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

#### 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.

#### 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.

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#### 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.

#### 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

Mentor
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/?i=((constant+of+gravitation)+*+(mass+of+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/?i=((constant+of+gravitation)+*+(mass+of+sun)*(mass+of+Earth))/(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.