# Space Septuplets

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Off in space you have a space station and two rogue "planets" that have broken away from their solar system.

The space station is a hollow shell of exotic material (negligible mass) and a has the same radius as Earth. It is spinning fast enough so that if you were "seatbelted" to the inside of the shell at the equator you would feel 1 "gee". (let's say there is a counter weight on the other side so you wouldn't wobble).

The first planet we'll call "Earth 1" as it has the same radius as Earth and the same mass. It is spinning at the same rate as the space station.

The second planet we'll call "Earth 2". It has the same radius as Earth, but twice the mass. It is also spinning at the same rate as the space station.

You have 7 "twins" who have volunteered for your experiment. Let's call them Archie, Bob, Chuck, Dave, Edgar, Frank and Gus.

A. Archie, stopwatch in hand is seat belted at the center of the space station

B. Bob is seat belted at the equator of the space station

C. Chuck is seat belted at the "North Pole" of Earth 1

D. Dave is seat belted at the equator of Earth 1

E Edgar is seatbelted at the North Pole of Earth 2

F. Frank is seat belted at the equator of Earth 2

Gus is given a 1 "gee" space pack and will take off and return in one year based on Archie's stopwatch. His job is to encircle the largest area possible in one year.

After one year, who has aged the most through least of the space septuplets, A, B, C, D, E, F, and G? They have a younger sister Hilda who's guess is ABCDEFG for no other reason than "that's the order they were born in".

What order would you put them in?

Are any the same?

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I will say that Archie has aged the most, as he is just sitting in one spot with no SR time dilation effect due to speed wrt the final inertial frame and no GR time dilation effect due to acceleration or being in a gravitational field. All the others have some time dilation effect/s.

So Hilda gets the first one right! (for the wrong reason)

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Archie shows no SR or Gravitional dilation

Bob only shows SR dilation (even though he is accelerating, this acceleration adds no additional time dilation.

Chuck shows only gravitational dilation, but it will be greater than the SR dilation shown by Bob.

Dave shows the same gravitational dilation as Chuck and in addition the SR dilation of Bob.

Edgar shows the only gravitational dilation but it is greater than the combined gravitational and SR dilation's shown by Dave.

Frank shows the gravitaional dilation of Edgar in addition to the SR dilation shown by Bob and Dave.

Gus shows only SR dilation due to his instantaneous speed at any given moment, but his average velocity will be high enough to show a greater time dilation than any of the others.

On a side note, an interesting result occurs if Earth 2 has both twice the mass and twice the Radius as Earth 1

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A, D, C and G, B and F, E, I thought, but then, I really know very little about GR or SR.

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in order of oldest to youngest: a,c,e,b,d,f,gus

because a (low speed, low gravity) and gus is experiencing the most (due to his speed) however the affects on the planets would be negligable and amount to only fractions of a second where gus would be significantly younger.

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There is a unit of kinematic dilation due to travelling at a speed v, the rotation speed of the station, and also a gravitational unit of dilation due to being in a gravitational well with an acceleration value of g. Let's call these KD and GD. Since kinematic dilation is v/2c2, this means that KD = GD/2

There is no kinematic effect on the planets, for pole vs. equator, as the effects of the equatorial bulge cancels this. All clocks on the geoid run at the same rate.

A has no dilation.

B has one unit of KD

C has one unit of GD

D has one unit of GD

E has two units of GD

F has two units of GD

G gets going reeaaally fast and has many units of KD.

So Hilda gets it all right, by accident (or by design of the problem) and allowing for the ties.

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There is a unit of kinematic dilation due to travelling at a speed v' date=' the rotation speed of the station, and also a gravitational unit of dilation due to being in a gravitational well with an acceleration value of g. Let's call these KD and GD. Since kinematic dilation is v/2c[sup']2[/sup], this means that KD = GD/2

There is no kinematic effect on the planets, for pole vs. equator, as the effects of the equatorial bulge cancels this. All clocks on the geoid run at the same rate.

A has no dilation.

B has one unit of KD

C has one unit of GD

D has one unit of GD

E has two units of GD

F has two units of GD

G gets going reeaaally fast and has many units of KD.

So Hilda gets it all right, by accident (or by design of the problem) and allowing for the ties.

I actually assumed no equatorial bulge (maybe we need exotic planet material as well or we end up with a pancake)

What difference would this make to the order?

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I actually assumed no equatorial bulge (maybe we need exotic planet material as well or we end up with a pancake)

What difference would this make to the order?

Due to the way the problem was set up, I also assumed no equatorial bulge. (so that Bob, Dave and Frank all traveled in circles of equal radius. )

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Archie shows no SR or Gravitional dilation

Bob only shows SR dilation (even though he is accelerating' date=' this acceleration adds no additional time dilation.

Chuck shows only gravitational dilation, but it will be greater than the SR dilation shown by Bob.

Dave shows the same gravitational dilation as Chuck and in addition the SR dilation of Bob.

Edgar shows the only gravitational dilation but it is greater than the combined gravitational and SR dilation's shown by Dave.

Frank shows the gravitaional dilation of Edgar in addition to the SR dilation shown by Bob and Dave.

Gus shows only SR dilation due to his instantaneous speed at any given moment, but his average velocity will be high enough to show a greater time dilation than any of the others.

On a side note, an interesting result occurs if Earth 2 has both twice the mass and twice the Radius as Earth 1[/quote']

Edited out my "brain cramp"

If Earth 3 shows up at twice the mass and twice the radius as Earth 1 with "Zach" at the North Pole and "Yuri" seat belted at the equator at what rate of spin does it become interesting as per your side note?

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These are two I find interesting. They are at the same speed and if you blindfold them they feel the same way. They cover the same distance. I would have thought they were "tied".
How do they end up feeling the same? Bob is traveling in a circle of Earth radius such that he feels 1g. Dave is traveling in the same circle around a planet of Earth mass, thus is essentially traveling at orbital speed and feels weightless. Perhaps you meant Bob and Frank? But even then, the local acceleration due to gravity or what you 'feel' locally does not determine what time dilation you will show.

If Earth 3 shows up at twice the mass and twice the radius as Earth 1 with "Zach" at the North Pole and "Yuri" seat belted at the equator at what rate of spin does it become interesting as per your side note?

The interesting part is that Zach and Chuck will show equal time dilation.(Though Zach will feel half the g-force that Chuck does.)

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How do they end up feeling the same? Bob is traveling in a circle of Earth radius such that he feels 1g. Dave is traveling in the same circle around a planet of Earth mass' date=' thus is essentially traveling at orbital speed and feels weightless. Perhaps you meant Bob and Frank? But even then, the local acceleration due to gravity or what you 'feel' locally does not determine what time dilation you will show.

The interesting part is that Zach and Chuck will show equal time dilation.(Though Zach will feel half the g-force that Chuck does.)[/quote']

OOPS! I might need "Xirb" to make my point. I will edit that post.

Xirb shows up in a space station that's twice the radius as Earth. He's seat belted at the equator and moving same speed as Bob so he feels half the force. He undergoes half the acceleration.

Are Bob and Xirb the same?

How fast is Zach moving?

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OOPS! I might need "Xirb" to make my point. I will edit that post.

Xirb shows up in a space station that's twice the radius as Earth. He's seat belted at the equator and moving same speed as Bob so he feels half the force. He undergoes half the acceleration.

Are Bob and Xirb the same?

in terms of time dilation' date=' yes.

How fast is Zach moving?

According to your post, Zach is sitting at the North pole of Earth 3, so he wouldn't be moving at all wrt Bob, who is sitting at the North pole of Earth 1.

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in terms of time dilation' date=' yes.

According to your post, Zach is sitting at the North pole of Earth 3, so he wouldn't be moving at all wrt[b'] Bob[/b], who is sitting at the North pole of Earth 1.

I assume you meant Chuck not Bob. I would have assumed Chuck's time dilation would be twice Zach's as only a gravitational component is present and Chuck's is twice Zach's. But this is not the case?

And there is no time dilation component due to acceleration in Bob's or Xirb's cases?

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I assume you meant Chuck not Bob.

To many names to keep them all straight:confused:

I would have assumed Chuck's time dilation would be twice Zach's as only a gravitational component is present and Chuck's is twice Zach's. But this is not the case?

No' date=' it is not. Gravitational time dilation is related to gravitational [i']potential[/i], not local strength of the field. The gravitational time dilation formula is:

$T = \frac{T'}{\sqrt{1-\frac{2GM}{Rc^2}}}$

Where M is the mass of the planet and R is the distance from the center of the planet. Note that if both M and R increase by the same factor, that factor cancels out of the equation.

On the other hand, the formula for acceleration due to gravity is:

$a = \frac{GM}{R^2}$

If we increase the radius and mass by the same factor, the resulting acceleration decreases by that factor.

And there is no time dilation component due to acceleration in Bob's or Xirb's cases?

No, there is not.

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To many names to keep them all straight:confused:No' date=' it is not. Gravitational time dilation is related to gravitational [i']potential[/i], not local strength of the field. The gravitational time dilation formula is:

$T = \frac{T'}{\sqrt{1-\frac{2GM}{Rc^2}}}$

Where M is the mass of the planet and R is the distance from the center of the planet. Note that if both M and R increase by the same factor, that factor cancels out of the equation.

On the other hand, the formula for acceleration due to gravity is:

$a = \frac{GM}{R^2}$

If we increase the radius and mass by the same factor, the resulting acceleration decreases by that factor.

No, there is not.

Interesting. Any thoughts/insights on why it is the way it is? Does acceleration only effect time dilation indirectly by changing the Kinematic component?

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I actually assumed no equatorial bulge (maybe we need exotic planet material as well or we end up with a pancake)

What difference would this make to the order?

That's a problem when you make reference to a planet. OK, if they are perfect speheres, then that breaks the tie between C & D and E & F. D sees slightly less time pass compared to C, and the same for F and E. So the order remains the same.

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Interesting. Any thoughts/insights on why it is the way it is?

If you look at the formula for Gravitational time dilation I gave' date=' you will note that it is the standard SR time dilation formula with [imath']\sqrt{\frac{2GM}{r}}[/imath] in place of v. This is the formula for escape velocity. This is the initial velocity an object would have to have in order to reach an infinite distance from mass M when starting at radius R from its center. The gravitational time dilation formula is for an observer an infinite distance from mass M. Thus one way of looking at it is that the time dilation is related to the energy that has to be given up climbing from radius R to infinity.

Does acceleration only effect time dilation indirectly by changing the Kinematic component?

As long as this acceleration is wrt to the observer, yes. If, however, the observer is in the accelerated frame itself, then said observer will see time dilation effects akin to that of Gravitational time dilation. Clocks lying in the direction of the acceleration will be determined as running fast and clocks lying in the opposite direction will be determined as running slow. How slow or fast these clocks run is determined by the magnitude of the acceleration and the distance to these clocks from the observer. (as measured on a line parallel to the direction of the acceleration.)

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Gravitational time dilation is related to gravitational potential' date=' not local strength of the field. The gravitational time dilation formula is:

[math']T = \frac{T'}{\sqrt{1-\frac{2GM}{Rc^2}}}[/math]

Where M is the mass of the planet and R is the distance from the center of the planet. Note that if both M and R increase by the same factor, that factor cancels out of the equation.

.

Does this formula hold as you approach R=0? As you "enter" a planet does the M for the formula decrease?

I'm assuming that if you were "hovering" at zero gravity in a small hollow sphere at the center of Earth 1 the time dilation correction factor would be zero. (T = T')

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That's a problem when you make reference to a planet[/b']. OK, if they are perfect speheres, then that breaks the tie between C & D and E & F. D sees slightly less time pass compared to C, and the same for F and E. So the order remains the same.

You still got full credit Professor, because you showed your work!

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Does this formula hold as you approach R=0? As you "enter" a planet does the M for the formula decrease?

You have to use a different formula for radii that lay inside the planet.

I'm assuming that if you were "hovering" at zero gravity in a small hollow sphere at the center of Earth 1 the time dilation correction factor would be zero. (T = T')

No. As you hover at the center of the Earth you are at a lower gravitational potential than you are at the surface. (you would have to do work against Earth's gravity to reach the surface) Since gravitational time dilation is related to difference in potential, this means a clock hovering at the center of the Earth would run slower than one on the surface. You still get T<T'.

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You have to use a different formula for radii that lay inside the planet.

No. As you hover at the center of the Earth you are at a lower gravitational potential than you are at the surface. (you would have to do work against Earth's gravity to reach the surface) Since gravitational time dilation is related to difference in potential' date=' this means a clock hovering at the center of the Earth would run slower than one on the surface. You still get T<T'.[/quote']

This seems consistent (I'd like to say it makes sense but I'm working against some previous misconceptions)

Thanks for the inputs.

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