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View Full Version : simple physics question can acceleration create equality in tidal forces?



tommac
2010-Apr-23, 03:26 PM
In single dimentional space i I have a gravitational source X, and 3 items accelerating towards X, at a distance, would the tidal forces in both directions be equal?

X----------------------------------------------------------------------------1----2----3

as 1,2,3 accelerate towards X ( lets say X is a black hole or neutron star )

you will have a situation where 2 will see 1 accelerate away and 2 will see 3 accelerate away in the opposite direction correct?

2 will see 1 accelerate away faster than 3 but at a distance would these be approximate say within a 1% margin of error?

To calculate the tidal acceleration do I take the difference of acceleration between X and 1 and X and 2 and that will produce the relative acceleration of 1 and 2?


To calculate the acceleration between a point and x would the formula be:

G( M + m ) / r^2

lets assume that M is much more massive than m so we have GM/r^2

so we should have:

GM(r1)^2 - GM/(r2)^2

that is the difference in acceleration correct?

GM ( 1/r1^2 - 1/r2^2 )

if r1= 100000 and r2 = 100001

then would

a=GM ( 1/10000000000 - 1/10000200001) =~ GM * 2 e -15


if r3 = 100002 then would
a =~ GM * 2 e -15

also ... so my question is per my caclulations is that the margin of error between the two accelerations is less than .0001% is that a correct assumption?

tusenfem
2010-Apr-23, 05:29 PM
My goodness man, why not just "solve" the equotion of motion, assuming they are starting from stand still and the get X(t) through the very simple equation that x(t) = x0 + 0.5 a t2.
What is the acceleration of "da big X"? Naturally you will find that it is a = G M / x2, so heyyyyyyy you cannot take the x(t) that I gave you, you have to solve the real equotion of motion d2x/dt2 = a(x,t).
Now if you have x(t) for object A, B and C then you just subtract A from B and B from C and you have your solution.
This is simple undergraduate physics, which you should be able to do, when you claim to read all those highly advanced papers.

tusenfem
2010-Apr-23, 05:31 PM
To calculate the acceleration between a point and x would the formula be:

G( M + m ) / r^2


And why is this equation WRONG!!!!!!!!!!!!!!!!!!??????????????????????

Ken G
2010-Apr-23, 06:21 PM
To calculate the tidal acceleration do I take the difference of acceleration between X and 1 and X and 2 and that will produce the relative acceleration of 1 and 2? Yes.


lets assume that M is much more massive than m so we have GM/r^2
Yes, you must assume that.


so we should have:

GM(r1)^2 - GM/(r2)^2

that is the difference in acceleration correct? Yes.


then would

a=GM ( 1/10000000000 - 1/10000200001) =~ GM * 2 e -15
Yes. The best way to do it is write r1 = r2 - dr, and expand 1/r12 as a Taylor series, obtaining 1/r22 * (1 + 2dr/r2). Then the tidal part is just the second term, which in your case shows the tidal acceleration is 2 parts in 105 times the acceleration at r2.


if r3 = 100002 then would
a =~ GM * 2 e -15
Yes, since r3-r2 differs from r1-r2 only in sign, so will the acceleration difference.


also ... so my question is per my caclulations is that the margin of error between the two accelerations is less than .0001% is that a correct assumption?
The difference between those accelerations is irrelevant, that's an even higher order effect than the tidal effect. The tidal effect is at the order 2 dr/r2, and the difference between r1 and r3 is just a difference in the direction of the tidal acceleration.

tommac
2010-Apr-23, 07:31 PM
Thank you for the help.

So just one slight question to 100% clear up my confusion, the force and acceleration between three distant points 12 vs 23
X----------------------------------------------1-------2-------3

is approximately the same right, just in the opposite direction, correct? And it is linear with delta r ( the distance between 1 and 2 and 2 and 3 ).


Yes.

Yes, you must assume that.
Yes.
Yes. The best way to do it is write r1 = r2 - dr, and expand 1/r12 as a Taylor series, obtaining 1/r22 * (1 + 2dr/r2). Then the tidal part is just the second term, which in your case shows the tidal acceleration is 2 parts in 105 times the acceleration at r2.
Yes, since r3-r2 differs from r1-r2 only in sign, so will the acceleration difference.

The difference between those accelerations is irrelevant, that's an even higher order effect than the tidal effect. The tidal effect is at the order 2 dr/r2, and the difference between r1 and r3 is just a difference in the direction of the tidal acceleration.

Ken G
2010-Apr-23, 08:42 PM
Yes.

tusenfem
2010-Apr-23, 09:09 PM
as long as you keep δr << R

tommac
2010-Apr-23, 09:15 PM
as long as you keep δr << R

Yes agreed.