Maths Help?

Hi there.

Erm, I've been playing some more with my gravitational simulation program,
and I've been trying to verify mathematically the results I've been
getting where I've run the program to simulate two masses* initially
at rest, 1 metre distant from each other, and with no other gravitational
field present.

It doesn't take Einstein to work out that the force on each changes with
the distance between them, so their acceleration changes too.

Being a lowly ESN slug in the field of maths, I'm too lame to work out the
maths to arrive at an equation for separation as a function of time, and
thereby calculate how long it takes for the masses to meet.

Obviously, given two masses a and b, the force on each is given by
F = G . m_a . m_b / r^2, and acceleration a_a = G . m_b / r^2, thanks to
Newton. But now, a_a and a_b depend strongly on r, so I can't use
s = u.t + 0.5 . a . t^2 to solve for the time required for the two to meet.

I _think_ that it's true to say that acceleration, being the 2nd derivative
of position w.r.t. time, is given by:

d^2(r)/dt^2 = 2.G.m_b/r^2,

so

r^2 . d^2(r)/dt^2 = 2.G.m_b

But I'm not even sure about this, and even if so, I'm far too useless at
maths to be able to take this any further. I really am stupid when it comes
to this! I assume some kind of integral over r is called for, intuitively,
at least, but words cannot express my ****wittedness in such matters.

Would someone be able to put me out of my dribbling, incompetent misery?
I'd like to get an equation to tell me the time taken for the masses to meet,
assuming negligible (zero) radius. If you could walk me through the maths
line by line, I'd really appreciate it. Remember, you're dealing with a
mathematical amoeba here!

FWIW, I _have_ googled for this, but maybe my search terms were crap, because
every page I found assumed constant gravitational acceleration, which is NOT
the case here.


Thanks in advance,


Martin

* Their mass is irrelevant, at least if they're equal; I haven't checked to
see whether the same is true for unequal masses yet.

By way of some kind of contribution "in return", for those who may be
interested, and who may yet be unaware, I discovered late this week, that
the definition of the kilogram has finally been changed so as to remove the
need for a physical artifact, thus: (Pasted from a post I made in another
forum)

The kilogram has FINALLY (as of 1999) been redefined so as to remove the
primary mass artifact, and is now defined, wait for it, thus: "The kilogram
is the mass of a body at rest whose equivalent energy equals the energy of
a collection of photons whose frequencies sum to 135 639 274 x 10^42 Hz."
That sure was a long time coming! BTW, don't believe me, believe NIST:

http://physics.nist.gov/TechAct.99/D...constants.html

Formerly, the kilogram standard was as follows: In 1889 the
international prototype kilogram was adopted as the standard for mass. The
prototype kilogram is a platinum-iridium cylinder with equal height and
diameter of 3.9 cm and slightly rounded edges. For a cylinder, these
dimensions present the smallest surface area to volume ratio to minimize
wear. The standard is carefully preserved in a vault at the International
Bureau of Weights and Measures and is used only on rare occasions. It
remains the standard today. The kilogram is [was - Martin] the only unit
still defined in terms of an arbitrary artifact instead of a natural
phenomenon.
--
M.A.Poyser Tel.: 07967 110890
Manchester, U.K. http://www.fleetie.demon.co.uk

Fleetie wrote:

r^2 . d^2(r)/dt^2 = 2.G.m_b



This is a non-linear differential equation. See if that helps.


ve the
primary mass artifact, and is now defined, wait for it, thus: "The kilogram
is the mass of a body at rest whose equivalent energy equals the energy of
a collection of photons whose frequencies sum to 135 639 274 x 10^42 Hz."
That sure was a long time coming! BTW, don't believe me, believe NIST:


Oh,,thats handy. How the hell do you realise it ?

An accurately balanced piece of stuff is much handier for the shops.

Steve

"Fleetie" escribió en el mensaje
...
| Hi there.
|
| Erm, I've been playing some more with my gravitational simulation program,
| and I've been trying to verify mathematically the results I've been
| getting where I've run the program to simulate two masses* initially
| at rest, 1 metre distant from each other, and with no other gravitational
| field present.
|
| It doesn't take Einstein to work out that the force on each changes with
| the distance between them, so their acceleration changes too.
|
| Being a lowly ESN slug in the field of maths, I'm too lame to work out the
| maths to arrive at an equation for separation as a function of time, and
| thereby calculate how long it takes for the masses to meet.
|
| Obviously, given two masses a and b, the force on each is given by
| F = G . m_a . m_b / r^2, and acceleration a_a = G . m_b / r^2, thanks to
| Newton. But now, a_a and a_b depend strongly on r, so I can't use
| s = u.t + 0.5 . a . t^2 to solve for the time required for the two to
meet.
|
| I _think_ that it's true to say that acceleration, being the 2nd
derivative
| of position w.r.t. time, is given by:
|
| d^2(r)/dt^2 = 2.G.m_b/r^2,
|
| so
|
| r^2 . d^2(r)/dt^2 = 2.G.m_b
|
| But I'm not even sure about this, and even if so, I'm far too useless at
| maths to be able to take this any further. I really am stupid when it
comes
| to this! I assume some kind of integral over r is called for, intuitively,
| at least, but words cannot express my ****wittedness in such matters.
|
| Would someone be able to put me out of my dribbling, incompetent misery?
| I'd like to get an equation to tell me the time taken for the masses to
meet,
| assuming negligible (zero) radius. If you could walk me through the maths
| line by line, I'd really appreciate it. Remember, you're dealing with a
| mathematical amoeba here!
|
| FWIW, I _have_ googled for this, but maybe my search terms were crap,
because
| every page I found assumed constant gravitational acceleration, which is
NOT
| the case here.
|
|
| Thanks in advance,
|
|
| Martin
|
| * Their mass is irrelevant, at least if they're equal; I haven't checked
to
| see whether the same is true for unequal masses yet.
|
| By way of some kind of contribution "in return", for those who may be
| interested, and who may yet be unaware, I discovered late this week, that
| the definition of the kilogram has finally been changed so as to remove
the
| need for a physical artifact, thus: (Pasted from a post I made in another
| forum)
|
| The kilogram has FINALLY (as of 1999) been redefined so as to remove the
| primary mass artifact, and is now defined, wait for it, thus: "The
kilogram
| is the mass of a body at rest whose equivalent energy equals the energy of
| a collection of photons whose frequencies sum to 135 639 274 x 10^42 Hz."
| That sure was a long time coming! BTW, don't believe me, believe NIST:
|
| http://physics.nist.gov/TechAct.99/D...constants.html
|
| Formerly, the kilogram standard was as follows: In 1889 the
| international prototype kilogram was adopted as the standard for mass. The
| prototype kilogram is a platinum-iridium cylinder with equal height and
| diameter of 3.9 cm and slightly rounded edges. For a cylinder, these
| dimensions present the smallest surface area to volume ratio to minimize
| wear. The standard is carefully preserved in a vault at the International
| Bureau of Weights and Measures and is used only on rare occasions. It
| remains the standard today. The kilogram is [was - Martin] the only unit
| still defined in terms of an arbitrary artifact instead of a natural
| phenomenon.
| --
| M.A.Poyser Tel.: 07967
110890
| Manchester, U.K.
http://www.fleetie.demon.co.uk
|
|

Hello...

Let's take the masses as very different, with the Earth's big mass Me and a
very small stone with much smaller mass Ms. The attraction force F on the
stone at a distance r would be G*(Me*Ms)/r^2.

But force is mass times acceleration, so, in general a=f/m, and in our case,
forgetting the smallish Ms,

a = f/m = (G*Me)/r^2

and a = v*dv/dr

equating...

v*dv/dr = (G*Me)/r^2

multiplying both sides by dr...

v*dv = (G*Me)*(dr/r^2)

integrating both sides, taking out the constants...

[INT from v1 to v2] v*dv = (G*Me)*[INT from r1 to r2] (dr/r^2)

Where v1 and v2 are the velocities at the distances r1 and r2. Now, after
some manipulations...

v2^2 - v1^2 = 2*G*Me*(1/r2 - 1/r1)

HTH

Rgds

Javier

"Steve Taylor"
Oh,,thats handy. How the hell do you realise it ?

An accurately balanced piece of stuff is much handier for the shops.

Steve

Well, I wouldn't worry yet; I think it's going to be a while
before grocers are legally required to sell veg by the Hertz.

Thanks for the replies so far. I'll have a detailed look through the
maths later this afternoon/evening, I think.


Martin
--
M.A.Poyser Tel.: 07967 110890
Manchester, U.K. http://www.fleetie.demon.co.uk

"J.S." escribió en el mensaje
...
|
| "Fleetie" escribió en el mensaje
| ...
| | Hi there.
| |
| | Erm, I've been playing some more with my gravitational simulation
program,
| | and I've been trying to verify mathematically the results I've been
| | getting where I've run the program to simulate two masses* initially
| | at rest, 1 metre distant from each other, and with no other
gravitational
| | field present.
| |
| | It doesn't take Einstein to work out that the force on each changes with
| | the distance between them, so their acceleration changes too.
| |
| | Being a lowly ESN slug in the field of maths, I'm too lame to work out
the
| | maths to arrive at an equation for separation as a function of time, and
| | thereby calculate how long it takes for the masses to meet.
| |
| | Obviously, given two masses a and b, the force on each is given by
| | F = G . m_a . m_b / r^2, and acceleration a_a = G . m_b / r^2, thanks to
| | Newton. But now, a_a and a_b depend strongly on r, so I can't use
| | s = u.t + 0.5 . a . t^2 to solve for the time required for the two to
| meet.
| |
| | I _think_ that it's true to say that acceleration, being the 2nd
| derivative
| | of position w.r.t. time, is given by:
| |
| | d^2(r)/dt^2 = 2.G.m_b/r^2,
| |
| | so
| |
| | r^2 . d^2(r)/dt^2 = 2.G.m_b
| |
| | But I'm not even sure about this, and even if so, I'm far too useless at
| | maths to be able to take this any further. I really am stupid when it
| comes
| | to this! I assume some kind of integral over r is called for,
intuitively,
| | at least, but words cannot express my ****wittedness in such matters.
| |
| | Would someone be able to put me out of my dribbling, incompetent misery?
| | I'd like to get an equation to tell me the time taken for the masses to
| meet,
| | assuming negligible (zero) radius. If you could walk me through the
maths
| | line by line, I'd really appreciate it. Remember, you're dealing with a
| | mathematical amoeba here!
| |
| | FWIW, I _have_ googled for this, but maybe my search terms were crap,
| because
| | every page I found assumed constant gravitational acceleration, which is
| NOT
| | the case here.
| |
| |
| | Thanks in advance,
| |
| |
| | Martin
| |
| | * Their mass is irrelevant, at least if they're equal; I haven't checked
| to
| | see whether the same is true for unequal masses yet.
| |
| | By way of some kind of contribution "in return", for those who may be
| | interested, and who may yet be unaware, I discovered late this week,
that
| | the definition of the kilogram has finally been changed so as to remove
| the
| | need for a physical artifact, thus: (Pasted from a post I made in
another
| | forum)
| |
| | The kilogram has FINALLY (as of 1999) been redefined so as to remove the
| | primary mass artifact, and is now defined, wait for it, thus: "The
| kilogram
| | is the mass of a body at rest whose equivalent energy equals the energy
of
| | a collection of photons whose frequencies sum to 135 639 274 x 10^42
Hz."
| | That sure was a long time coming! BTW, don't believe me, believe NIST:
| |
| | http://physics.nist.gov/TechAct.99/D...constants.html
| |
| | Formerly, the kilogram standard was as follows: In 1889 the
| | international prototype kilogram was adopted as the standard for mass.
The
| | prototype kilogram is a platinum-iridium cylinder with equal height and
| | diameter of 3.9 cm and slightly rounded edges. For a cylinder, these
| | dimensions present the smallest surface area to volume ratio to minimize
| | wear. The standard is carefully preserved in a vault at the
International
| | Bureau of Weights and Measures and is used only on rare occasions. It
| | remains the standard today. The kilogram is [was - Martin] the only unit
| | still defined in terms of an arbitrary artifact instead of a natural
| | phenomenon.
| | --
| | M.A.Poyser Tel.: 07967
| 110890
| | Manchester, U.K.
| http://www.fleetie.demon.co.uk
| |
| |
|
| Hello...
|
| Let's take the masses as very different, with the Earth's big mass Me and
a
| very small stone with much smaller mass Ms. The attraction force F on the
| stone at a distance r would be G*(Me*Ms)/r^2.
|
| But force is mass times acceleration, so, in general a=f/m, and in our
case,
| forgetting the smallish Ms,
|
| a = f/m = (G*Me)/r^2
|
| and a = v*dv/dr
|
| equating...
|
| v*dv/dr = (G*Me)/r^2
|
| multiplying both sides by dr...
|
| v*dv = (G*Me)*(dr/r^2)
|
| integrating both sides, taking out the constants...
|
| [INT from v1 to v2] v*dv = (G*Me)*[INT from r1 to r2] (dr/r^2)
|
| Where v1 and v2 are the velocities at the distances r1 and r2. Now, after
| some manipulations...
|
| v2^2 - v1^2 = 2*G*Me*(1/r2 - 1/r1)
|
| HTH
|
| Rgds
|
| Javier

But... you wanted the time...

Well, I believe we can get it from the velocities, as v^2 = r^2/t^2

So the left side becomes ((r2^2)/(t^2))-((r1^2)/(t^2)), or, leaving t^2 as
the common denominator, ((r2^2)-(r1^2))/t^2

whence:

t^2 = ((r2^2) - (r1^2))/2*G*Me(1/r2 - 1/r1)

IIDGIW / If I didn't get it wrong...

Rgds

Javier

But... you wanted the time...

Well, I believe we can get it from the velocities, as v^2 = r^2/t^2

So the left side becomes ((r2^2)/(t^2))-((r1^2)/(t^2)), or, leaving t^2 as
the common denominator, ((r2^2)-(r1^2))/t^2

whence:

t^2 = ((r2^2) - (r1^2))/2*G*Me(1/r2 - 1/r1)

IIDGIW / If I didn't get it wrong...

Rgds

Javier

Hi.

Firstly, here, have some scissors: 8

Secondly, thanks. But, being the unworthy dufus that I am, I'm still too
thick to get it. I think you mean that r1 and r2 are two separations;
an initial separation and a final separation?

Well I'm asking about the case wherein the two masses (of negligible
radius) end up meeting, i.e. r2=0. So the maths blows up. Yet in
real life (apart from the fact that negligible radius isn't possible),
it is possible for the two to meet.

When r2=0, your equation becomes:

t^2 = -r1^2 / 2*G*Me(1/0 - 1/r1)

Or, if we restrict ourselves to finite cases wherein r2 is merely
"very small", then as r2 approaches zero, then

t^2 ~= -r1^2 / 2*G*Me*(BIG NUMBER)

So it seems as r2 approaches zero, t^2 also approaches zero.

I am definitely misunderstanding something here, but I told you
I have all the flair for maths of a decaying rat corpse.


Martin
--
M.A.Poyser Tel.: 07967 110890
Manchester, U.K. http://www.fleetie.demon.co.uk

JRS: In article ,
seen in news:uk.sci.astronomy, Fleetie
posted at Sat, 20 Dec 2003 20:11:06 :-

Would someone be able to put me out of my dribbling, incompetent misery?
I'd like to get an equation to tell me the time taken for the masses to meet,
assuming negligible (zero) radius. If you could walk me through the maths
line by line, I'd really appreciate it. Remember, you're dealing with a
mathematical amoeba here!

Read "Jupiter Five", by ACC. Does it for very different masses, though.

Also read my Gravity Web pages.

The general situation was treated by Kepler & Newton, et al.


The case of one of two equal masses starting one metre apart is the same
as the case of one mass starting half a metre from a fixed body of a
quarter of the mass, by symmetry.

The time to meet should therefore be 0.5^2.5 of the orbital period,
which is a factor of about
0.176776695296636881100211090526212259821208984422 118509147

Check that reasoning though, I've not paid much attention to equal
masses before. More figures are available if needed.



You wrote :

The kilogram has FINALLY (as of 1999) been redefined so as to remove the
primary mass artifact, and is now defined, wait for it, thus: "The
kilogram is the mass of a body at rest whose equivalent energy equals
the energy of a collection of photons whose frequencies sum to 135 639
274 x 10^42 Hz." That sure was a long time coming! BTW, don't believe
me, believe NIST:

http://physics.nist.gov/TechAct.99/D...constants.html


NIST is not internationally authoritative, though it should if
interpreted correctly be reliable. Have you found that at BIPM, for
example? If it is indeed internationally standardised, I'm surprised
that I don't recall seeing mention in, for example, "Physics World".

Note how the Americans dislike referring to standards bodies elsewhere -
they've been talking about Lead seconds (absence of, lately) as if the
responsibility was theirs.

Artefact.

--
© John Stockton, Surrey, UK. Turnpike v4.00 MIME. ©
Web URL:http://www.merlyn.demon.co.uk/ - FAQqish topics, acronyms & links;
some Astro stuff via astro.htm, gravity0.htm; quotes.htm; pascal.htm; &c, &c.
No Encoding. Quotes before replies. Snip well. Write clearly. Don't Mail News.

"Fleetie" escribió en el mensaje
...
| But... you wanted the time...
|
| Well, I believe we can get it from the velocities, as v^2 = r^2/t^2
|
| So the left side becomes ((r2^2)/(t^2))-((r1^2)/(t^2)), or, leaving t^2
as
| the common denominator, ((r2^2)-(r1^2))/t^2
|
| whence:
|
| t^2 = ((r2^2) - (r1^2))/2*G*Me(1/r2 - 1/r1)
|
| IIDGIW / If I didn't get it wrong...
|
| Rgds
|
| Javier
|
| Hi.
|
| Firstly, here, have some scissors: 8
|
| Secondly, thanks. But, being the unworthy dufus that I am, I'm still too
| thick to get it. I think you mean that r1 and r2 are two separations;
| an initial separation and a final separation?



**************
Hello...

Yes, r2 and r1 were the initial and final (after the fall, I mean)
separations of the stone from the Earth... I made a test with the velocity,
for a fall from 100000 km and got 2823 m/s, while the figure would be 44271
m/s with a constant gravity... But now that I'm checking my scribblings, I
see there's some minus sign lacking in the expression, because the radicand
is negative... I believe, however, that the result (2823) is right... (after
that 'correction'...)
**************



|
| Well I'm asking about the case wherein the two masses (of negligible
| radius) end up meeting, i.e. r2=0. So the maths blows up. Yet in
| real life (apart from the fact that negligible radius isn't possible),
| it is possible for the two to meet.
|
| When r2=0, your equation becomes:
|
| t^2 = -r1^2 / 2*G*Me(1/0 - 1/r1)
|
| Or, if we restrict ourselves to finite cases wherein r2 is merely
| "very small", then as r2 approaches zero, then
|
| t^2 ~= -r1^2 / 2*G*Me*(BIG NUMBER)
|
| So it seems as r2 approaches zero, t^2 also approaches zero.
|



**************

Same 'problem' as above, probably. I got 35414 seconds (9,84 hours) for the
fall time from 100000 km, against 1,25 hours for the same fall under a
constant field. Sounds plausible...

**************




| I am definitely misunderstanding something here, but I told you
| I have all the flair for maths of a decaying rat corpse.
|



***************

Mine is worse, but I keep trying... Not sure if it's very wise to be so
stubborn...

Regards

Javier

***************


|
| Martin
| --
| M.A.Poyser Tel.: 07967
110890
| Manchester, U.K.
http://www.fleetie.demon.co.uk
|
|

JRS: In article , seen in
news:uk.sci.astronomy, Dr John Stockton posted
at Sun, 21 Dec 2003 17:36:48 :-

Fleetie wrote :

The kilogram has FINALLY (as of 1999) been redefined so as to remove the
primary mass artifact, and is now defined, wait for it, thus: "The
kilogram is the mass of a body at rest whose equivalent energy equals
the energy of a collection of photons whose frequencies sum to 135 639
274 x 10^42 Hz." That sure was a long time coming! BTW, don't believe
me, believe NIST:

http://physics.nist.gov/TechAct.99/D...constants.html


NIST is not internationally authoritative, though it should if
interpreted correctly be reliable. Have you found that at BIPM, for
example? If it is indeed internationally standardised, I'm surprised
that I don't recall seeing mention in, for example, "Physics World".


That page actually says :

Redefinition of the Kilogram. Motivated by recent NIST advances in
determining the Planck constant h using a moving-coil watt balance, B.N.
Taylor and P.J. Mohr considered the question of redefining the kilogram
in such a way that the value of h would be fixed, thereby allowing a
watt balance to be used to directly calibrate unknown standards of mass
[see Metrologia 36, 63 (1999)]. The proposed definition, analogous to
the present definition of the meter which has the effect of fixing the
value of the speed of light in vacuum c, is "The kilogram is the mass of
a body at rest whose equivalent energy equals the energy of a collection
of photons whose frequencies sum to 135 639 274 x 10^42 Hz." Based on
the equations E = mc^2 and E = h , this definition implies h = 6.626 068
9 ... x 10^-34 J s. Such a definition would eliminate the last material
artifact from the SI and would allow any laboratory in the world that
was so inclined to realize the unit of mass in the SI.


That is, very clearly, a mere proposal. The kilogram has not been
redefined, although it is likely that it will be before very long.

"The kilogram is the unit of mass; it is equal to the mass of the
international prototype of the kilogram." remains TRUE.


No redefinition of the SI units can be described as FINAL, since a
meeting of the appropriate authorities cannot bind its successors. It
is conceivable that new physics might imply new relationships between
quantities, of a nature that the numbers (such as 6.626E-34, 299E6,
etc.) were mathematically interrelated; or that a new experimental
technique might make it possible to build a perfect diamond kilogram by
assembling a counted number of C12 atoms. Those would require new
definitions of fundamental constants.


And that Webster fellow is unreliable on spelling still.

--
© John Stockton, Surrey, UK. / ©
Web URL:http://www.merlyn.demon.co.uk/ - FAQish topics, acronyms, & links.
Correct = 4-line sig. separator as above, a line precisely "-- " (SoRFC1036)
Do not Mail News to me. Before a reply, quote with "" or " " (SoRFC1036)

Artefact.

Hmm, dictionary.com tells me both are acceptable, but dictionary.com
has American spellings. I did wonder when I typed it the first few
times recently, but "artifact" "looked" right. My English-Chinese
dictionary has "artefact", and I can't be bothered digging out the
OED CD-ROM I have, which for some reason I can't seem to make install
on my hard disk; it always seems to want the CD to be in the drive,
which seems pretty stupid to me.

Oh, strike the above; I just did bother.

OED (which I regard as THE final authority; others may differ) lists
"artifact" as "artifact n. var. of ARTEFACT.". So, I maintain it's an
acceptable spelling, though perhaps not the "primary" spelling of the
word.

I'm afraid that I don't know what "the orbital period" would be in
the situation you describe, so your ratio isn't any use to me at the
moment; that's not meant to sound disparaging; merely an admission of
ignorance.


Martin
--
M.A.Poyser Tel.: 07967 110890
Manchester, U.K. http://www.fleetie.demon.co.uk