Orbit change due to hypothetical mass change

In summary, the conversation discusses the effect of a time-dependent central mass (such as the Earth's increasing mass) on the distance between two objects with different masses. The theory is based on the orbit equation and the parameters of angular momentum, standard gravitational parameter, and eccentricity. The resulting conclusion is that a change in the central mass would lead to a proportional change in the distance between the two objects. However, the actual reason for the Earth-moon distance increasing is mainly due to tidal friction and geological measurements show a discrepancy between predicted and observed values. The discussion also points out uncertainties and discrepancies in measurements and theories.
  • #1
meteor84
6
0
Hi everyone,

I want to discuss the following question:

What would happen to the distance between two objects with masses [itex] M [/itex] and [itex] m [/itex] with [itex] M >> m [/itex] if the central mass [itex] M [/itex] was time-dependent (e.g. the moon's orbit if the mass of the Earth would increase slowly)?

My theory goes like this:

The orbit equation in the above case can be written as

[itex]r = \frac{l^2}{m^2 \; \gamma} \frac{1}{1+e}, \ \ \ \ \ (1)[/itex]where r is the separation distance between the two bodies and \theta is the angle that [itex]\mathbf{r}[/itex] makes with the axis of periapsis (also called the true anomaly). The parameter [itex]l[/itex] is the angular momentum of the orbiting body about the central body, and is equal to [itex]mr^2\dot{\theta}[/itex].
The parameter [itex]\gamma[/itex] is the constant for which [itex]\gamma/r^2[/itex] equals the acceleration of the smaller body (for gravitation, [itex]\gamma[/itex] is the standard gravitational parameter, GM). The parameter e is the eccentricity of the orbit.
(See: http://en.wikipedia.org/wiki/Orbit_equation)

For making it easier, as a first approach, I would like to look at the special case of a circular orbit ([itex] e = 0 [/itex]) in which equation (1) reduces to:

[itex]
r = \frac{l^2}{m^2 \; \gamma}. \ \ \ \ \ (2)
[/itex]

From equation (2) and the definitions of [itex]l[/itex] and [itex]\gamma[/itex] it follows that another valid expression for r is:

[itex]
r = \sqrt[3]{\frac{G \; M}{(d\theta / dt)^2} }. \ \ \ \ \ (3)
[/itex]Since the orbital period is [itex] T = \frac{2 \pi r}{v} [/itex] and the tangential velocity [itex]v[/itex] should be unaffected by a change of the central mass it follows that [itex] \frac{d \theta}{d t} \propto r^{-1} [/itex].

Together with equation (3) I come to the conclusion that the following relation is valid:


[itex]
r \propto G \; M
[/itex]

and since G is a constant:[itex]
r \propto M.
[/itex]So a change of for example [itex] \frac{\Delta M}{M} = 10^{-11} [/itex] in the mass of the central body should lead to the same relative change in the distance r.

Maybe this is one possible explanation for the 3.8 cm per year increase of the distance between the Earth and the moon, but I'm not sure if my derivation is correct.

Cheers,
S.
 
Last edited:
Astronomy news on Phys.org
  • #2
meteor84 said:
Since the orbital period is [itex] T = \frac{2 \pi r}{v} [/itex] and the tangential velocity [itex]v[/itex] should be unaffected by a change of the central mass it follows that [itex] \frac{d \theta}{d t} \propto r^{-1} [/itex].
Actually, I think the tangential velocity of the moon would be affected by a change in mass of the earth: it would change slightly as a radial component of velocity would be introduced briefly and then resume a new tangential speed in a new orbit.

Also I think it's more complicated than what you have outlined since the earth-moon system has a constant angular momentum associated with it. So changing the mass of the Earth would change the angular momentum of the system as well.

And this is the actual reason the moon-earth distance is increasing: the Earth's rotation is slowing down because of tidal friction (not increasing mass) and this transfers to the moon which increases it's orbit accordingly.
 
  • #3
paisiello2 said:
And this is the actual reason the moon-earth distance is increasing: the Earth's rotation is slowing down because of tidal friction (not increasing mass) and this transfers to the moon which increases it's orbit accordingly.

This is one reason the moon-earth distance is increasing. But new measurements show that it can't be the only reason.

Tidal friction explains only 2.9 cm/yr, but the Lunar Laser Ranging Experiment (LLRE) reports 3.8 cm/yr. So 0.9 cm/yr remain unexplained.

(See: http://www.lpi.usra.edu/meetings/lpsc2013/pdf/2436.pdf)
 
  • #4
Your reference cites uncertainties that overlap in the values which suggests that maybe the discrepancy is just experimental error.

Also, the 3.8cm number represents current measurements while the 2.9cm is based on geological measurements from 300 million years ago. It's possible that the rate of change has not been constant over that time period.
 
Last edited:
  • #5
paisiello2 said:
Your reference cites uncertainties that overlap in the values which suggests that maybe the discrepancy is just experimental error.

For the LLRE measurements I can't see such an overlap.
And I think the LLRE measurements are more reliable than the data taken from eclipse observations are.

paisiello2 said:
It's possible that the rate of change has not been constant over that time period.

Yes, of course this is possible.
But there is still the fact that for present values we have a discrepancy between the modeled value and the most exact measurement of about 30 %. So I think the search for additional explanations makes sense.
 
  • #6
Your reference cites the LLRE measurement as 3.82+/-0.7cm/yr.

It also cites the tidal measurements as 2.9+/-0.6cm/yr.

So it looks like there is an overlap.
 
Last edited:
  • #7
paisiello2 said:
Your reference cites the LLRE measurement as 3.82+/-0.7cm/yr.

Ok, I see your point.
The reference has obviously an error, because in the first lines the uncertainty of the LLRE measurements is 0.07 cm/yr (point before the zero) and later on it is 0.7 cm/yr.

But still the 2.91 cm/yr which the detailed numerical simulation predicts is outside of the 3.82+/-0.7cm/yr.
In the reference they write:
LLRE's laser light differs by over 12 [itex] \delta [/itex].
 
  • #8
It should probably be noted here, that discussions of personal theories are not allowed on PF as per the rules.
Similarly, papers sourced need to be from peer-reviewed publications. Conference proceedings usually do not fall into that category, and I can't find the article mentioned in post #3 as being published in any of the accepted journals.
 
  • #9
I've closed this thread.

First, a minor correction:
paisiello2 said:
Your reference cites the LLRE measurement as 3.82+/-0.7cm/yr.
The measurements from the Lunar Laser Ranging Experiment yield a recession rate of 3.82±0.07 cm/year, not 3.82±0.7.

The current rate is high compared to the average over the last 620 million years, 2.17±0.31 cm/year (G.Williams, "Geological constraints on the Precambrian history of Earth's rotation and the Moon's orbit". [Reviews of Geophysics 38 (1): 37–60 (2000).)

I don't know where the paper cited in post #3 got it's numbers, particularly the value from "detailed numerical simulations."

The value of 2.17±0.31 cm/year is an average rate. The lunar recession rate strongly depends on the layout of the Earth's continents. The recession rate was low during periods of supercontinents and when the continents allowed a freer tidal flow. The recession rate is high right now because the Americas and Africa/Eurasia create two huge barriers to a free tidal flow. There's more tidal friction now than there has been in the past, so the recession rate is higher now than it has been on average.This site is not the place for speculative nonsense such as an expanding Earth. This thread is closed.
 

1. What is meant by "orbit change due to hypothetical mass change"?

"Orbit change due to hypothetical mass change" refers to the potential alteration of an object's orbit around another body, such as a planet or star, caused by a change in the mass of either the orbiting object or the central body. This concept is often studied in astronomy and astrophysics to better understand the dynamics of celestial bodies.

2. How does changing the mass of an object affect its orbit?

According to Kepler's laws of planetary motion, an object's orbit is determined by its mass and the mass of the body it is orbiting. Therefore, changing the mass of either the orbiting object or the central body will result in a change in the orbit. This can lead to changes in the orbital period, shape, and distance of the orbiting object.

3. Can the hypothetical mass change of a planet or star cause significant changes in the orbits of other objects in its system?

Yes, a change in the mass of a central body can have a significant impact on the orbits of other objects in its system. For example, a decrease in the mass of a star could cause its planets to move closer to it, resulting in shorter orbital periods and potentially changing the habitability of those planets.

4. Are there any real-life examples of orbit changes caused by a hypothetical mass change?

Yes, there are several known cases where the mass of a celestial body has changed, resulting in observable changes in the orbits of other objects. One example is the asteroid 21 Lutetia, which has experienced a decrease in mass due to collisions, resulting in changes in its orbit around the Sun.

5. How do scientists study the effects of hypothetical mass changes on orbital dynamics?

Scientists use computer simulations and mathematical models to study the effects of hypothetical mass changes on orbital dynamics. They also observe and analyze real-life examples, as well as conduct experiments in controlled environments, to better understand the relationship between mass and orbit in different scenarios.

Similar threads

  • Special and General Relativity
2
Replies
43
Views
2K
  • Special and General Relativity
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
974
  • Advanced Physics Homework Help
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
11
Views
1K
  • Astronomy and Astrophysics
Replies
18
Views
3K
  • Introductory Physics Homework Help
10
Replies
335
Views
7K
  • Astronomy and Astrophysics
Replies
11
Views
2K
  • Astronomy and Astrophysics
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
144
Back
Top