Insights Either the Sun Is Getting Smaller or Gravity Is Getting Weaker

[klotza]

Greg Bernhardt submitted a new PF Insights post

Either the Sun Is Getting Smaller or Gravity Is Getting Weaker

Continue reading the Original PF Insights Post.

 

[fresh_42]

Very interesting. There's a minor LaTeX error after "77,000,000,000".
And last but not least a good reference to quote on the FAQ "Do constants change over time or space?"

 

[JMz]

Greg Bernhardt submitted a new PF Insights post

Either the Sun Is Getting Smaller or Gravity Is Getting Weaker
View attachment 223610

To one point: The Solar System -- the orbital radii of the planets and other bodies -- could be getting larger as the Sun's mass slowly shrinks. However, it could also be getting smaller, due to friction from the solar wind. (Also radiation pressure, in the case of tiny bodies, especially dust.) The getting-larger effect should apply equally to all bodies, and the increase should be a simple proportionality with distance. But the getting-smaller effect(s) should be greater at smaller solar distances and for less dense bodies. Which effect is dominant, and for which bodies?

 

[Greg Bernhardt]

Really enjoyed this Insight!

 

[mfb]

Very interesting. There's a minor LaTeX error after "77,000,000,000".
And last but not least a good reference to quote on the FAQ "Do constants change over time or space?"

There is also one at "10²6 Watts"

> Together, the predicted mass loss rate is consistent with the measured decreasing gravitational parameter given its uncertainty.
So... we expect a mass loss, and we see the effect of the mass loss in the expected amount. Nothing surprising, unlike the topic would suggest.

 

[Drakkith]

Some would hope the Earth's gravity is getting weaker. It would certainly make people feel better after the end of year holiday dinners.

 

[Brazedowl]

To one point: The Solar System -- the orbital radii of the planets and other bodies -- could be getting larger as the Sun's mass slowly shrinks. However, it could also be getting smaller, due to friction from the solar wind. (Also radiation pressure, in the case of tiny bodies, especially dust.) The getting-larger effect should apply equally to all bodies, and the increase should be a simple proportionality with distance. But the getting-smaller effect(s) should be greater at smaller solar distances and for less dense bodies. Which effect is dominant, and for which bodies?

Well isn't the momentum of the solar wind directed, for the most part, directly away from the sun? So it would contribute to an expansion overall, yes? Perhaps when the charged particles interact with magnetic fields of (some) planetary bodies and they're deflected? But still seems like an "outward" directed force.

 

[mfb]

A uniform outwards force does not increase orbits, it gives them a different period. It acts like a slightly lower (but constant) solar mass. What is left is some sort of friction, which decreases the orbital radius of objects over time.

 

[Chronos]

Isn't sixty parts per quadrillion decrease in gravity per year smaller than the uncertainty in the gravitational constant given by CODATA?

 

[mfb]

Yes. So what?
What astronomers measure is not the gravitational constant, but the product of the gravitational constant and the mass of objects, GM. This product can be measured much more precisely than either G or M.

 

[JMz]

Yes. So what?
What astronomers measure is not the gravitational constant, but the product of the gravitational constant and the mass of objects, GM. This product can be measured much more precisely than either G or M.

Yes. So in principle, the title of this thread is misleading: The product is decreasing, but (a) it could be either or (b) it could be both or (c) one could be increasing slightly but the other decreasing more.

 

[fluidistic]

I'm curious if they took into account the momentum of the EM field (or photons) created by the Sun exerted on the planets (and the black body radiation of the planets and cosmic background blackbody radiation). By intuition, even though Mercury is a small planet, it's very close to the Sun and might be the most affected to it.

 

[JMz]

I'm curious if they took into account the momentum of the EM field (or photons) created by the Sun exerted on the planets (and the black body radiation of the planets and cosmic background blackbody radiation). By intuition, even though Mercury is a small planet, it's very close to the Sun and might be the most affected to it.

This is covered by @mfb's post #8 here (even though that was stated in the context of the solar wind).

 

[Chronos]

It is very difficult to draw any meaningful conclusions about individual factor variance based on the product of individual factors across cosmological time scales.

 

[fluidistic]

A uniform outwards force does not increase orbits, it gives them a different period. It acts like a slightly lower (but constant) solar mass. What is left is some sort of friction, which decreases the orbital radius of objects over time.

Does this mean that if the Sun was, say, so shiny that the outward force more than counteract the gravitational force between the Earth and the Sun, the Earth would still have its orbit intact?

I do not understand how the Earth could stay with the same orbit if the photons hitting them from the Sun were so much more numerous than they are now that the outward force would be so great that there should not be any stable orbit. Am I missing something?

 

[mfb]

Does this mean that if the Sun was, say, so shiny that the outward force more than counteract the gravitational force between the Earth and the Sun, the Earth would still have its orbit intact?

No, of course not. These particles are not in an orbit at all. That only affects particles smaller than ~500 nm, however.

 

[ensign_nemo]

Has anyone calculated the loss of energy from the gravitational waves emitted by the Sun and Jupiter, or the Sun and the other gas giants?

For the Sun and Jupiter,

m1 = 1.989 × 10^30 kg (Sun) ~ 2 × 10^30 kg
m2 = 1.898 × 10^27 kg (Jupiter) = 0.0009543 m1 ~ 2 × 10^27 kg
R = 778.5 million km = 7.785 x 10^11 m ~ 8 x 10^11 m

The best reference I could find with a quick Google search was:

http://www.physics.usu.edu/Wheeler/GenRel2013/Notes/GravitationalWaves.pdf

but I'm not familiar enough with the formulas to be confident that I'm applying them correctly.

 

[mfb]

About 200 W for Earth/Sun, a bit more for Jupiter. Wikipedia has formulas where you can just plug in masses, distances and eccentricities (use e=0 for a good approximation).

 

[TZT]

Some of you likely already know some of the following, but it was a very interesting point, about the Sun, when I heard it.

Jupiter is massive, and so is the Sun. Given the mass of Jupiter it has great total gravitational force. That force also is exerted noticeably on the Sun itself.
Due to the gravitational force of Jupiter it actually affects the Sun's position in our solar system, such that the Sun is not in a constant epicenter where all the planets form their orbits around that constant point. Rather, Jupiter's gravity pulls on the Sun such that the Sun also has an "orbit", and it's orbit is around an imaginary point that the sun orbits around. So, the Sun itself is also pulled towards Jupiter as Jupiter is pulled towards the sun, and their respective gravity creates both Jupiters's orbit as well as giving the Sun an "orbit" that is around the "center" of the solar system.

Very cool. And it demonstrates the massive gravitational power of Jupiter.

Question:
As the Sun's mass lessens, and with that it's gravitation force lessens, then at what point in the future will Jupiter's gravity have a greater effect on the planets between it and the Sun? IOW, at some future point, will Jupiter's gravitational effect become stronger on the Earth as the Sun's gravitational effect becomes lesser? And, when, or if, that happens, will the Earth's orbit change? Will the effect of Jupiter's gravitational pull slow the Earth's orbit around the sun? Or, can it create an effect of a slow orbit during certain Earth to Jupiter orbiting positions, and then speed up the orbit during different relative positions?

This of course is predicated on the idea that the Sun's mass is lessened due mainly to fusion and mass coronal ejections, while Jupiter's mass remains constant over this same period.

 

[PeterDonis]

Due to the gravitational force of Jupiter it actually affects the Sun's position in our solar system, such that the Sun is not in a constant epicenter where all the planets form their orbits around that constant

This is true of all the planets, not just Jupiter. Jupiter's effect is by far the largest, but I don't think it's the only detectable one. The strictly correct statement is that all of the objects in the Solar system orbit its barycenter, which can be thought of as the "center of mass" of the system as a whole, and this barycenter does not coincide with the center of any of the objects, including the Sun.

As the Sun's mass lessens, and with that it's gravitation force lessens, then at what point in the future will Jupiter's gravity have a greater effect on the planets between it and the Sun?

Never. The rate of mass loss of the Sun is much, much too small. The sun is over 1100 times as massive as Jupiter, and the mass loss rate is roughly 6 parts in $10^{14}$ per year, so even after 5 billion years more, when the Sun is expected to become a red giant, the Sun will have lost only about 1 part in 100,000 of its current mass, so it will still be over 1100 times as massive as Jupiter.

 

[TZT]

This is true of all the planets, not just Jupiter. Jupiter's effect is by far the largest, but I don't think it's the only detectable one. The strictly correct statement is that all of the objects in the Solar system orbit its barycenter, which can be thought of as the "center of mass" of the system as a whole, and this barycenter does not coincide with the center of any of the objects, including the Sun.
Never. The rate of mass loss of the Sun is much, much too small. The sun is over 1100 times as massive as Jupiter, and the mass loss rate is roughly 6 parts in $10^{14}$ per year, so even after 5 billion years more, when the Sun is expected to become a red giant, the Sun will have lost only about 1 part in 100,000 of its current mass, so it will still be over 1100 times as massive as Jupiter.

Soooo, you're sayin there's a chance!? LOL.

Thanks for the answer.

 

[TZT]

This is true of all the planets, not just Jupiter. Jupiter's effect is by far the largest, but I don't think it's the only detectable one. The strictly correct statement is that all of the objects in the Solar system orbit its barycenter, which can be thought of as the "center of mass" of the system as a whole, and this barycenter does not coincide with the center of any of the objects, including the Sun.

Well now, of course Jupiter's gravity affects all the planets, as all the planets can affect each other. But now, that wasn't the point, and that's not what I was asking or talking about.

[Moderator's note: rest of post deleted as it was off topic.]

 

[PeterDonis]

that's not what I was asking or talking about.

It appears that I did address what you were asking and talking about. So the rest of the argument about this subthread is pointless, and I am deleting it.

 

[mfb]

Both Jupiter and Saturn contribute notably to the position of the Sun relative to the barycenter of the Solar System. Saturn has a smaller mass but the larger distance means it still plays a relevant role.

Wikipedia has an image - the 12 year period comes from Jupiter and the perturbation of that cycle comes from Saturn, with smaller contributions from other planets.

 

[Utkarsh Jha]

Very interesting. There's a minor LaTeX error after "77,000,000,000".
And last but not least a good reference to quote on the FAQ "Do constants change over time or space?"

Yes it is interesting. In fact I thought to link between dark energy and this phenomenon by the information from the link about amazing facts about the universe and god- https://dailyjunkies.com/15-amazing-facts-about-the-universe-and-god/.If you could link the phenomenon it would be greatful

 

[Utkarsh Jha]

Both Jupiter and Saturn contribute notably to the position of the Sun relative to the barycenter of the Solar System. Saturn has a smaller mass but the larger distance means it still plays a relevant role.

Wikipedia has an image - the 12 year period comes from Jupiter and the perturbation of that cycle comes from Saturn, with smaller contributions from other planets.

Never heard about this before but its interesting

 

[ensign_nemo]

The article states that the quantity that was measured was the distance from the Earth to Mercury:

"By very precisely tracking the distance to MESSENGER from Earth, the collaboration was able to measure the change in the Earth-Mercury distance over the several years of observation. This amounted to about two meters, and this was the first nonzero measurement of the changing gravitational parameter of the sun (keeping in mind the distance between them is at least 77,000,000,000 meters). They found that the sun’s gravity was getting proportionally weaker by 6.13±1.47×10−146.13±1.47×10−14 per year (61 parts per quadrillion), and thus the solar system is slowly expanding as well."

However, we know that tidal effects from friction will gradually slow down the orbit of a small body orbiting around a larger body, even if the small body is a solid mass such as the Earth's moon.

For example, quoting from https://phys.org/news/2015-09-moon.html :

"Gravitational coupling between the moon and the bulge nearest the moon acts as a torque on Earth's rotation, draining angular momentum and rotational kinetic energy from Earth's spin. In turn, angular momentum is added to the moon's orbit, accelerating it, which lifts the moon into a higher orbit with a longer period.

As a result of this, the distance between Earth and moon is increasing, and Earth's spin is slowing down. Measurements from lunar ranging experiments with laser reflectors (which were left behind during the Apollo missions) have found that the moon's distance to Earth increases by 38 mm (1.5 in) per year."

The equations for orbital mechanics always assume that there is conservation of energy in the system, but over long periods of time tiny effects such as heat dissipated by tidal friction make this assumption incorrect.

Has anyone estimated the effects of tidal friction on the Sun and the eight major planets?

Mercury would experience a greater effect because it's deeper in the gravitational field of the Sun, but the fact that Earth has oceans and an atmosphere and Mercury does not might well create much more friction. Liquids and gases dissipate tidal energy faster than solids. IOW, the Sun-Earth distance might increase faster than the Sun-Mercury distance, thus explaining the increasing gap between the two planets without requiring any changes to the way that gravity works.

The fact that Earth also has a large Moon would also affect the amount of dissipation of the total kinetic energy of the Earth, although that may not affect the distance from Mercury to Earth directly.

 

[ensign_nemo]

After further thought, if the Earth is moving away from the Sun over over long periods of time, that might also help to solve the "faint young Sun" paradox, as it implies that Earth was closer to the Sun three or four billion years ago and thus received more sunshine. Same idea would apply to a wet early Mars.

https://en.wikipedia.org/wiki/Faint_young_Sun_paradox

 

[ensign_nemo]

We've had probes in orbit for extended periods of time around Mercury, Venus, Mars, Jupiter, and Saturn.

Has anyone tried to use the data accumulated by these probes to measure the change in distance between Earth and each of these planets over long period of time?

I've only seen data for the change in the Earth-Mercury distance. Why can't we do the same for the other four planets?

 

[mfb]

The tidal effects are negligible.

The size of the tides is proportional to mass/distance³. For tides on Earth the Sun has a weaker effect than the Moon already because the larger distance wins over the larger mass. For changes in Earth's orbit, you are now looking into feedback of these tides on the Earth/Sun gravitational interaction - so you get another factor 1/distance³, another factor of a few million relative to the Earth's influence on the Moon. In addition you want to change the orbit of Earth, not the orbit of the Moon. Another factor 80 from the larger mass. Completely negligible.
Mercury is closer to the Sun but it is still negligible for it.

This paper has a formula to estimate the effect: $\langle \dot a \rangle = -\frac{4}{3} n a^{-4} \hat s \left( (1+23e^2) + 7 e^2 D)\right)$ where $D=\frac{\hat p}{2 \hat s}$ and $\hat s = \frac{9 k_0 m_1}{4 Q_0 m_0}R_0^5$ and $\hat p = \frac{9 k_1 m_0}{2 Q_1 m_1}R_1^5$. Here index 0 is the star, index 1 is the planet, m and R are mass and radius, k is the Love number and Q is the dissipation function. e is the orbital eccentricity. n is not described but as far as I understand it is the orbital frequency. It has to have units of 1/time, in the worst case I get it wrong by a factor 2 pi.

For Earth Wikipedia has a value of the Love number - about 0.3. Constraints on energy dissipation in the Earth's body tide from satellite tracking and altimetry (PDF) tells us Q=280 for the Earth. I didn't find values for the Sun. Q in the Solar System (PDF) gives $\geq 10^4$ for the gas giants, I'll use that value. Note that larger values lead to smaller tides. Let's use k=1, this is just an order of magnitude estimate anyway.
For Earth e=0.017 and n=1/year and a=1 AU.

Let's plug that in:
$\hat p$ = 1.7*10³⁷ m⁵ for Earth
$\hat s$ = 1.1*10³⁵ m⁵ for Sun
D = 77

Finally $\langle \dot a \rangle = -\frac{4}{3 year} (1 AU)^{-4} \cdot 1.1 \cdot 10^{35} m^5 \cdot \left( (1+23\cdot(0.017)^2) + 7 (0.017)^2 \cdot 77)\right)$
$\langle \dot a \rangle = -3.4\cdot 10^{-10} \frac{m}{year}$

It is probably optimistic to assume that Mercury has the same k and Q as Earth. In that case $\hat p = 2.5\cdot 10^{36} m^5$ and $\hat s = 6\cdot 10^{33} m^5$ and $\langle \dot a \rangle = -0.18 \frac{\mu m}{year}$. Way too small to measure.

 

[ensign_nemo]

Just as a sanity check, I took the figures given in your reference for the total energy dissipated by tidal forces:

"P = 110 ± 25 GW. For comparison this is 22 times smaller than the M% dissipation rate in the oceans (Cartwright & Ray 1991), but an order of magnitude greater than the rate in the atmosphere (Platzman 1991)."

https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20000072440.pdf

I multiplied 110 GW by 23, to combine losses from solid and liquid friction, which gives an estimate of the total amount of energy dissipated by all tidal forces on Earth:

http://www.wolframalpha.com/input/?i=110+*+10^9+*23+W

This gives 2.53 TW, about the same as total electrical power consumption on Earth.

I then used 1/2 * (G * Mass_earth * Mass_sun / Orbital radius of Earth):

http://www.wolframalpha.com/input/?...sun)*(mass+of+Earth))/(2*(astronomical+unit))

This gives 2.649 x 10^33 Joules for the orbital kinetic energy of Earth. Wolfram helpfully adds that this figure is 98% of their number for the "Earth's orbital kinetic energy around the Sun", so it looks right.

Divide the second number by the first number:

http://www.wolframalpha.com/input/?.../(2*(astronomical+unit))+/+(110+*+10^9+*23+W)

The result is that the Earth will stop orbiting the Sun due to its loss of orbital kinetic energy from tidal friction in a mere 33 trillion years.

If we rather cavalierly assume constant tidal friction over time, the age of the Earth is 4.5 billion years, so we have lost about

http://www.wolframalpha.com/input/?i=(age+of+the+earth)+/+33+trillion+years

or 1.38 * 10 ^ -4 = 0.000138 of the initial kinetic energy.

Call it about one percent of one percent.

That's also tossing all lunar tidal friction into the equation, which is not accurate for our solar distance calculation, as that mostly slows down the moon, rather than the Earth.

So, yeah ... it's not a big factor.