<p>There is a fair degree of uncertainty on how far the moon was in the past, and consequently how fast it will recede in the future. Here is a graph</p>
<p><a href="https://i.sstatic.net/uNao2.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/uNao2.png" alt="enter image description here" /></a></p>
<p><sub>Earth-Moon distance evolution through geologic time. The model curves assume different lunar recession rates and apply Eq. (3) of Walker and Zahnle (1986). LLR = Lunar Laser Ranging lunar recession is from Dicky et al. (1994); the Roche Limit is the distance from Earth that the Moon is pulled apart, and equals 2.5 Earth radii (not shown), and occurs at 1.6 Ga for the LLR lunar recession rate. Greyed text and symbols are models: WA (a-d) = model from the Milankovitch Calculator of Waltham (2015). B&amp;L (a-c) = model of Berger and Loutre (1994). TD = ocean and solid Earth tidal dissipation model of Webb (1982). Bold black text and symbols are interpreted data: EL = Australian tidalite data from the Elatina Sandstone interpreted by Williams (2000). BC (a-b) = North American tidalite data from the Big Cottonwood Formation interpreted by (a) Sonett and Chan (1998) and by (b) Sonnet et al. (1996a, b). XL = cyclostratigraphic data from the Chinese Xiamaling Formation interpreted by Meyers and Malinverno (2018). WW (a-b) = BIF laminae sequences in the Australian Weeli Wolli Formation interpreted by (a) Walker and Zahnle (1986) and by (b) Williams (1989). Red bold text and symbol: DGM </sub></p>
<p><a href="https://www.researchgate.net/figure/Earth-Moon-distance-evolution-through-geologic-time-The-model-curves-assume-different_fig6_259842883" rel="nofollow noreferrer">Source</a></p>
<p>Currently the moon is receding at 3.82 cm per year, but this value is generally understood to be anomalously high (due to resonance in the oceans resulting in greater transfer of angular momentum).  But at this rate, the moon will recede 191000 km over the next five billion years, putting it at 575000km from Earth, and so a little above your value of 525000.  But this estimate is an over estimate.  The true rate of recession is likely to be less than half that (so about 100,000 km) and the final position of the moon is likely to be less than 525000km.</p>

CC BY-SA 4.0
2024-01-14T18:29:25.727Z
The Sun will become a red giant in about 7.5 billion years.
ProfRob
CC BY-SA 4.0
2024-01-14T18:43:22.580Z
Not only is the current recession rate highly anomalously high (which plate tectonics will fix in a few tens of millions of years), it most likely will be a perpetually and ridiculously anomalously low in another half a billion to a billion years in the future when the Sun's ever increasing luminosity makes the Earth's oceans boil away.
David Hammen
CC BY-SA 4.0
2024-01-14T18:45:14.103Z
Well that gives the moon a little more time then.  I'm confident in giving the answer "maybe".  Our models of lunar recession are not good enough to give a certain Yay or Nay.
James K
CC BY-SA 4.0
2024-01-17T20:21:14.837Z
So the lunar recession rate is tied to what earth's continental coastlines look like, and without a solid idea of how the continents will change in the next several million years, we can't predict how tidal drag will change -- is that accurate?
Darth Pseudonym
CC BY-SA 4.0
2024-01-14T15:24:12.530Z
2024-01-14T15:24:12.530Z
James K
<p>The moon's gravity produces two tidal bulges that (1) slow down Earth's rotation due to friction and (2) Earth's rotation pushes the tidal bulges so that they are always a bit ahead of the moon. This lifts the moon a bit whereby kinetic energy is transformed in potential energy and the moon is slowed down to the orbital velocity of its new distance (so that centrifugal and centripetal force balance). And this happens continuously. If you want to calculate it through it's complicated.</p>
<p>But we can estimate if the final distance is larger than 525,000 km. The usual argument goes with conservation of the system's angular momentum.</p>
<p>Below is Python code to calculate the new distance of the moon after the angular momentum of the Earth rotating around itself is added to the angular momentum of the moon orbiting Earth. So the idea is to check what the new distance is assuming Earth has slowed down completely.</p>
<pre><code>import numpy as np    
# radius and mass
Me = 5.972E24   # mass of Earth
Mm = 1/81*Me    # mass of Moon
Re = 6371E3     # Radius Earth
Rme = 384400E3  # distance Earth - moon

# Periods
Tm = 27*24*3600+7*3600+43*60 # period of moon around Earth (27 days, 7 hours etc...)
Te = 23*3600 + 56*60 # Earth day (siderial 23 hours, 56 mins)

# speed of moon around earth
v = 2*np.pi*Rme/Tm
print(v*3.6) # 3.6 for m/s to km/h

# angular momentum
Ie = 2/5*Me*Re**2  # moment of inertia of solid sphere (earth)
we = 2*np.pi/Te    # angular velocity earth

Le = Ie*we         # angular momentum spinning Earth
Lme = Mm*v*Rme     # angular momentum of moon orbiting Earth
print(Le)
print(Lme)

# if all L go into Lme... 
(Keppler 3 is needed as both a new v and distance Rme are in the equation of moon orbiting Earth)
gamma = 6.6743E-11 # gravitational constant for Keppler3 below
# Lme = Mm*2*pi*Rme**2 / Tm           # v in Lme
# Tm**2 = 4*pi**2/(gamma M) * Rme**3  # using Keppler 3
tmp = 4*np.pi**2/(gamma * (Me + Mm))  # just for making equations shorter
# Lme = Mm*2*pi*Rme**2 /sqrt(tmp)* Rme**-3/2  # just info transforming equations
# Lme = Mm*2*pi*sqrt(Rme)/sqrt(tmp)   # just info transforming equations
L = Lme + Le
Rme_new = (L/(Mm*2*np.pi)*np.sqrt(tmp))**2
print(Rme_new/1000)                   # final distance in km
</code></pre>
<p>In this code, first, mass, radius and distance are defined. Then, the speed <span class="math-container">$v_{orbit}$</span> of the moon around Earth is calculated because the angular momentum <span class="math-container">$L_{me}$</span> of the moon orbiting Earth is <span class="math-container">$L_{me} = M_m \cdot v_{orbit} \cdot R_{me}$</span> with <span class="math-container">$R_{me}$</span> being the distance of Earth and moon. For the rotating Earth's angular momentum the moment of inertia <span class="math-container">$I_e = 2/5 \cdot M_e \cdot R_e^2$</span> of a solid sphere is used. Finally, the angular momenta are added and solved for the new distance the moon has after all angular momentum is now in orbiting Earth. As the new <span class="math-container">$v_{orbit}$</span> and the new distance <span class="math-container">$R_{me}$</span> are in that equation (<span class="math-container">$L_{me} = M_m \cdot v_{orbit} \cdot R_{me}$</span>), Keppler's third law is used connecting new period and new distance, so that we can solve the equation for the new distance <span class="math-container">$R_{me}$</span>. The result is approx 593000 km.</p>
<p>Note, we neglected that the moon is also rotating around itself, but once per month, so this contribution is small. Also we assumed circular orbits and that the Earth stopped entirely (in reality Earth will still rotate around itself once per &quot;new month&quot; after facing the same side to the moon in the end, like the moon already does today). So this result is an upper estimate, but it's likely that 525,000 km is achieved. However, it doesn't say anything about how long it will take...</p>

CC BY-SA 4.0
2024-01-14T19:23:05.637Z
2024-01-14T19:23:05.637Z
Charles Tucker 3
<p>As most people know, the Moon is receding from Earth at a rate of a few centimeters per year. This rate is not constant; as the Moon gets farther away, the tides it raises will be weaker and thus will cause it to recede slower.</p>
<p>According to my calculations, the Earth-Moon barycenter will lie outside Earth when the Moon is at a distance of roughly 525,000km (assuming a circular orbit).</p>
<p>So the question is: Before the Sun turns into a red giant and messes with the Earth-Moon system due to drag and mass loss, will the Moon have receded far enough that the Earth-Moon barycenter lies outside either body? This would be easily answered by a graph showing the Moon's distance over time, but such a thing eludes me.</p>

<the-moon><orbital-mechanics><tidal-forces>
Will the Moon recede far enough that the Earth-Moon barycenter lies outside of Earth's radius before the Sun becomes a red giant?
user267545
# Will the Moon recede far enough that the Earth-Moon barycenter lies outside of Earth's radius before the Sun becomes a red giant?

(Asked by: username_0 on 2024-01-14)

As most people know, the Moon is receding from Earth at a rate of a few centimeters per year. This rate is not constant; as the Moon gets farther away, the tides it raises will be weaker and thus will cause it to recede slower.

According to my calculations, the Earth-Moon barycenter will lie outside Earth when the Moon is at a distance of roughly 525,000km (assuming a circular orbit).

So the question is: Before the Sun turns into a red giant and messes with the Earth-Moon system due to drag and mass loss, will the Moon have receded far enough that the Earth-Moon barycenter lies outside either body? This would be easily answered by a graph showing the Moon's distance over time, but such a thing eludes me.

**Tags:** the-moon, orbital-mechanics, tidal-forces


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## 2 answers

### Answer by username_1 on 2024-01-14. Score: 3

There is a fair degree of uncertainty on how far the moon was in the past, and consequently how fast it will recede in the future. Here is a graph

[![enter image description here](https://i.sstatic.net/uNao2.png)](https://i.sstatic.net/uNao2.png)

Earth-Moon distance evolution through geologic time. The model curves assume different lunar recession rates and apply Eq. (3) of Walker and Zahnle (1986). LLR = Lunar Laser Ranging lunar recession is from Dicky et al. (1994); the Roche Limit is the distance from Earth that the Moon is pulled apart, and equals 2.5 Earth radii (not shown), and occurs at 1.6 Ga for the LLR lunar recession rate. Greyed text and symbols are models: WA (a-d) = model from the Milankovitch Calculator of Waltham (2015). B&L (a-c) = model of Berger and Loutre (1994). TD = ocean and solid Earth tidal dissipation model of Webb (1982). Bold black text and symbols are interpreted data: EL = Australian tidalite data from the Elatina Sandstone interpreted by Williams (2000). BC (a-b) = North American tidalite data from the Big Cottonwood Formation interpreted by (a) Sonett and Chan (1998) and by (b) Sonnet et al. (1996a, b). XL = cyclostratigraphic data from the Chinese Xiamaling Formation interpreted by Meyers and Malinverno (2018). WW (a-b) = BIF laminae sequences in the Australian Weeli Wolli Formation interpreted by (a) Walker and Zahnle (1986) and by (b) Williams (1989). Red bold text and symbol: DGM

[Source](https://www.researchgate.net/figure/Earth-Moon-distance-evolution-through-geologic-time-The-model-curves-assume-different_fig6_259842883)

Currently the moon is receding at 3.82 cm per year, but this value is generally understood to be anomalously high (due to resonance in the oceans resulting in greater transfer of angular momentum). But at this rate, the moon will recede 191000 km over the next five billion years, putting it at 575000km from Earth, and so a little above your value of 525000. But this estimate is an over estimate. The true rate of recession is likely to be less than half that (so about 100,000 km) and the final position of the moon is likely to be less than 525000km.

> **Comments:**
> **username_2**: The Sun will become a red giant in about 7.5 billion years.  
> **username_3**: Not only is the current recession rate highly anomalously high (which plate tectonics will fix in a few tens of millions of years), it most likely will be a perpetually and ridiculously anomalously low in another half a billion to a billion years in the future when the Sun's ever increasing luminosity makes the Earth's oceans boil away.  
> **username_1**: Well that gives the moon a little more time then.  I'm confident in giving the answer "maybe".  Our models of lunar recession are not good enough to give a certain Yay or Nay.  
> **username_4**: So the lunar recession rate is tied to what earth's continental coastlines look like, and without a solid idea of how the continents will change in the next several million years, we can't predict how tidal drag will change -- is that accurate?  

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### Answer by username_5 on 2024-01-14. Score: 0

The moon's gravity produces two tidal bulges that (1) slow down Earth's rotation due to friction and (2) Earth's rotation pushes the tidal bulges so that they are always a bit ahead of the moon. This lifts the moon a bit whereby kinetic energy is transformed in potential energy and the moon is slowed down to the orbital velocity of its new distance (so that centrifugal and centripetal force balance). And this happens continuously. If you want to calculate it through it's complicated.
But we can estimate if the final distance is larger than 525,000 km. The usual argument goes with conservation of the system's angular momentum.
Below is Python code to calculate the new distance of the moon after the angular momentum of the Earth rotating around itself is added to the angular momentum of the moon orbiting Earth. So the idea is to check what the new distance is assuming Earth has slowed down completely.
import numpy as np    
# radius and mass
Me = 5.972E24   # mass of Earth
Mm = 1/81*Me    # mass of Moon
Re = 6371E3     # Radius Earth
Rme = 384400E3  # distance Earth - moon

# Periods
Tm = 27*24*3600+7*3600+43*60 # period of moon around Earth (27 days, 7 hours etc...)
Te = 23*3600 + 56*60 # Earth day (siderial 23 hours, 56 mins)

# speed of moon around earth
v = 2*np.pi*Rme/Tm
print(v*3.6) # 3.6 for m/s to km/h

# angular momentum
Ie = 2/5*Me*Re**2  # moment of inertia of solid sphere (earth)
we = 2*np.pi/Te    # angular velocity earth

Le = Ie*we         # angular momentum spinning Earth
Lme = Mm*v*Rme     # angular momentum of moon orbiting Earth
print(Le)
print(Lme)

# if all L go into Lme... 
(Keppler 3 is needed as both a new v and distance Rme are in the equation of moon orbiting Earth)
gamma = 6.6743E-11 # gravitational constant for Keppler3 below
# Lme = Mm*2*pi*Rme**2 / Tm           # v in Lme
# Tm**2 = 4*pi**2/(gamma M) * Rme**3  # using Keppler 3
tmp = 4*np.pi**2/(gamma * (Me + Mm))  # just for making equations shorter
# Lme = Mm*2*pi*Rme**2 /sqrt(tmp)* Rme**-3/2  # just info transforming equations
# Lme = Mm*2*pi*sqrt(Rme)/sqrt(tmp)   # just info transforming equations
L = Lme + Le
Rme_new = (L/(Mm*2*np.pi)*np.sqrt(tmp))**2
print(Rme_new/1000)                   # final distance in km

In this code, first, mass, radius and distance are defined. Then, the speed $v_{orbit}$ of the moon around Earth is calculated because the angular momentum $L_{me}$ of the moon orbiting Earth is $L_{me} = M_m \cdot v_{orbit} \cdot R_{me}$ with $R_{me}$ being the distance of Earth and moon. For the rotating Earth's angular momentum the moment of inertia $I_e = 2/5 \cdot M_e \cdot R_e^2$ of a solid sphere is used. Finally, the angular momenta are added and solved for the new distance the moon has after all angular momentum is now in orbiting Earth. As the new $v_{orbit}$ and the new distance $R_{me}$ are in that equation ($L_{me} = M_m \cdot v_{orbit} \cdot R_{me}$), Keppler's third law is used connecting new period and new distance, so that we can solve the equation for the new distance $R_{me}$. The result is approx 593000 km.
Note, we neglected that the moon is also rotating around itself, but once per month, so this contribution is small. Also we assumed circular orbits and that the Earth stopped entirely (in reality Earth will still rotate around itself once per "new month" after facing the same side to the moon in the end, like the moon already does today). So this result is an upper estimate, but it's likely that 525,000 km is achieved. However, it doesn't say anything about how long it will take...


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