If a body begins accelerating at 9.8 s/s how long would it take to travel 27,000 light years(assuming c restraint is removed)?

Is there an opposite to an inverse square equation.

I’ll add further questions as they occur to me.

Postpocelipse said:

If a body begins accelerating at 9.8 s/s how long would it take to travel 27,000 light years(assuming c restraint is removed)?

Is there an opposite to an inverse square equation.

I’ll add further questions as they occur to me.

d = vt x 1/2at ²

where
d = distance
v = initial velocity
a = acceleration
t = time

Assuming the body starts at rest (v = 0) then the equation simplifies to

d = 1/2at ²

27,000 ly = 2.55434267 × 10 ²⁰ m

therefore

2.55434267 × 10 ²⁰ = 1/2*9.81*t ²

solve for t (in seconds)

thank you d_s.

remember what I said about needing to learn the fundamentals…

kinematics is high school stuff…

you’ll never be able to properly explain yourself if you can’t solve even simple problems.

diddly-squat said:

2.55434267 × 10 ²⁰ = 1/2*9.81*t ²

solve for t (in seconds)

OK

7.2164 billion seconds, or about 228.67 years

diddly-squat said:

remember what I said about needing to learn the fundamentals…

kinematics is high school stuff…

you’ll never be able to properly explain yourself if you can’t solve even simple problems.

Yes fair enough. I don’t have a bad head for maths I just found trying to absorb complex equation fundamentals only confused the relativity exercise I had developed. I figured out the source of that confusion by finding the initial resolution within this exercise. There are further resolutions to it but now that I have defined the initial resolution I can start understanding the mathematical path to the greater solutions within it.

The Rev Dodgson said:

diddly-squat said:

2.55434267 × 10 ²⁰ = 1/2*9.81*t ²

solve for t (in seconds)

OK

7.2164 billion seconds, or about 228.67 years

I included this just to start getting my bearings on linear calculations. It also gives me something mathematical to compare to the restrictions relativity places against acceleration. Thanks for your time on it

A question related to my query regarding a mirror equation to an inverse square. Is terminal velocity mathematically defined in respect to c?

Postpocelipse said:

A question related to my query regarding a mirror equation to an inverse square. Is terminal velocity mathematically defined in respect to c?

only if by ‘c’ you mean drag coefficient

diddly-squat said:

Postpocelipse said:

A question related to my query regarding a mirror equation to an inverse square. Is terminal velocity mathematically defined in respect to c?

only if by ‘c’ you mean drag coefficient

hmmmm. I don’t think that would be what I am looking for.

What is the significance of 27,000 years?

Other than being 30^3.

The Rev Dodgson said:

diddly-squat said:

2.55434267 × 10 ²⁰ = 1/2*9.81*t ²

solve for t (in seconds)

OK

7.2164 billion seconds, or about 228.67 years

including leap years?

The Rev Dodgson said:

What is the significance of 27,000 years?

Other than being 30^3.

Arbitrary figure I chose because it is the approximate distance to the centre of the galaxy. Mostly wanted a big number.

stumpy_seahorse said:

The Rev Dodgson said:

diddly-squat said:

2.55434267 × 10 ²⁰ = 1/2*9.81*t ²

solve for t (in seconds)

OK

7.2164 billion seconds, or about 228.67 years

including leap years?

He’s an engineer .. Close enough

stumpy_seahorse said:

including leap years?

Yes :)

The Rev Dodgson said:

stumpy_seahorse said:

including leap years?

Yes :)

how else would you get there without slowing down?

Postpocelipse said:

If a body begins accelerating at 9.8 s/s how long would it take to travel 27,000 light years(assuming c restraint is removed)?

Is there an opposite to an inverse square equation.

I’ll add further questions as they occur to me.

228.67 years

A further question following on from that is:

If a body begins accelerating at 9.8 m/s^2 how long would it take to travel 27,000 light years (constant force so acceleration reduces as approaching c, but taking into account Fitzgerald contraction so distance is reduced)?

mollwollfumble said:

Postpocelipse said:

If a body begins accelerating at 9.8 s/s how long would it take to travel 27,000 light years(assuming c restraint is removed)?

Is there an opposite to an inverse square equation.

I’ll add further questions as they occur to me.

228.67 years

A further question following on from that is:

If a body begins accelerating at 9.8 m/s^2 how long would it take to travel 27,000 light years (constant force so acceleration reduces as approaching c, but taking into account Fitzgerald contraction so distance is reduced)?

I’m betting I won’t have an answer first but you just gave me excellent stuff to look up. :)

Fitzgerald contraction is something I’ve been looking for a name to. :)

Postpocelipse said:

Fitzgerald contraction is something I’ve been looking for a name to. :)

In the instance a particle is accelerating toward a massive body at terminal velocity, how is the space of the massive body modelled from the frame of reference of the accelerating particle? I’ve been picturing it as the mass of the massive body flattening out as a 2d plane in opposition to direction of travel. At terminal velocity this would be completely flat. Accelerate past terminal velocity and the edges of the 2d plane begin to curve toward the particle. Enough acceleration and the 2d plane representing the total centre of momentum flips so that the accelerating particle has centre of mass?

I haven’t been able to find anything that covers this scenario. If I could approximate a reference I might have some idea of the maths I am trying to understand.

In the instance a particle is accelerating toward a massive body at terminal velocity…

how can it be accelerating if it is at terminal velocity?

JudgeMental said:

In the instance a particle is accelerating toward a massive body at terminal velocity…

how can it be accelerating if it is at terminal velocity?

Sorry. Another instance of my appropriating terminology incorrectly. Gravetic constant assuming no atmosphere seems a bit wieldy.

Gravetic constant assuming no atmosphere seems a bit wieldy.

yeah, best to keep the bull simple that way you’ll be able to keep up.

Postpocelipse said:

Postpocelipse said:

Fitzgerald contraction is something I’ve been looking for a name to. :)

In the instance a particle is accelerating toward a massive body at terminal velocity, how is the space of the massive body modelled from the frame of reference of the accelerating particle? I’ve been picturing it as the mass of the massive body flattening out as a 2d plane in opposition to direction of travel. At terminal velocity this would be completely flat. Accelerate past terminal velocity and the edges of the 2d plane begin to curve toward the particle. Enough acceleration and the 2d plane representing the total centre of momentum flips so that the accelerating particle has centre of mass?

I haven’t been able to find anything that covers this scenario. If I could approximate a reference I might have some idea of the maths I am trying to understand.

Um, reference. As a book I like “Subtle is the lord, the science and life of Albert Einstein” – a shockingly bad title for an excellent book.

For web reference, try this webpage with equations and calculator http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html

Postpocelipse said:

Postpocelipse said:

Fitzgerald contraction is something I’ve been looking for a name to. :)

In the instance a particle is accelerating gravetically toward a massive body without intervening atmosphere, how is the space of the massive body modelled from the frame of reference of the accelerating particle? I’ve been picturing it as the mass of the massive body flattening out as a 2d plane in opposition to direction of travel. At 9.8 s/s this would be completely flat. Accelerate past 9.8 s/s and the edges of the 2d plane begin to curve toward the particle. Enough acceleration and the 2d plane representing the total centre of momentum flips so that the accelerating particle has centre of mass?

I haven’t been able to find anything that covers this scenario. If I could approximate a reference I might have some idea of the maths I am trying to understand.

As fixed as I can manage till I look up some stuff on gravetic acceleration.

JudgeMental said:

Gravetic constant assuming no atmosphere seems a bit wieldy.

yeah, best to keep the bull simple that way you’ll be able to keep up.

that advice is free right?

mollwollfumble said:

Postpocelipse said:

Postpocelipse said:

Fitzgerald contraction is something I’ve been looking for a name to. :)

In the instance a particle is accelerating toward a massive body at terminal velocity, how is the space of the massive body modelled from the frame of reference of the accelerating particle? I’ve been picturing it as the mass of the massive body flattening out as a 2d plane in opposition to direction of travel. At terminal velocity this would be completely flat. Accelerate past terminal velocity and the edges of the 2d plane begin to curve toward the particle. Enough acceleration and the 2d plane representing the total centre of momentum flips so that the accelerating particle has centre of mass?

I haven’t been able to find anything that covers this scenario. If I could approximate a reference I might have some idea of the maths I am trying to understand.

Um, reference. As a book I like “Subtle is the lord, the science and life of Albert Einstein” – a shockingly bad title for an excellent book.

For web reference, try this webpage with equations and calculator http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html

ta. Sorry about including terminal velocity in this. A long time ago I had the meaning understood but somewhere along the line I think I just decided that terminal velocity is a good name for gravetic acceleration.

Postpocelipse said:

In the instance a particle is accelerating gravetically toward a massive body without intervening atmosphere, how is the space of the massive body modelled from the frame of reference of the accelerating particle? I’ve been picturing it as the mass of the massive body flattening out as a 2d plane in opposition to direction of travel. At 9.8 s/s this would be completely flat. Accelerate past 9.8 s/s and the edges of the 2d plane begin to curve toward the particle. Enough acceleration and the 2d plane representing the total centre of momentum flips so that the accelerating particle has centre of mass?

I think I’m trying to understand Fitzgerald contraction in a gravetic scenario

mollwollfumble said:

Postpocelipse said:

If a body begins accelerating at 9.8 s/s how long would it take to travel 27,000 light years(assuming c restraint is removed)?

Is there an opposite to an inverse square equation.

I’ll add further questions as they occur to me.

228.67 years

A further question following on from that is:

If a body begins accelerating at 9.8 m/s^2 how long would it take to travel 27,000 light years (constant force so acceleration reduces as approaching c, but taking into account Fitzgerald contraction so distance is reduced)?

I suspect that the answer in shipboard time is still 228.67 years, or close to it, even taking into account the limit of the speed of c.

My hand-waving argument runs as follows.
Let gamma = square root (1/(1-v^2/c^2))
Using relativistic mass, m = m_0 * gamma
F = m * (d^2 x/dt^2)
F / m_0 = (d^2 x/dt^2) * gamma
But F / m0 is a constant so d^2 x/dt^2 is proportional to 1 / gamma
Integrating once dx / dt is proportional to 1 / gamma
Integrating a second time x is proportional to 1 / gamma
So x * gamma is a constant
But from Fitzgerald contraction the un-contracted distance (ie. 27,000 light years) is x * gamma which is a constant.

So gamma drops out of the equation and Newtonian physics applies, which gives 228.67 years.

Do you see where I cheated, slightly? I ignored the time integration of 1 / gamma. But 1 / gamma is going to be close to zero when v approaches c so I’m banking on the time integration of 1 / gamma being small.

mollwollfumble said:

mollwollfumble said:

Postpocelipse said:

If a body begins accelerating at 9.8 s/s how long would it take to travel 27,000 light years(assuming c restraint is removed)?

Is there an opposite to an inverse square equation.

I’ll add further questions as they occur to me.

228.67 years

A further question following on from that is:

If a body begins accelerating at 9.8 m/s^2 how long would it take to travel 27,000 light years (constant force so acceleration reduces as approaching c, but taking into account Fitzgerald contraction so distance is reduced)?

I suspect that the answer in shipboard time is still 228.67 years, or close to it, even taking into account the limit of the speed of c.

My hand-waving argument runs as follows.
Let gamma = square root (1/(1-v^2/c^2))
Using relativistic mass, m = m_0 * gamma
F = m * (d^2 x/dt^2)
F / m_0 = (d^2 x/dt^2) * gamma
But F / m0 is a constant so d^2 x/dt^2 is proportional to 1 / gamma
Integrating once dx / dt is proportional to 1 / gamma
Integrating a second time x is proportional to 1 / gamma
So x * gamma is a constant
But from Fitzgerald contraction the un-contracted distance (ie. 27,000 light years) is x * gamma which is a constant.

So gamma drops out of the equation and Newtonian physics applies, which gives 228.67 years.

Do you see where I cheated, slightly? I ignored the time integration of 1 / gamma. But 1 / gamma is going to be close to zero when v approaches c so I’m banking on the time integration of 1 / gamma being small.

I couldn’t show that you cheated mathematically or logically. You have given me another way to look at that gravity acceleration question though. That is refreshing. Been trying to figure out how to find another aprroximation for it for a while. :)

Holy shit. I think you have given me exactly what I was looking for when I started this thread. At least the basic approach that has to be given to it. Now I just have to get my head round your handwaving explanation. :D

mollwollfumble said:

Do you see where I cheated, slightly? I ignored the time integration of 1 / gamma. But 1 / gamma is going to be close to zero when v approaches c so I’m banking on the time integration of 1 / gamma being small.

Yes. Cheating should be justified. Point illustration is forgiveable. I won’t ask you to justify that with physical measurement of the question. No thank you very much but I would very much prefer you handwave and argue……

:D night Moll

> Do you see where I cheated, slightly?

Trying to get a much more accurate answer by integrating the equations in Excel. Unfortunately, so far the solution goes unstable at velocities above 99.99987% of the speed of light. Will try again later.

mollwollfumble said:

> Do you see where I cheated, slightly?

Trying to get a much more accurate answer by integrating the equations in Excel. Unfortunately, so far the solution goes unstable at velocities above 99.99987% of the speed of light. Will try again later.

pfff. If it only goes unstable at 99.99987% then I’ll just stay within aaaawww .00 something 1 until that unstable range is well clear and defined in all consequential definitives. That should give me some nice clear time to contemplate the intervening %‘s. Wave on good sir….

:D

mollwollfumble said:

> Do you see where I cheated, slightly?

Trying to get a much more accurate answer by integrating the equations in Excel. Unfortunately, so far the solution goes unstable at velocities above 99.99987% of the speed of light. Will try again later.

By that speed the spacecraft has travelled 6.3 light years (distance measured at starting point) in 1.521 years (time measured on spacecraft).

mollwollfumble said:

mollwollfumble said:

> Do you see where I cheated, slightly?

Trying to get a much more accurate answer by integrating the equations in Excel. Unfortunately, so far the solution goes unstable at velocities above 99.99987% of the speed of light. Will try again later.

By that speed the spacecraft has travelled 6.3 light years (distance measured at starting point) in 1.521 years (time measured on spacecraft).

So radius and mass??

Postpocelipse said:

So radius and mass??

Just absorbing. No need to detail.

http://newt.phys.unsw.edu.au/einsteinlight/index.html

JudgeMental said:

http://newt.phys.unsw.edu.au/einsteinlight/index.html

For tonight knowing not to drop anything heavier than a fart should keep me sleeping. Especially if I’m going to let it hit the ground. :/

Postpocelipse said:

JudgeMental said:

http://newt.phys.unsw.edu.au/einsteinlight/index.html

For tonight knowing not to drop anything heavier than a fart should keep me sleeping. Especially if I’m going to let it hit the ground. :/

Khan.com

mollwollfumble said:

mollwollfumble said:

> Do you see where I cheated, slightly?

Trying to get a much more accurate answer by integrating the equations in Excel. Unfortunately, so far the solution goes unstable at velocities above 99.99987% of the speed of light. Will try again later.

By that speed the spacecraft has travelled 6.3 light years (distance measured at starting point) in 1.521 years (time measured on spacecraft).

I hadn’t defined why they were using the term radiative forcing in the climate change question. Cheers for the wave off Moll.

:D

mollwollfumble said:

mollwollfumble said:

Postpocelipse said:

If a body begins accelerating at 9.8 s/s how long would it take to travel 27,000 light years(assuming c restraint is removed)?

Is there an opposite to an inverse square equation.

I’ll add further questions as they occur to me.

228.67 years

A further question following on from that is:

If a body begins accelerating at 9.8 m/s^2 how long would it take to travel 27,000 light years (constant force so acceleration reduces as approaching c, but taking into account Fitzgerald contraction so distance is reduced)?

I suspect that the answer in shipboard time is still 228.67 years, or close to it, even taking into account the limit of the speed of c.

My hand-waving argument runs as follows.
Let gamma = square root (1/(1-v^2/c^2))
Using relativistic mass, m = m_0 * gamma
F = m * (d^2 x/dt^2)
F / m_0 = (d^2 x/dt^2) * gamma
But F / m0 is a constant so d^2 x/dt^2 is proportional to 1 / gamma
Integrating once dx / dt is proportional to 1 / gamma
Integrating a second time x is proportional to 1 / gamma
So x * gamma is a constant
But from Fitzgerald contraction the un-contracted distance (ie. 27,000 light years) is x * gamma which is a constant.

So gamma drops out of the equation and Newtonian physics applies, which gives 228.67 years.

Do you see where I cheated, slightly? I ignored the time integration of 1 / gamma. But 1 / gamma is going to be close to zero when v approaches c so I’m banking on the time integration of 1 / gamma being small.

OK, I’ve done a stable integration of the equations on Excel. Result is the spacecraft flies 27,000 light years (distance measured at starting point) in close to 1,736 shipboard years. – Unless I’ve made a slip-up in the calculation.

So that’s longer than the Newtonian 229 years, but still well short of 27,000 years.

With constant force and an initial acceleration of 9.8 m/s^2, and ignoring mass loss due to fuel usage, the crossover point is 16.9. It takes the spacecraft 16.9 years to travel 16.9 light years.

mollwollfumble said:

With constant force and an initial acceleration of 9.8 m/s^2, and ignoring mass loss due to fuel usage, the crossover point is 16.9. It takes the spacecraft 16.9 years to travel 16.9 light years.

That is a very low deflection threshold. Without a fuel requirement that speaks volumes.

The expression “terminal velocity” has two meanings in physics, but I get the feeling by your usage that you’re not referring to either of those. Usually, “terminal velocity” refers to the highest velocity attainable by an object in free fall in an atmosphere.

However, “terminal velocity” is sometimes used to refer to the highest velocity attainable by an object falling from an arbitrarily large distance towards a planet or star. If the planet has no atmosphere this is equal to the escape velocity.

And I don’t know what you mean by “gravetic acceleration”. Do you mean gravitational acceleration? Or perhaps a constant acceleration that is equal in magnitude to the gravitational acceleration on the Earth’s surface.

Anyway, you may be interested in some of the info on this page: The Relativistic Rocket

PM 2Ring said:

The expression “terminal velocity” has two meanings in physics, but I get the feeling by your usage that you’re not referring to either of those. Usually, “terminal velocity” refers to the highest velocity attainable by an object in free fall in an atmosphere.

However, “terminal velocity” is sometimes used to refer to the highest velocity attainable by an object falling from an arbitrarily large distance towards a planet or star. If the planet has no atmosphere this is equal to the escape velocity.

And I don’t know what you mean by “gravetic acceleration”. Do you mean gravitational acceleration? Or perhaps a constant acceleration that is equal in magnitude to the gravitational acceleration on the Earth’s surface.

Gravitational acceleration is what I intended. Mollwollfumble has provided the boundary equations I’ve been looking for very neatly. I’m hoping the equation he has assembled has been considered before now. If it hasn’t Moll rocks.

Moll I have it this way..Your original equation providing 98.1% = stationary test providing confinement equivalent to local centre of momentum

Equation providing 16.9yrs = freefall test provides centre of momentum confinement equivalent to path of supernova collapse

Fuel mass defines confinement of distribution to core with resolution.

Just roughly. A lot going on here so thank you very much for putting that together. My brain cramp is dealt with. Interested to see what else you make of it.

Moll the equations you have constructed represent my approach to isolating centre of momentum by measuring radial velocity. This is what I intended when I constructed the equation C = (i)e when -1 = m

If you don’t mind Moll could I get your email address? I just found something I’d like to run by you off the record.

Postpocelipse said:

Moll I have it this way..Your original equation providing 98.1% = stationary test providing confinement equivalent to local centre of momentum

Equation providing 16.9yrs = freefall test provides centre of momentum confinement equivalent to path of supernova collapse

Fuel mass defines confinement of distribution to core with resolution.

Just roughly. A lot going on here so thank you very much for putting that together. My brain cramp is dealt with. Interested to see what else you make of it.

Was still trying to clarify my non maths questions regarding the equation you illustrated. I’ll put that behind me I think. I can at least identify that the numbers you have crunched are my approach to defining the confinement of radial velocity.

what a fuck around it has been for me to get one little math exercise out of my head. wonder how long the stupid cramp is going to twitch.

:) I can just skip from one subject to another and never lose my place now. This is great. :)