gravity proportionally speaking

John E. Franklin · 2008-02-15 12:48:07

The classic gravity between
2 objects is proportional to both
masses treated separately and
inversely to the distance
between them treated separately
from each side. This results in
multiplying the 2 masses on the
top and the 2 equal distances in
the denominator getting the
well know equation:

F = k m1 m2 / d**2, where
** means ^ and k is a constant
that converts the units to Newtons.

Interestly, if we imagine the system
getting larger or smaller in total
proportion as the masses are objects
in 3-D, then you see that the
force of gravity is proportional to the
size of the system to the 4th power.

Simply substitute d1**3 for m1 and
d2**3 for m2. Now d1 and d2 is the
average diameter size for the masses,
by taking the cube root of the masses.

You can't really make a system larger or
smaller in everyway because then the
size of the subatomic particles would
change size too, but I am ignoring this,
and simply making the system bigger or
smaller by using more material.

So d**3 d**3 / d**2 = d**4 = gravity, in ratios anyway.

So why is gravity between two objects based on
the size of the system to the 4th power??

Seems strange to me?? I would have guessed the 3rd power.
But it is a weird world I guess.

Last edited by John E. Franklin (2008-02-15 12:50:17)

John E. Franklin · 2008-02-15 16:13:54

I just layed down to sleep, but got up to tell you
that if Two big planets were X diameters apart and
two tiny planets were X diameters apart, and
they started at rest, and then accelerated
into each other, then both systems would
collide in the exact same time!!!
This is because F = ma .
=~= means proportional...
** means to the power of
Again F = ma
So d**4 =~= d**3 a,
so a =~= d.

To check, I thought I was wrong for a second due to
the time squared in d = (1/2)a t**2, but it is correct.

a is acceleration
v = (a)(t)
d = (1/2)(a)(t**2)
If d and a are both 4, t for time is the same
length of time as if
d and a are both 1, or other identical values.
So that's awesome.
The planets collide at the same time, even
though the systems are on different
size scales, as long as the planets are
bigger just as the distance is in between
them!!!!
I hesitate, but still say... Neat Huh??

JaneFairfax · 2008-02-16 00:37:20

John E. Franklin wrote:

To check, I thought I was wrong for a second due to
the time squared in d = (1/2)a t**2, but it is correct.
a is acceleration
v = (a)(t)
d = (1/2)(a)(t**2)

You canNOT use those equations! The acceleration is not constant.

You must use a differential equation to describe the interaction:

(The minus sign is because x (the distance between the two bodies) is decreasing with respect to t.)

Last edited by JaneFairfax (2008-02-16 04:47:05)

JaneFairfax · 2008-02-16 04:13:50

Hey. The differential equation above is rather interesting. A particular solution is

Does it remind you of anything in particular? YES! Keplers third law of planetary motion, which says that the cube of the distance of a planet from the Sun is proportional to the square of its period of revolution!

luca-deltodesco · 2008-02-16 04:22:26

nice ;P

John E. Franklin · 2008-02-16 04:36:38

Thanks for pointing out the acceleration is not constant.
I will have to learn more math in the future to
try to continue these ideas.
I wonder how I should proceed?
For a system that is larger than another,
the initial acceleration is also larger in
a linear way, if that part is right.
I'll keep thinking and learn more math in the
future decades. Thanks!!
So even though I don't know how to
get the actual time now that the
acceleration keeps changing, I am
going to make a wild guess that the
two systems still hit at the same time, but
I really have to start learning higher mathematics,
so I can do amazing things!!

Last edited by John E. Franklin (2008-02-16 04:39:43)

JaneFairfax · 2008-02-16 05:08:02

Oh, by the way the particular solution above is not a particular solution to your problem (in which one mass accelerates towards another and eventually colliding into it). Its just an interesting solution that appears to be how Keplers third law can be derived from Newtons more general law of gravitation.

From this site, Ive worked out that the general solution to the differential equation above is

Unfortunately I dont know how to evaluate the LHS integral.

John E. Franklin · 2008-02-16 06:49:49

Wow! What's

and ?

JaneFairfax · 2008-02-16 07:29:54

Theyre arbitrary constants. Since the differential equation is second order, you would expect the general solution to have two arbitrary constants.

Actually, Ive discovered a mistake Ive made. As one body accelerates towards the other, the other is not stationary but will also be accelerating towards the first body hence the distance x is actually decreasing faster than normal!

Well, we can assume by symmetry that the distance between the two bodies is decreasing twice as fast as the distance between either body and a fixed point between them. Thus the differential equation of motion should be

And the general solution would be

By the way, the integral is doable, but appears to be rather complicated: http://www.artofproblemsolving.com/Foru … p?t=188861

Last edited by JaneFairfax (2008-02-16 11:08:44)

JaneFairfax · 2008-02-16 11:12:35

Now, you want to calculate the time to impact. Lets simplify a bit and set the first arbitrary constant 0. The the differential equation of motion becomes

Thus

If the initial separation of the two bodies is D, then

There you are, then. Thats the expresson for the time it takes for two bodies each of mass M separated by distance D to come to blows by virtue of their mutual gravitation. Thus the time to impact is inversely proportional to the square root of mass, and the square of the time to impact is proportional to the cube of the initial distance apart.

John E. Franklin · 2008-02-16 19:34:05

It is wonderful to have someone that can do
higher maths around. You know, this is a
really nice answer because, talking proportionally
again, say the system is twice as large, including
the planets, and the distances between.
Then the mass in denominator will be eight
times (2 cubed) larger, and the numerator
will be 8 times larger too, all inside of your
square root sign, so it looks like a larger
system will have the same time to collide,
providing all your wonderful work is
correct.

Math Is Fun Forum

#1 2008-02-15 12:48:07

gravity proportionally speaking

#2 2008-02-15 16:13:54

Re: gravity proportionally speaking

#3 2008-02-16 00:37:20

Re: gravity proportionally speaking

#4 2008-02-16 04:13:50

Re: gravity proportionally speaking

#5 2008-02-16 04:22:26

Re: gravity proportionally speaking

#6 2008-02-16 04:36:38

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#7 2008-02-16 05:08:02

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#8 2008-02-16 06:49:49

Re: gravity proportionally speaking

#9 2008-02-16 07:29:54

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#10 2008-02-16 11:12:35

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#11 2008-02-16 19:34:05

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