The Unusual Force of Gravity on the Moon

[dougettinger]

Why is the Moon's orbit held in its place by the Sun's gravity force and not the Earth's gravity force? In fact, the Moon is the only satellite whose parent planet's gravity does not hold it in orbit around itself.

[Amir]

i didn't know that. is it mentioned any where or you are just asking about it?
it looks so odd! if it's held in it's orbit by sun, why it move in that path around earth? and why it's period of orbiting it equal to earth's while they have different masses?

[dougettinger]

i didn't know that. is it mentioned any where or you are just asking about it?
it looks so odd! if it's held in it's orbit by sun, why it move in that path around earth? and why it's period of orbiting it equal to earth's while they have different masses?

The Moon and Earth are in sychronise orbits with the Moon forming a wave motion because it has less mass.

Compute the force of gravity between the Moon and Earth and the Moon and Sun. The force of gravity between the Sun and Moon is significantly greater.

The orbital path of the Moon around the Sun is everywhere concave toward the Sun unlike any other natural satellite in the solar system.

These conditions are known. My question is - why?

[Amir]

i hadn't noticed that until now. yes, that's right. i just computed the forces, what i got was that the force of gravity between sun and moon is ~2.3 times greater than the force of gravity between earth and moon.
now you are asking why and i can't say anything else but "just because of random distances".
as you said moon is mainly orbiting sun, but it's also too near to earth to avoid it's gravitation field, so there's no other way but orbiting earth too! -couldn't it be that simple?

[neufer]

The Moon and Earth are in sychronise orbits with the Moon forming a wave motion because it has less mass.

Compute the force of gravity between the Moon and Earth and the Moon and Sun.
The force of gravity between the Sun and Moon is significantly greater.

The orbital path of the Moon around the Sun is everywhere concave toward the Sun unlike any other natural satellite in the solar system.

These conditions are known. My question is - why?

The moon is the furthest known natural satellite of the earth

The furthest known natural satellites of all of the gas giants share this same property
that their orbital paths around the Sun are everywhere concave toward the Sun
(although, just barely in the case of Saturn).

[hstarbuck]

Interesting stuff here. Also, the Sun's pull on the Earth is ~18 times that as the Moon's pull on the Earth yet the Moon has a greater effect on the Earth's tides. I am thinking that the difference in force from the front side to the back side of the Earth is greater for the Moon's pull but I have not done this calculation. Basically if you think of an inverse square graph and go out further into the field than the difference in force is less for same distance (Earth's diameter).

If the moon and Earth were removed from the Solar system and placed in empty space with the same velocities and distance a month would still be a month right? I calculate 27.4 some odd days with formula from my physics book. So what does any of this mean? What effect does it have?

[Chris Peterson]

Why is the Moon's orbit held in its place by the Sun's gravity force and not the Earth's gravity force? In fact, the Moon is the only satellite whose parent planet's gravity does not hold it in orbit around itself.

Because. Really, that's the answer. It's just a question of orbital mechanics. I'm not at all sure the Moon is the only satellite in the Solar System that has this property, but it doesn't matter. For any star-planet-moon system you can construct scenarios like this, simply by selecting the right combination of masses and orbital distances. That determines which gravity source is dominant from the standpoint of the moon.

[neufer]

the Sun's pull on the Earth is ~18 times that as the Moon's pull on the Earth yet the Moon has a greater effect on the Earth's tides. I am thinking that the difference in force from the front side to the back side of the Earth is greater for the Moon's pull but I have not done this calculation. Basically if you think of an inverse square graph and go out further into the field than the difference in force is less for same distance (Earth's diameter).

The Sun's pull on the Earth is a lot more than ~18 times that as the Moon's pull on the Earth. (Maybe more like 180 times.)

But the difference in force from the front side to the back side of the Earth is indeed greater for the Moon's pull as you state.

Tidal forces drop off with the cube of the distance not the square. Since volume also increases with the cube of the distance and the moon just covers the sun during a solar eclipse it turns out that the difference between lunar tides & solar tides is just due to their different densities.

The moon is 2.4 times denser than the sun on average; hence, it's tidal effect is 2.4 times larger.

If the moon and Earth were removed from the Solar system and placed in empty space with the same velocities and distance a month would still be a month right? I calculate 27.4 some odd days with formula from my physics book. So what does any of this mean? What effect does it have?

Under that situation the moon does take a few minutes shorter time to orbit the earth than it currently does (due to the orbiting around the sun).

If such a 'free range' moon were out at a distance of 1.5 million km then it would take about 211 days to orbit the earth.

However, artificial satellites out at 1.5 million km in the L1 & L2 Lagrange points actually take 365 days to orbit the earth due to the influence of the sun.

[hstarbuck]

The Sun's pull on the Earth is a lot more than ~18 times that as the Moon's pull on the Earth. (Maybe more like 180 times.)

Drat those powers of ten--I was in a rush when I did the calculation. But hey, power of ten, factor of 2, or factor of pi off, don't I still get partial credit if I show my work?

Anyway, a couple minutes difference isn't much so I--my opinion--would say that the Earth's gravitational effects on the moon dominate over the Sun's for relevance to my world--even though they are smaller.

[Amir]

The Sun's pull on the Earth is a lot more than ~18 times that as the Moon's pull on the Earth. (Maybe more like 180 times.)

the Sun's pull on the Earth is ~18 times that as the Moon's pull on the Earth

which formula did you use to calculate that?

[hstarbuck]

which formula did you use to calculate that?

Just Newton's Law for Universal Gravitation using Sun's mass and 1 AU for Sun's acceleration at Earth's distance and Moon's mass and average distance of Earth to Moon (center to center). It actually came out to 170 times more for Sun's acceleration. I left out the Earth's mass because it cancels out--the accelerations work just as well. Did I make another mistake?--very possible. a = Gm/r²

[Amir]

Did I make another mistake?--very possible.

no, i was just asking. i got the same number.

[dougettinger]

The Moon and Earth are in sychronise orbits with the Moon forming a wave motion because it has less mass.

Compute the force of gravity between the Moon and Earth and the Moon and Sun.
The force of gravity between the Sun and Moon is significantly greater.

The orbital path of the Moon around the Sun is everywhere concave toward the Sun unlike any other natural satellite in the solar system.

These conditions are known. My question is - why?

The moon is the furthest known natural satellite of the earth

The furthest known natural satellites of all of the gas giants share this same property
that their orbital paths around the Sun are everywhere concave toward the Sun
(although, just barely in the case of Saturn).

I wish to have more details about this fact. Are you saying that only the farthest natural satellites (spheroidal in shape) have this characteristic of concave shaped orbits pointed toward the Sun for the outer planets? So then all the other satellites do not have this characteristic. Is this correct? I was told that only the Moon had this characteristic and I could be wrong. Is there a certain relationship between the planet's orbit size and the satellite's orbit size that creates this characteristic?

[dougettinger]

Why is the Moon's orbit held in its place by the Sun's gravity force and not the Earth's gravity force? In fact, the Moon is the only satellite whose parent planet's gravity does not hold it in orbit around itself.

Because. Really, that's the answer. It's just a question of orbital mechanics. I'm not at all sure the Moon is the only satellite in the Solar System that has this property, but it doesn't matter. For any star-planet-moon system you can construct scenarios like this, simply by selecting the right combination of masses and orbital distances. That determines which gravity source is dominant from the standpoint of the moon.

Let's return to orbital mechanics for a moment. Yes, the Moon indeed is the only natural satellite with this property. Issac Asimov computed these forces for all the known major satellites in one of his astronomy books. If the Earth is removed from the Earth-Moon system the Moon should remain in its same orbit around the Sun. If an outer parent planet is removed from its system of satellites, do these satellites remain close to their parent's orbit around the Sun disregarding influences between the satellites themselves? I believe the outer planet satellites would drastically deviate from their parent planet's orbital diameter and period.

I am trying to arrive at the conclusion that the Moon is indeed an anomalous satellite in the solar system. But how much anomalous is it? And why?

[neufer]

The moon is the furthest known natural satellite of the earth

The furthest known natural satellites of all of the gas giants share this same property
that their orbital paths around the Sun are everywhere concave toward the Sun
(although, just barely in the case of Saturn).

I wish to have more details about this fact. Are you saying that only the farthest natural satellites (spheroidal in shape) have this characteristic of concave shaped orbits pointed toward the Sun for the outer planets? So then all the other satellites do not have this characteristic. Is this correct? I was told that only the Moon had this characteristic and I could be wrong. Is there a certain relationship between the planet's orbit size and the satellite's orbit size that creates this characteristic?

Now, I never said anything about the satellites being spheroidal in shape.

The criteria for a circular low inclination satellite orbit is that:

• (R/r) > (P/p)²

where:
R = the planet's orbit size
r = the satellite's orbit size
P = the planet's orbital period
p = the satellite's orbital period
------------------------------------------

• Neptune / Neso example:
  
  _R = 4,503 Gm _______ P = 60,190 days
  _r = 49.285 Gm ______ p = 9741 days
  (R/r) = 91.4 ________ (P/p)² = 38.2

Note, however, like many outer moons, Neso follows a highly inclined & highly eccentric orbit.
Ergo, I can't guarantee that this doesn't affect the situation though I doubt it.

In any event, it is certainly proof that the gas planets are at least capable of having outer moons that satisfy your criteria.

[neufer]

Issac Asimov computed these forces for all the known major satellites in one of his astronomy books.

It is certainly true for all major satellites.

If the Earth is removed from the Earth-Moon system the Moon should remain in its same orbit around the Sun.

Not EXACTLY the same orbit...but close

If an outer parent planet is removed from its system of satellites, do these satellites remain close to their parent's orbit around the Sun disregarding influences between the satellites themselves? I believe the outer planet satellites would drastically deviate from their parent planet's orbital diameter and period.

The criteria for a satellite being unlikely to leave the solar system
if the parent planet is removed is a relatively mild one:

• (R/r) > (P/p)

where:
R = the planet's orbit size
r = the satellite's orbit size
P = the planet's orbital period
p = the satellite's orbital period
------------------------------------------

• Neptune / Triton example:
  
  _R = 4,503 Gm _________ P = 60,190 days
  _r = 0.355 Gm _________ p = 5.877 days
  (R/r) = 12,700 ________ (P/p) = 10,242

Triton probably would not leave the solar system
although all six moons interior to Triton certainly would.

None of the Galilean moons would leave the solar system.

[dougettinger]

The moon is the furthest known natural satellite of the earth

The furthest known natural satellites of all of the gas giants share this same property
that their orbital paths around the Sun are everywhere concave toward the Sun
(although, just barely in the case of Saturn).

I wish to have more details about this fact. Are you saying that only the farthest natural satellites (spheroidal in shape) have this characteristic of concave shaped orbits pointed toward the Sun for the outer planets? So then all the other satellites do not have this characteristic. Is this correct? I was told that only the Moon had this characteristic and I could be wrong. Is there a certain relationship between the planet's orbit size and the satellite's orbit size that creates this characteristic?

Now, I never said anything about the satellites being spheroidal in shape.

The criteria for a circular low inclination satellite orbit is that:

• (R/r) > (P/p)²

where:
R = the planet's orbit size
r = the satellite's orbit size
P = the planet's orbital period
p = the satellite's orbital period
------------------------------------------

• Neptune / Neso example:
  
  _R = 4,503 Gm _______ P = 60,190 days
  _r = 49.285 Gm ______ p = 9741 days
  (R/r) = 91.4 ________ (P/p)² = 38.2

Note, however, like many outer moons, Neso follows a highly inclined & highly eccentric orbit.
Ergo, I can't guarantee that this doesn't affect the situation though I doubt it.

In any event, it is certainly proof that the gas planets are at least capable of having outer moons that satisfy your criteria.

Art, thank you for your equation that satisfies my criteria for an orbit being at all times concave toward the Sun. I would be interested in its derivation. My math skills are limited.

However, you did cut hairs lengthwise which is understandable. You picked an irregular satellite that is retrograde with highly inclined and elliptical orbit and considered to be part of a collisional scenario. Neso is also the farthest satellite from the Sun except maybe Charon of Pluto but then Pluto is not a planet. I was really applying my criteria (which I can now represent by an equation - WOW) to normal satellites in aproximately co-planar, prograde orbits, considered to be formed at roughly the same time as the planets. Also, these satellites would be massive enough to form spheroid shapes and not be irregular. I apologize for not being clear about those facts.

[dougettinger]

Issac Asimov computed these forces for all the known major satellites in one of his astronomy books.

It is certainly true for all major satellites.

If the Earth is removed from the Earth-Moon system the Moon should remain in its same orbit around the Sun.

Not EXACTLY the same orbit...but close

If an outer parent planet is removed from its system of satellites, do these satellites remain close to their parent's orbit around the Sun disregarding influences between the satellites themselves? I believe the outer planet satellites would drastically deviate from their parent planet's orbital diameter and period.

The criteria for a satellite being unlikely to leave the solar system
if the parent planet is removed is a relatively mild one:

• (R/r) > (P/p)

where:
R = the planet's orbit size
r = the satellite's orbit size
P = the planet's orbital period
p = the satellite's orbital period
------------------------------------------

• Neptune / Triton example:
  
  _R = 4,503 Gm _________ P = 60,190 days
  _r = 0.355 Gm _________ p = 5.877 days
  (R/r) = 12,700 ________ (P/p) = 10,242

Triton probably would not leave the solar system
although all six moons interior to Triton certainly would.

None of the Galilean moons would leave the solar system.

Thanks again for your second equation and an example. Is this equation derived from Kepler's third law? It is somewhat different from the first equation. Let me try to explain this equation in words. If a planet is removed from a particular planet-satellite system in our solar system, the satellite will escape the Sun's gravity field if the given parameters are used in the subject equation and the ratio of R/r is less than P/p.

The conclusion that can be made from your example is that the Moon is little affected by a missing Earth while a regular satellite in the Jovian system would be greatly affected even though it may not escape the Sun's gravity field. Is this correct? I really would appreciate a reference for your equations. Perhaps you derived these equations yourself.

[neufer]

The criteria for a circular low inclination satellite orbit is that:

• (R/r) > (P/p)²

where:
R = the planet's orbit size
r = the satellite's orbit size
P = the planet's orbital period
p = the satellite's orbital period
------------------------------------------

• Neptune / Neso example:
  
  _R = 4,503 Gm _______ P = 60,190 days
  _r = 49.285 Gm ______ p = 9741 days
  (R/r) = 91.4 ________ (P/p)² = 38.2

Note, however, like many outer moons, Neso follows a highly inclined & highly eccentric orbit.
Ergo, I can't guarantee that this doesn't affect the situation though I doubt it.

In any event, it is certainly proof that the gas planets are at least capable of having outer moons that satisfy your criteria.

Art, thank you for your equation that satisfies my criteria for an orbit being at all times concave toward the Sun. I would be interested in its derivation. My math skills are limited.

Let's start with your own criteria that "the force of gravity between the Sun and moon
is significantly greater than between the moon and it's parent planet:

• Basically {equation 1} :
  (M_sun / R²_planet) > (m_planet / r²_moon)
  
  but Kepler's third law (plus a little Newton) tells us that:
  constant x Mass_parent = (R³_satellite / Period²_satellite)
  
  substituting these masses into equation 1 gets rid of the Sun's Mass:
  (R_planet / Period²_planet ) > (r_moon / period²_moon)
  or (R_planet / r_moon) > (Period_planet / period_moon)²
  ...............................................................
  the other (less restrictive) criteria just requires that
  the moon's orbital velocity is small vis-a-vis the parent planet's orbital velocity:
  (2π R_planet / Period_planet ) > (2π r_moon / period_moon)
  or (R_planet / r_moon) > (Period_planet / period_moon)

[dougettinger]

The criteria for a circular low inclination satellite orbit is that:

• (R/r) > (P/p)²

where:
R = the planet's orbit size
r = the satellite's orbit size
P = the planet's orbital period
p = the satellite's orbital period
------------------------------------------

• Neptune / Neso example:
  
  _R = 4,503 Gm _______ P = 60,190 days
  _r = 49.285 Gm ______ p = 9741 days
  (R/r) = 91.4 ________ (P/p)² = 38.2

Note, however, like many outer moons, Neso follows a highly inclined & highly eccentric orbit.
Ergo, I can't guarantee that this doesn't affect the situation though I doubt it.

In any event, it is certainly proof that the gas planets are at least capable of having outer moons that satisfy your criteria.

Art, thank you for your equation that satisfies my criteria for an orbit being at all times concave toward the Sun. I would be interested in its derivation. My math skills are limited.

Let's start with your own criteria that "the force of gravity between the Sun and moon
is significantly greater than between the moon and it's parent planet:

• Basically {equation 1} :
  (M_sun / R²_planet) > (m_planet / r²_moon)
  
  but Kepler's third law (plus a little Newton) tells us that:
  constant x Mass_parent = (R³_satellite / Period²_satellite)
  
  substituting these masses into equation 1 gets rid of the Sun's Mass:
  (R_planet / Period²_planet ) > (r_moon / period²_moon)
  or (R_planet / r_moon) > (Period_planet / period_moon)²
  ...............................................................
  the other (less restrictive) criteria just requires that
  the moon's orbital velocity is small vis-a-vis the parent planet's orbital velocity:
  (2π R_planet / Period_planet ) > (2π r_moon / period_moon)
  or (R_planet / r_moon) > (Period_planet / period_moon)

Art, thanks for deriving this equation using Kepler and Newton. I understand also the the second equation, but do not grasp the logic or geometry of this equation that supports my criteria of everywhere concave toward the Sun. I have a mental block. Could you possible explain this equation in a graphical manner? Asimov explained that he laid out the orbital path on a drawing which I confirmed myself.

I applied your second equation to the Jovian satellites. The equation indicates that the Moon is no exception in having this orbital characteristic.

[makc]

how about we compute gravitational force pulling the earth towards galaxy center?

[Chris Peterson]

how about we compute gravitational force pulling the earth towards galaxy center?

It's a bit of a messy calculation due to the distributed mass of the Milky Way. You could work backwards, starting with our ~250 million year orbital period. You can also get a sense of the magnitude of the force by considering our orbital speed is 220 km/s, and our escape velocity with respect to the Milky Way is only 525 km/s. So we're not all that closely bound- it wouldn't take much in terms of a closely passing star to send us on our way (not that this is a very likely scenario, however).

[neufer]

...our orbital speed is 220 km/s, and our escape velocity with respect to the Milky Way is only 525 km/s. So we're not all that closely bound- it wouldn't take much in terms of a closely passing star to send us on our way (not that this is a very likely scenario, however).

If a passing star came close enough to our sun to give Sol a kick of > 100 km/s, I would think that being ejected from the Milky Way would be the last of our problems.