<p><em>(Disclaimer: As I already pointed out in a comment to the question above, I never did a calculation with $H_0$ before and I might be utterly, horrible wrong with my interpretation.)</em></p>

<p>If you completely ignore the slowly changing orbit of earth and only take expansion of space into account and assume the Hubble-parameter to be pretty constant in the timeframe of 1 My, we can calculate the difference of the orbital period of earth using Keppler's third law [3]:</p>

<p>$T = 2\pi\sqrt(a^3/GM)$</p>

<p>for</p>

<p>$a = 1.4959789*10^{11} m$ (semi-major axis of earth today) [1]<br>
$G = 6.67*10^{-11} Nm^2/kg^2$ (gravitational constant)<br>
$M = 1.988435*10^{30} kg$ (mass sun) [1]</p>

<p>We also assume: $H_0 = 2.3*10^{-18} s^-1$ [2] (Hubble parameter then and today in SI-units) which basically means "in every second a meter get $2.3*10^{-18} m$ longer".</p>

<p>Instead of taking the length of an (siderial) orbital period of earth from some source, let's calculate it manually first and take it as a reference.</p>

<p>$T_{today} = 2 \pi \sqrt((1.4959789*10^{11}m)^3/(6.67*10^{-11} Nm^2/kg^2 * 1.988435*10^{30} kg))$ = 365 days 8 hours 56 minutes 13.45 seconds</p>

<p>Pretty close and a good reference for more calculations.</p>

<p>Now, what was earth's semi-major axis 1 million years ago, only taking into account a constant $H_0$?</p>

<p>$x - (2.3*10^{-18} s^-1 * 1 My * x) = 1.4959789*10^{11} m$<br>
Solving for $x$ leads to $x = 1.49598*10^{11} m$.
<br>(Sorry for the lousy precision; I only have Wolfram Alpha at my hands right now.)</p>

<p>The old semi-major axis is a little smaller. Using Keppler's law again we can calculate the orbital period again:</p>

<p>$T_{old} = 2 \pi \sqrt((1.496*10^{11} m)^3/(6.67*10^{-11} Nm^2/kg^2 * 1.988435*10^{30} kg))$ = 365 days 8 hours 56 minutes 48.26 seconds</p>

<p>So, subtracting both times from another we can say that <em>1 My ago the year was indeed 34.81 seconds shorter</em>.</p>

<p>However. This probably doesn't mean much; the orbit changes slightly over time anyway; the Hubble-parameter is not considered a constant any more, it changes slightly over time; and while this was an interesting question I don't trust my interpretation much and hope that someone else who's more qualified than me could enlighten the question better than I ever could.</p>

<p>(I hope I didn't botch anything somewhere. I need more coffee.)</p>

<p>[1] Source: Wolfram Alpha<br>
[2] Source for Hubble-parameter in SI-units taken from the German Wikipedia: <a href="http://de.wikipedia.org/wiki/Hubble-Konstante#Definition" rel="nofollow">http://de.wikipedia.org/wiki/Hubble-Konstante#Definition</a><br>
[3] <a href="http://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body" rel="nofollow">http://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body</a></p>

CC BY-SA 3.0
2013-12-18T11:42:46.157Z
Well do you have any any idea what would be the length of the day at that time
SpringLearner
CC BY-SA 3.0
2013-12-18T12:11:30.600Z
Uh, not right now. If the orbit changes, angular momentum has to remain constant, so something will change. Need to think of that later.
Alexander Janssen
CC BY-SA 3.0
2013-12-18T15:41:42.620Z
I don't think that cosmic expansion has anything to do with changes in the year length. (see comment to the question)
Francesco Montesano
CC BY-SA 3.0
2013-12-18T15:53:50.233Z
@FrancescoMontesano I'm not convinced either. I mean my whole idea would only be true if mass is somehow attached to expanding space - but wouldn't that mean that space itself excerts a force on mass? It makes my head hurt and I'm absolutely not qualified. However I found it fun to give it some time thinking about it.
Alexander Janssen
CC BY-SA 3.0
2013-12-18T16:04:46.480Z
@AlexanderJanssen: it's complicated for a comment. I wouldn't say the expansion exerts a force. It's more like dragging whatever is in the universe. But when gravitation attraction between two masses becomes strong enough, they begin to disentangle from the expansion and when they reach equilibrium their reciprocal motion becomes (mostly) independent from what happens outside the system (although some parameters of their status might be influenced by the expansion status when they decoupled)
Francesco Montesano
CC BY-SA 3.0
2013-12-18T16:07:49.973Z
@FrancescoMontesano It somehow makes sense. Thanks!
Alexander Janssen
CC BY-SA 3.0
2013-12-18T11:18:17.783Z
2014-04-11T14:05:58.900Z
2014-04-11T14:05:58.900Z
Alexander Janssen
<p>The Hubble expansion has no bearing whatsoever on the length of the year. This is because the whole Milky Way galaxy (and in fact most galaxies, if not all, and even local groups) has decoupled from the Hubble flow long ago. In fact, it could only form after it decoupled. Note that M31, our sister galaxy, is in fact falling onto the Milky Way rather than receding (as the Hubble flow would imply), demonstrating that the whole of the Local Group (of galaxies) is decoupled from the Hubble flow.</p>

<p>What happens is that any over-density expands at less than the Hubble rate and thereby grows. Galaxies (and larger structures) form from small relative over-densities that eventually grow large enough to withstand the overall expansion and instead collapse under their own gravity to form bound objects, such as galaxy clusters, galaxies, star clusters, and stars. This implies that the Hubble flow has no bearing on the inner dynamics of such systems.</p>

<p>Of course, the number of days in a year was higher in the past than today, but that is only because the Earth is spinning down (due to tidal friction with the Moon), so that the days become longer.</p>

<p>If anything has had an effect on the semi-major axis of the Earth orbit (and hence on its period), then that is gravitational interactions with the other planets. However, weak interactions (secular perturbations) can only alter the orbital eccentricity and leave the semi-major axis unaltered.</p>

<p>Finally, there is a tiny effect from the Sun loosing mass (to the Solar wind). The period of any orbiting bodies is proportional to $M_\odot^{-1/2}$.</p>

CC BY-SA 3.0
2013-12-19T04:06:07.753Z
Thanks for the answer,well alexander also said the same thing but I wanted to know how much it is longer than today
SpringLearner
CC BY-SA 3.0
2013-12-21T10:15:53.027Z
I've found a german site (http://www.scilogs.de/relativ-einfach/astronomisches-grundwissen-9/) explaining (translation mine): "It's also important to know, how expansion influences bound systems - for example, a galaxy, our solar system or even the atoms we're made of with their nucleii and their bound electrons. The short answer: it doesn't. If bonding is strong enough, it won't be changed by cosmic expansion. Only starting with length-scales, where cosmos has about the same density - means no bigger concentrations of mass here and mass deficits there - cosmic expansion may work unhindered."
Alexander Janssen
CC BY-SA 3.0
2013-12-23T19:40:55.667Z
@Envite: 1) The fact that cosmological expansion is not dynamically important for galaxy-group scale systems does not imply that it does not exhibit any influence at all, as one could understand from your post. Your answer would have been so much better if you could specify the upper bound on the influence of global expansion on the length of the year, 2) M31 is not coming towards us due to gravitational forces, it just happens to have its proper velocity directed at us.
Alexey Bobrick
CC BY-SA 3.0
2013-12-25T13:46:42.733Z
@AlexeyBobrick I disagree regarding M31. Very likely the local group of galaxies (of which the Milky Way and M31 are the main masses) is gravitationally bound. Hence their velocities are not random. M31 and the Milky Way are likely to merge (and form an elliptical galaxy) in a few $10^9$ years.
Walter
CC BY-SA 3.0
2013-12-25T13:59:46.853Z
@Walter: Your argument is not sound enough. Imagine a stellar cluster, which is also bound. The stars there move with proper velocities of about a few km/s. However, if you just take two stars at a parsec distance and launch them at km/s velocity randomly, they will be unbound. Similarly, M31 just happens to have its peculiar velocity directed at the Milky Way. Do you have any other arguments?
Alexey Bobrick
CC BY-SA 3.0
2013-12-26T17:08:43.057Z
@AlexeyBobrick (1) the difference here is that M31 and the Milky Way (MW) dominate the total mass of the local group, so you should not compare to a star cluster, but a binary star plus some planets. (2) In this case, if the relative velocity of M31 wrt MW is receding, gravitational forces will eventually overcome it and result in the observed approaching velocity. Note also that recent determinations of M31s proper motion (and hence transverse velocity) yielded a value much smaller than the radial velocity). There is some research literature on this (search for "local group timing argument").
Walter
CC BY-SA 3.0
2013-12-26T18:49:32.970Z
@Walter: Well, I got relative escape velocity for MW and M31 of $100-150 \textrm{km/s}$, accounting for their DM halos. In the meanwhile, their relative radial velocity is of the same order. So, it is rather speculative to say, whether MW-Andromeda interaction is determining the merger or not. I guess one other good reference to look at is just the BT book, which is rather much agreeing with you here.
Alexey Bobrick
CC BY-SA 3.0
2013-12-27T19:36:36.767Z
@AlexeyBobrick *I got relative escape velocity for MW and M31 of 100−150km/s* where/how did you get this number? It sounds much too small.
Walter
CC BY-SA 3.0
2013-12-27T20:42:53.787Z
@Walter: Which ones do you get/can refer to? My estimate is for MW+M31 in isolation at 1Mpc distance and having both $10^{12}M_{\odot}$ in mass.
Alexey Bobrick
CC BY-SA 3.0
2013-12-18T17:10:27.770Z
2013-12-25T13:44:11.220Z
2013-12-25T13:44:11.220Z
Walter
<p>We know that the universe is gradually expanding and this indirectly means that the gravitational force between sun, earth, planets and other stars (roughly anything in the universe) is gradually decreasing as gravitational force is indirectly proportional to square of distance between the objects.</p>

<p>So I think this also effects to length of the year. If yes then is it possible to know how days does 1 year had 1 million year back?</p>

CC BY-SA 3.0
2013-12-18T08:48:26.560Z
If you say days, you you mean the length of today's day, or the number of rotations the Earth had back then?
Mr Lister
CC BY-SA 3.0
2013-12-18T08:53:20.460Z
@MrLister Well number of days at that time
SpringLearner
CC BY-SA 3.0
2013-12-18T11:09:00.613Z
I might have an answer to that, 1 million years ago the year was 34.81 seconds shorter if you only take expansion into account, but I'm not entirely convinced about my intepretation of the Hubble-parameter (I never did a calculation with that). Although I might be completely wrong, should I post my answer anyway? I worked some time on it but then decided after some thinking that I might have completely misinterpreted the meaning of $H_0$.
Alexander Janssen
CC BY-SA 3.0
2013-12-18T11:12:16.397Z
@AlexanderJanssen yes why not post as answer.As per rules of this site if your post is useful then your answer will be upvoted(whether it completely answers or not)
SpringLearner
CC BY-SA 3.0
2013-12-18T11:13:18.947Z
@JqueryLearner Alright then, gimme a minute to pull up my notes.
Alexander Janssen
CC BY-SA 3.0
2013-12-18T14:57:26.317Z
I am not sure than the space "slipping under Earth's feet" would move it farther from the Sun, being Earth gravitationally bounded to Sun. If this were true, we should see expansion inside galaxies by comparing far (thus old) galaxies with near (thus recent) ones, and I have no notice of that.
Envite
CC BY-SA 3.0
2013-12-18T15:40:33.660Z
The solar system (and in fact the Galaxy) is decoupled from cosmic expansion. Any change in the length of the year depends only on local dynamics. Unless the gravitational constant changes with time, but this is an other problem.
Francesco Montesano
CC BY-SA 3.0
2013-12-18T17:18:32.733Z
@JqueryLearner The number of days in a year and the length of the year are two **completely different** things (because the length of a day changes with time).
Walter
CC BY-SA 3.0
2013-12-19T11:26:25.187Z
@FrancescoMontesano please make an answer from that
Envite
CC BY-SA 3.0
2013-12-20T10:17:53.457Z
@Envite: I don't think that I have an answer. Besides Walter already expanded my comments
Francesco Montesano
<solar-system><gravity><newtonian-gravity>
What was the length of year 1 million years back?
SpringLearner
# What was the length of year 1 million years back?

(Asked by: username_0 on 2013-12-18)

We know that the universe is gradually expanding and this indirectly means that the gravitational force between sun, earth, planets and other stars (roughly anything in the universe) is gradually decreasing as gravitational force is indirectly proportional to square of distance between the objects.

So I think this also effects to length of the year. If yes then is it possible to know how days does 1 year had 1 million year back?

**Tags:** solar-system, gravity, newtonian-gravity

> **Comments:**
> **username_1**: If you say days, you you mean the length of today's day, or the number of rotations the Earth had back then?  
> **username_0**: @MrLister Well number of days at that time  
> **username_2**: I might have an answer to that, 1 million years ago the year was 34.81 seconds shorter if you only take expansion into account, but I'm not entirely convinced about my intepretation of the Hubble-parameter (I never did a calculation with that). Although I might be completely wrong, should I post my answer anyway? I worked some time on it but then decided after some thinking that I might have completely misinterpreted the meaning of $H_0$.  
> **username_0**: @AlexanderJanssen yes why not post as answer.As per rules of this site if your post is useful then your answer will be upvoted(whether it completely answers or not)  
> **username_2**: @JqueryLearner Alright then, gimme a minute to pull up my notes.  
> **username_3**: I am not sure than the space "slipping under Earth's feet" would move it farther from the Sun, being Earth gravitationally bounded to Sun. If this were true, we should see expansion inside galaxies by comparing far (thus old) galaxies with near (thus recent) ones, and I have no notice of that.  
> **username_4**: The solar system (and in fact the Galaxy) is decoupled from cosmic expansion. Any change in the length of the year depends only on local dynamics. Unless the gravitational constant changes with time, but this is an other problem.  
> **username_5**: @JqueryLearner The number of days in a year and the length of the year are two **completely different** things (because the length of a day changes with time).  
> **username_3**: @FrancescoMontesano please make an answer from that  
> **username_4**: @username_3: I don't think that I have an answer. Besides username_5 already expanded my comments  

---

## 2 answers

### Answer by username_5 on 2013-12-18. Score: 15 | Accepted answer

The Hubble expansion has no bearing whatsoever on the length of the year. This is because the whole Milky Way galaxy (and in fact most galaxies, if not all, and even local groups) has decoupled from the Hubble flow long ago. In fact, it could only form after it decoupled. Note that M31, our sister galaxy, is in fact falling onto the Milky Way rather than receding (as the Hubble flow would imply), demonstrating that the whole of the Local Group (of galaxies) is decoupled from the Hubble flow.

What happens is that any over-density expands at less than the Hubble rate and thereby grows. Galaxies (and larger structures) form from small relative over-densities that eventually grow large enough to withstand the overall expansion and instead collapse under their own gravity to form bound objects, such as galaxy clusters, galaxies, star clusters, and stars. This implies that the Hubble flow has no bearing on the inner dynamics of such systems.

Of course, the number of days in a year was higher in the past than today, but that is only because the Earth is spinning down (due to tidal friction with the Moon), so that the days become longer.

If anything has had an effect on the semi-major axis of the Earth orbit (and hence on its period), then that is gravitational interactions with the other planets. However, weak interactions (secular perturbations) can only alter the orbital eccentricity and leave the semi-major axis unaltered.

Finally, there is a tiny effect from the Sun loosing mass (to the Solar wind). The period of any orbiting bodies is proportional to $M\_\odot^{-1/2}$.

> **Comments:**
> **username_0**: Thanks for the answer,well alexander also said the same thing but I wanted to know how much it is longer than today  
> **username_2**: I've found a german site (http://www.scilogs.de/relativ-einfach/astronomisches-grundwissen-9/) explaining (translation mine): "It's also important to know, how expansion influences bound systems - for example, a galaxy, our solar system or even the atoms we're made of with their nucleii and their bound electrons. The short answer: it doesn't. If bonding is strong enough, it won't be changed by cosmic expansion. Only starting with length-scales, where cosmos has about the same density - means no bigger concentrations of mass here and mass deficits there - cosmic expansion may work unhindered."  
> **username_6**: @username_3: 1) The fact that cosmological expansion is not dynamically important for galaxy-group scale systems does not imply that it does not exhibit any influence at all, as one could understand from your post. Your answer would have been so much better if you could specify the upper bound on the influence of global expansion on the length of the year, 2) M31 is not coming towards us due to gravitational forces, it just happens to have its proper velocity directed at us.  
> **username_5**: @AlexeyBobrick I disagree regarding M31. Very likely the local group of galaxies (of which the Milky Way and M31 are the main masses) is gravitationally bound. Hence their velocities are not random. M31 and the Milky Way are likely to merge (and form an elliptical galaxy) in a few $10^9$ years.  
> **username_6**: @username_5: Your argument is not sound enough. Imagine a stellar cluster, which is also bound. The stars there move with proper velocities of about a few km/s. However, if you just take two stars at a parsec distance and launch them at km/s velocity randomly, they will be unbound. Similarly, M31 just happens to have its peculiar velocity directed at the Milky Way. Do you have any other arguments?  
> **username_5**: @AlexeyBobrick (1) the difference here is that M31 and the Milky Way (MW) dominate the total mass of the local group, so you should not compare to a star cluster, but a binary star plus some planets. (2) In this case, if the relative velocity of M31 wrt MW is receding, gravitational forces will eventually overcome it and result in the observed approaching velocity. Note also that recent determinations of M31s proper motion (and hence transverse velocity) yielded a value much smaller than the radial velocity). There is some research literature on this (search for "local group timing argument").  
> **username_6**: @username_5: Well, I got relative escape velocity for MW and M31 of $100-150 \textrm{km/s}$, accounting for their DM halos. In the meanwhile, their relative radial velocity is of the same order. So, it is rather speculative to say, whether MW-Andromeda interaction is determining the merger or not. I guess one other good reference to look at is just the BT book, which is rather much agreeing with you here.  
> **username_5**: @AlexeyBobrick *I got relative escape velocity for MW and M31 of 100−150km/s* where/how did you get this number? It sounds much too small.  
> **username_6**: @username_5: Which ones do you get/can refer to? My estimate is for MW+M31 in isolation at 1Mpc distance and having both $10^{12}M_{\odot}$ in mass.  

---

### Answer by username_2 on 2013-12-18. Score: 9

*(Disclaimer: As I already pointed out in a comment to the question above, I never did a calculation with $H\_0$ before and I might be utterly, horrible wrong with my interpretation.)*

If you completely ignore the slowly changing orbit of earth and only take expansion of space into account and assume the Hubble-parameter to be pretty constant in the timeframe of 1 My, we can calculate the difference of the orbital period of earth using Keppler's third law [3]:

$T = 2\pi\sqrt(a^3/GM)$

for

$a = 1.4959789\*10^{11} m$ (semi-major axis of earth today) [1]  
$G = 6.67\*10^{-11} Nm^2/kg^2$ (gravitational constant)  
$M = 1.988435\*10^{30} kg$ (mass sun) [1]

We also assume: $H\_0 = 2.3\*10^{-18} s^-1$ [2] (Hubble parameter then and today in SI-units) which basically means "in every second a meter get $2.3\*10^{-18} m$ longer".

Instead of taking the length of an (siderial) orbital period of earth from some source, let's calculate it manually first and take it as a reference.

$T\_{today} = 2 \pi \sqrt((1.4959789\*10^{11}m)^3/(6.67\*10^{-11} Nm^2/kg^2 \* 1.988435\*10^{30} kg))$ = 365 days 8 hours 56 minutes 13.45 seconds

Pretty close and a good reference for more calculations.

Now, what was earth's semi-major axis 1 million years ago, only taking into account a constant $H\_0$?

$x - (2.3\*10^{-18} s^-1 \* 1 My \* x) = 1.4959789\*10^{11} m$  
Solving for $x$ leads to $x = 1.49598\*10^{11} m$.
  
(Sorry for the lousy precision; I only have Wolfram Alpha at my hands right now.)

The old semi-major axis is a little smaller. Using Keppler's law again we can calculate the orbital period again:

$T\_{old} = 2 \pi \sqrt((1.496\*10^{11} m)^3/(6.67\*10^{-11} Nm^2/kg^2 \* 1.988435\*10^{30} kg))$ = 365 days 8 hours 56 minutes 48.26 seconds

So, subtracting both times from another we can say that *1 My ago the year was indeed 34.81 seconds shorter*.

However. This probably doesn't mean much; the orbit changes slightly over time anyway; the Hubble-parameter is not considered a constant any more, it changes slightly over time; and while this was an interesting question I don't trust my interpretation much and hope that someone else who's more qualified than me could enlighten the question better than I ever could.

(I hope I didn't botch anything somewhere. I need more coffee.)

[1] Source: Wolfram Alpha  
[2] Source for Hubble-parameter in SI-units taken from the German Wikipedia: <http://de.wikipedia.org/wiki/Hubble-Konstante#Definition>  
[3] <http://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body>

> **Comments:**
> **username_0**: Well do you have any any idea what would be the length of the day at that time  
> **username_2**: Uh, not right now. If the orbit changes, angular momentum has to remain constant, so something will change. Need to think of that later.  
> **username_4**: I don't think that cosmic expansion has anything to do with changes in the year length. (see comment to the question)  
> **username_2**: @FrancescoMontesano I'm not convinced either. I mean my whole idea would only be true if mass is somehow attached to expanding space - but wouldn't that mean that space itself excerts a force on mass? It makes my head hurt and I'm absolutely not qualified. However I found it fun to give it some time thinking about it.  
> **username_4**: @AlexanderJanssen: it's complicated for a comment. I wouldn't say the expansion exerts a force. It's more like dragging whatever is in the universe. But when gravitation attraction between two masses becomes strong enough, they begin to disentangle from the expansion and when they reach equilibrium their reciprocal motion becomes (mostly) independent from what happens outside the system (although some parameters of their status might be influenced by the expansion status when they decoupled)  
> **username_2**: @FrancescoMontesano It somehow makes sense. Thanks!  

---

