero.

I'll have to see if this is a true result or some calculation error.

It seems to be a direct result of the arccosine part for that comes out to 1. It doesn't make sense to have this extra bit to me for there are no angles involved.

Edited June 18, 2015 by Robittybob1

[Janus]

In another test of the rates of attraction I increased the density to 10 kg/m^3 the time to impact was reduced by about a 1/3rd.

I wonder if the time to impact is the time for the spheres to touch their outer edges.

 

I should be able to test this if I make the gap = 0 then T should be zero as well.

Strangely enough when the gap is zero the time is exactly 1 so that is a bit odd for it should also be zero.

I'll have to see if this is a true result or some calculation error.

It seems to be a direct result of the arccosine part for that comes out to 1. It doesn't make sense to have this extra bit to me for there are no angles involved.

If the gap is zero, then x and r ( the ending and starting distances between the centers of the objects) are equal and thus x/r is 1. The arccos of 1 is 0. Also in the other half of the denominator, we have a factor of 1-x/r, which when x=r equals 0. So in the denominator, we get 0+0, giving an answer of t=0. If you are getting an answer of 1, you are doing something wrong. The arccos is a result of the calculus used to derive the equation. When you integrate certain functions like -1/sqrt(1-x^2) you get an inverse trigonometric function as the result (with the angle measured in radians).

 

 

Here's a chart comparing the fall times of A-B and C-D. It plots the difference in fall times as the gap distance starts at zero and increases. (C-D fall time minus A-B fall time).

 

 

The difference starts at zero (when the gap is zero and both fall times are zero. Then as the gap distance increases, at first the A-B fall time is going to be shorter than the C-D fall time. This difference will increase for a while and then start to decrease until eventually the C-D fall time becomes shorter and then stays that way.

Edited June 18, 2015 by Janus

[Robittybob1]

So do you think that the density of the balls has no effect on the free fall time? I was getting faster falling with increased density, but your graph is "difference" in fall times. Maybe the difference in times stays the same but that seems unlikely.

What is your understanding on that please?

 

I did have the Excel referring to the wrong cell and now when the gap is zero I'm getting the answer of zero time. Thanks for that.

Edited June 18, 2015 by Robittybob1

[Janus]

So do you think that the density of the balls has no effect on the free fall time? I was getting faster falling with increased density, but your graph is "difference" in fall times. Maybe the difference in times stays the same but that seems unlikely.

What is your understanding on that please?

 

I did have the Excel referring to the wrong cell and now when the gap is zero I'm getting the answer of zero time. Thanks for that.

Well of course density is going to be factor. After all, higher density for a sphere of a given size means a higher mass for the sphere. Since the fall time in the equation is inversely proportional to the square of the sum of the masses, a higher total mass of the system leads to a shorter fall time. But quite frankly, this is a pretty obvious and uninteresting result. If we just assume an equal density for all the objects involved we don't need to bother with the actual value of that density.( what I actually did was assume a density such that B had a mass(M) was just the right value to make GM=1.

 

It isn't the actual numerical values of the fall times that are of any interest either. It is how the fall times compare relative to each other that are interesting. And even the magnitude of the difference isn't important. It is the pattern of where the fall times for one pair of objects is the lesser at one distance and becomes the greater at another.

 

The relative radii of A and C play a role in just where the crossover takes place, the larger the difference, the larger the gap distance where the cross over.

 

Of course you could assume different densities for the objects, but this would remove the clean relationship between the radii of the objects involved and their masses. (for instance, by adjusting the relative densities of A and C you can make the fall time of either pair larger, smaller or equal to the fall time of the other pair at any gap distance.)

[StringJunky]

Well of course density is going to be factor. After all, higher density for a sphere of a given size means a higher mass for the sphere. Since the fall time in the equation is inversely proportional to the square of the sum of the masses, a higher total mass of the system leads to a shorter fall time. But quite frankly, this is a pretty obvious and uninteresting result. If we just assume an equal density for all the objects involved we don't need to bother with the actual value of that density.( what I actually did was assume a density such that B had a mass(M) was just the right value to make GM=1.

 

It isn't the actual numerical values of the fall times that are of any interest either. It is how the fall times compare relative to each other that are interesting. And even the magnitude of the difference isn't important. It is the pattern of where the fall times for one pair of objects is the lesser at one distance and becomes the greater at another.

 

The relative radii of A and C play a role in just where the crossover takes place, the larger the difference, the larger the gap distance where the cross over.

 

Of course you could assume different densities for the objects, but this would remove the clean relationship between the radii of the objects involved and their masses. (for instance, by adjusting the relative densities of A and C you can make the fall time of either pair larger, smaller or equal to the fall time of the other pair at any gap distance.)

Yes, I wanted to keep the differences and number of parameters very simple and relative. Your answers are expressed in the way I wanted to see it. Thanks.

[md65536]

It isn't the actual numerical values of the fall times that are of any interest either. It is how the fall times compare relative to each other that are interesting. And even the magnitude of the difference isn't important. It is the pattern of where the fall times for one pair of objects is the lesser at one distance and becomes the greater at another.

 

The relative radii of A and C play a role in just where the crossover takes place, the larger the difference, the larger the gap distance where the cross over.

But a cross-over isn't guaranteed, is it?

Let Mdiff = M_B - M_A

The radius of A (and thus also the final distance x) decreases proportional to the cube root of Mdiff,

while the total mass of the system decreases linearly wrt. Mdiff.

Wouldn't that require that whatever other variables you use (still assuming fixed density of all masses), you can choose the difference in the masses to be small enough that the CD system beats the AB system for any gap size?

 

It seems the converse is not true... if you set the variables so that the AB system wins, it looks like you can make CD win just by increasing the start distance r enough (if AB can win at all, then a crossover can be found by just increasing the gap)???

 

 

Or that is... if you choose variables so that AB wins, then simply increase the gap and CD will eventually win.

If you choose variables so that CD wins with negligible gap, you need to decrease the mass of A to make AB win, as adjusting the gap won't change the outcome.

 

Perhaps you might be able to choose a low enough density so that CD wins regardless of both gap and mass of A???

Edited June 19, 2015 by md65536

[Robittybob1]

Yes, I wanted to keep the differences and number of parameters very simple and relative. Your answers are expressed in the way I wanted to see it. Thanks.

Can you explain why there is a crossover?

[md65536]

Can you explain why there is a crossover?

It is because the difference in radius (r_diff = r_A - r_C) is more significant relative to a small gap than to a large one.

 

Let's say you set your variables such that AB and CD touch at the same time. CD has greater mass which is balanced by AB having a shorter distance between the centers of the masses.

If you decrease the gap, r_diff becomes proportionally larger relative to the gap, and the distance between the masses becomes more important, and AB touches first.

If you increase the gap, r_diff becomes less significant relative to the gap, and the differences between the masses becomes more important and CD touches first.

[Robittybob1]

It is because the difference in radius (r_diff = r_A - r_C) is more significant relative to a small gap than to a large one.

 

Let's say you set your variables such that AB and CD touch at the same time. CD has greater mass which is balanced by AB having a shorter distance between the centers of the masses.

If you decrease the gap, r_diff becomes proportionally larger relative to the gap, and the distance between the masses becomes more important, and AB touches first.

If you increase the gap, r_diff becomes less significant relative to the gap, and the differences between the masses becomes more important and CD touches first.

I see you really understand it well.

[DimaMazin]

I see you really understand it well.

I think we'd understand that better if we have graph of t_AB/t_CD

That is interesting that big pig-iron ball can fall slower than small pig-iron ball at the same very small gap of a fall.

Edited June 21, 2015 by DimaMazin

[md65536]

I see you really understand it well.

I don't think so, I've just been trying to figure out the implications of only parts of the math, and not even checking to see if the guesses work out, so I've been making mistakes until someone else figures out more math.

 

 

Anyway I decided to at least test some values using the maths posted. Trying out a few different values, it seems that only choosing the radius of ball A relative to ball B is enough to make the CD system win, regardless of the other values???

 

It seems that if r_A is greater than (or equal to???) half of r_B, then the CD system will win no matter what the gap is. If it's less than half, then there will be a crossover (ie. there are small gaps for which the AB system wins, and larger gaps for which CD wins). I don't trust my abilities and don't (yet) want to try solving it algebraically... does anyone find the same results?

[StringJunky]

Thinking further on this scenario, I think It might be better for me to understand and simplifying by removing the variable of the difference in volume/radii:

 

Let A be the same volume as B,C and D. but only 5% of the mass of B,C and D. This way, the centres and extremities are equal distances. The principle of equivalence says that they will touch at the same time. Is this correct? Everything else about the setup remains the same as before in a vacuum; air resistance is irrelevant.

 

Edited; Altered question for clarity.

Edited June 23, 2015 by StringJunky

[DimaMazin]

Thinking further on this scenario, I think It might be better, for simplicity and removal of the variable of the difference in volume, a modification:

 

Let A be the same volume as B,C and D. but only 5% of the mass of B,C and D. This way, the centres and extremities are equal distances. The principle of equivalence says that they will touch at the same time. Is this correct? Everything else about the setup remains the same as before in a vacuum; air resistance is irrelevant.

No. Then t_AB/t_CD>1

C and D have more force than A and B proportionally to their masses.

Edited June 23, 2015 by DimaMazin

[Robittybob1]

No. Then t_AB/t_CD>1

C and D have more force than A and B proportionally to their masses.

Might be a good one to test out the formula above.

 

C - D beat A - B and regardless of the gap size and the ratio of the times always equals 1.38013. Now that is rather weird!

[imatfaal]

Thinking further on this scenario, I think It might be better for me to understand and simplifying by removing the variable of the difference in volume/radii:

 

Let A be the same volume as B,C and D. but only 5% of the mass of B,C and D. This way, the centres and extremities are equal distances. The principle of equivalence says that they will touch at the same time. Is this correct? Everything else about the setup remains the same as before in a vacuum; air resistance is irrelevant.

 

Edited; Altered question for clarity.

 

 

No it is not correct. The lighter ball accelerates at the same speed as a heavy ball would in the same position; but the light ball attracts its partner heavy ball proportionally less. The closing time is higher.

 

This is the weak equivalence principle

All test particles at the alike spacetime point in a given gravitational field will undergo the same acceleration, independent of their properties, including their rest mass

But by making ball A light you are changing the gravitational field that ball B experiences - it is no longer constant and you cannot use the above principle.

Might be a good one to test out the formula above.

 

C - D beat A - B and regardless of the gap size and the ratio of the times always equals 1.38013. Now that is rather weird!

 

No weird at all. Look at Janus' equation - all that is varying is the mass. The ratio of the times is the ratio of the roots of the masses

 

sqrt(2)/sqrt(1.05) = 1.38013