Earth, Sun, Moon Orbit

[Natko]

Why does the Earth orbit the Sun, but the Moon pull on Earth's tides? Earth orbits the sun, proving that the Sun has a greater gravitational pull on the Earth than the Moon. However, the Moon control's Earth's tides (unlike the Sun), which proves the opposite-the Moon has a greater gravitational pull than the Sun.

So which one (Sun or Moon) have a greater gravitational pull on the Earth.

 

[tiny-tim]

Welcome to PF!

Hi Natko! Welcome to PF!

What matters is the gradient of the gravitational field …

the rate at which the gravity increases or decreaes with height.

Since the field is proportional to 1/r², its gradient is proportional to 1/r³.

The Sun's mass is so much greater than the Moon's mass that the tiny 1/r² factor still leaves the Sun winning on ordinary gravitational strength.

But the even tinier 1/r³ factor for tidal effects leaves the Moon winning (though not by much)!

 

[Natko]

Hi Natko! Welcome to PF!

What matters is the gradient of the gravitational field …

the rate at which the gravity increases or decreaes with height.

Since the field is proportional to 1/r², its gradient is proportional to 1/r³.

The Sun's mass is so much greater than the Moon's mass that the tiny 1/r² factor still leaves the Sun winning on ordinary gravitational strength.

But the even tinier 1/r³ factor for tidal effects leaves the Moon winning (though not by much)!

Thanks for the quick reply!
Can you explain this in more understandable terms as I am only in Grade 9 please, especially what you mean by "gradient," "field," and the "1/r3 factor".

 

[tiny-tim]

Hi Natko!

For "field", read "force" instead.

"Gradient" is how fast the force changes as you go away from the Sun (or Moon).

You probably know that the force of gravity is proportional to 1/r², where r is the distance from the Sun (or Moon) …

if you draw a graph of y = 1/x², you see that it flattens out pretty fast as x gets larger

but its slope (the gradient) flattens out even faster still.

(i'm going to bed now :zzz: … would anyone else like to carry on from there? )

 

[Drakkith]

Thanks for the quick reply!
Can you explain this in more understandable terms as I am only in Grade 9 please, especially what you mean by "gradient," "field," and the "1/r3 factor".

The gradiant is the difference in the strength of gravity between two distances away from an object. For example, the force of gravity on your head is less than it is on your feet.

Both the Moon and the Earth orbit the Sun, which is probably not how you've heard it before. But it's true! Since the gradiant of the Sun's gravity has fallen off so much out hear at Earth, the water on each side doesn't feel much of a difference. However, while the Moon is much less massive than the Sun, it is also much much closer. As tiny-tim pointed out, the gradiant falls off faster than the strength. So the different sides of the Earth feel a much greater difference in the strength of gravity from the Moon.

What happens is that the Moon pulls the near side the strongest, then the middle of the Earth a little less, and then the opposite side even less. The Moon pulls the water on the near side towards it leading to a high tide there, then it pulls the rest of the Earth AWAY from the water on the other side, leading to a high tide there as well, with low tides in between.

 

[D H]

Thanks for the quick reply!
Can you explain this in more understandable terms as I am only in Grade 9 please, especially what you mean by "gradient," "field," and the "1/r3 factor".

The Earth as a whole is accelerating toward the moon. This acceleration is given by

$$a_{\text{earth}} = \frac{GM_{\text{moon}}}{R^2}$$

where G is the universal gravitational constant, M_moon, and R is the distance between the center of the moon and the center of the earth. Now imagine a drop of water on the surface of the Earth directly on the line from the Earth to the moon. This drop of water is a bit closer to the moon than is the center of the earth. Denoting the radius of the Earth with a lower case r, the distance between the drop and the moon is R-r. This shorter distance means the drop experiences a slightly greater acceleration than does the Earth as a whole:

$$a_{\text{drop}} = \frac{GM_{\text{moon}}}{(R-r)^2}$$

Factoring R out of R-r yields $R-r=R(1-r/R)$. Thus another way to write the acceleration of the drop is

$$a_{\text{drop}} = \frac{GM_{\text{moon}}}{R^2}\frac1{(1-r/R)^2}$$

The acceleration of the drop relative to the Earth as a whole is simply the difference between these:

$$\begin{aligned} a_{\text{rel}} &= \frac{GM_{\text{moon}}}{(R-r)^2} - \frac{GM_{\text{moon}}}{R^2} \\ &= \frac{GM_{\text{moon}}}{R^2}\frac1{(1-r/R)^2} - \frac{GM_{\text{moon}}}{R^2} \\ &= \frac{GM_{\text{moon}}}{R^2}\left(\frac1{(1-r/R)^2}-1\right) \end{aligned}$$

The first term in parentheses can be rewritten as

$$\frac1{(1-r/R)^2} = 1+2\frac r R + 3\left(\frac r R\right)^2 + 4\left(\frac r R\right)^3 + \cdots$$

With this, the acceleration of the drop relative to the Earth becomes

$$\begin{aligned} a_{\text{rel}} &= \frac{GM_{\text{moon}}}{R^2}\left(\frac1{(1-r/R)^2}-1\right) \\ &= \frac{GM_{\text{moon}}}{R^2} \left(2\frac r R + 3\left(\frac r R\right)^2 + 4\left(\frac r R\right)^3 + \cdots\right) \end{aligned}$$
Because the distance to the moon or the sun is much greater than the radius of the earth, the first term is going to be the dominant one. Thus

$$a_{\text{rel}} \approx \frac{GM_{\text{moon}}}{R^2} 2\frac r R = \frac{2GM_{\text{moon}}r}{R^3}$$

So, 1/R³. Substitute the moon for the sun and the same kind of inverse cube relation will hold.

 

[Natko]

The gradiant is the difference in the strength of gravity between two distances away from an object. For example, the force of gravity on your head is less than it is on your feet.

Both the Moon and the Earth orbit the Sun, which is probably not how you've heard it before. But it's true! Since the gradiant of the Sun's gravity has fallen off so much out hear at Earth, the water on each side doesn't feel much of a difference. However, while the Moon is much less massive than the Sun, it is also much much closer. As tiny-tim pointed out, the gradiant falls off faster than the strength. So the different sides of the Earth feel a much greater difference in the strength of gravity from the Moon.

What happens is that the Moon pulls the near side the strongest, then the middle of the Earth a little less, and then the opposite side even less. The Moon pulls the water on the near side towards it leading to a high tide there, then it pulls the rest of the Earth AWAY from the water on the other side, leading to a high tide there as well, with low tides in between.

What are your sources for the Earth is pulled away from the water?

 

[Natko]

The Earth as a whole is accelerating toward the moon. This acceleration is given by

$$a_{\text{earth}} = \frac{GM_{\text{moon}}}{R^2}$$

where G is the universal gravitational constant, M_moon, and R is the distance between the center of the moon and the center of the earth. Now imagine a drop of water on the surface of the Earth directly on the line from the Earth to the moon. This drop of water is a bit closer to the moon than is the center of the earth. Denoting the radius of the Earth with a lower case r, the distance between the drop and the moon is R-r. This shorter distance means the drop experiences a slightly greater acceleration than does the Earth as a whole:

$$a_{\text{drop}} = \frac{GM_{\text{moon}}}{(R-r)^2}$$

Factoring R out of R-r yields $R-r=R(1-r/R)$. Thus another way to write the acceleration of the drop is

$$a_{\text{drop}} = \frac{GM_{\text{moon}}}{R^2}\frac1{(1-r/R)^2}$$

The acceleration of the drop relative to the Earth as a whole is simply the difference between these:

$$\begin{aligned} a_{\text{rel}} &= \frac{GM_{\text{moon}}}{(R-r)^2} - \frac{GM_{\text{moon}}}{R^2} \\ &= \frac{GM_{\text{moon}}}{R^2}\frac1{(1-r/R)^2} - \frac{GM_{\text{moon}}}{R^2} \\ &= \frac{GM_{\text{moon}}}{R^2}\left(\frac1{(1-r/R)^2}-1\right) \end{aligned}$$

The first term in parentheses can be rewritten as

$$\frac1{(1-r/R)^2} = 1+2\frac r R + 3\left(\frac r R\right)^2 + 4\left(\frac r R\right)^3 + \cdots$$

With this, the acceleration of the drop relative to the Earth becomes

$$\begin{aligned} a_{\text{rel}} &= \frac{GM_{\text{moon}}}{R^2}\left(\frac1{(1-r/R)^2}-1\right) \\ &= \frac{GM_{\text{moon}}}{R^2} \left(2\frac r R + 3\left(\frac r R\right)^2 + 4\left(\frac r R\right)^3 + \cdots\right) \end{aligned}$$
Because the distance to the moon or the sun is much greater than the radius of the earth, the first term is going to be the dominant one. Thus

$$a_{\text{rel}} \approx \frac{GM_{\text{moon}}}{R^2} 2\frac r R = \frac{2GM_{\text{moon}}r}{R^3}$$

So, 1/R³. Substitute the moon for the sun and the same kind of inverse cube relation will hold.

This seems very credible, but I cannot understand anything. Sorry.

 

[tiny-tim]

Hi Natko!

(just got up … :zzz:)

Imagine a spaceship in space, pointing away from the Sun (it can be orbiting, or falling, it doesn't matter).

The gravity on the bottom of the spaceship is stronger than the gravity on the top.

If the spaceship was stretchy, this difference in force would stretch it.

The closer the spaceship is to the Sun, the bigger the difference, and so the bigger the stretching force.

Near a black hole, the difference is very strong, and the extreme stretching there (of spaceships or of astronauts) is popularly known as "spaghettification".

Of course, physicists don't call it "stretching force", they call it "tidal force".

Let's calculate how much the stretching force is.

Suppose the spaceship has length L, and suppose its centre is at distance R from the centre of the Sun.

Then the gravity on the bottom of the spaceship is proportional to 1/(R - L/2)²,

and the gravity on the top of the spaceship is proportional to 1/(R + L/2)² (slightly less).

The difference is 1/(R - L/2)² - 1/(R + L/2)²

= ((R + L/2)² - (R - L/2)²)/((R + L/2)²(R - L/2)²)

= 2RL/(R² - L²/4)²

which if L is very much less than R, is very nearly 2RL/R⁴, or 2L/R³.

So the difference in force on a spaceship of length L is 2L/R³ …

ie the stretching force per length is proportional to 1/R³.

Water is stretchy (well, it flows, which has the same effect ), and so the water round the Earth is stretched away from the centre of the Earth, in line with the Sun.

This stretching ("tidal") force is proportional to 1/R³.

There is of course also a stretching force cause by the Moon, proportional to 1/R_Moon³, where R_Moon is the distance to the Moon.

The Sun is about 400 times further from us than the Moon is (R/R_Moon ~ 400),

so (Sun's tidal force)/(Moon's tidal force) is about 400 times smaller than (Sun's gravity)/(Moon's gravity). ​

 

[Cosmo Novice]

The Earth as a whole is accelerating toward the moon. This acceleration is given by

$$a_{\text{earth}} = \frac{GM_{\text{moon}}}{R^2}$$

where G is the universal gravitational constant, M_moon, and R is the distance between the center of the moon and the center of the earth. Now imagine a drop of water on the surface of the Earth directly on the line from the Earth to the moon. This drop of water is a bit closer to the moon than is the center of the earth. Denoting the radius of the Earth with a lower case r, the distance between the drop and the moon is R-r. This shorter distance means the drop experiences a slightly greater acceleration than does the Earth as a whole:

$$a_{\text{drop}} = \frac{GM_{\text{moon}}}{(R-r)^2}$$

Factoring R out of R-r yields $R-r=R(1-r/R)$. Thus another way to write the acceleration of the drop is

$$a_{\text{drop}} = \frac{GM_{\text{moon}}}{R^2}\frac1{(1-r/R)^2}$$

The acceleration of the drop relative to the Earth as a whole is simply the difference between these:

$$\begin{aligned} a_{\text{rel}} &= \frac{GM_{\text{moon}}}{(R-r)^2} - \frac{GM_{\text{moon}}}{R^2} \\ &= \frac{GM_{\text{moon}}}{R^2}\frac1{(1-r/R)^2} - \frac{GM_{\text{moon}}}{R^2} \\ &= \frac{GM_{\text{moon}}}{R^2}\left(\frac1{(1-r/R)^2}-1\right) \end{aligned}$$

The first term in parentheses can be rewritten as

$$\frac1{(1-r/R)^2} = 1+2\frac r R + 3\left(\frac r R\right)^2 + 4\left(\frac r R\right)^3 + \cdots$$

With this, the acceleration of the drop relative to the Earth becomes

$$\begin{aligned} a_{\text{rel}} &= \frac{GM_{\text{moon}}}{R^2}\left(\frac1{(1-r/R)^2}-1\right) \\ &= \frac{GM_{\text{moon}}}{R^2} \left(2\frac r R + 3\left(\frac r R\right)^2 + 4\left(\frac r R\right)^3 + \cdots\right) \end{aligned}$$
Because the distance to the moon or the sun is much greater than the radius of the earth, the first term is going to be the dominant one. Thus

$$a_{\text{rel}} \approx \frac{GM_{\text{moon}}}{R^2} 2\frac r R = \frac{2GM_{\text{moon}}r}{R^3}$$

So, 1/R³. Substitute the moon for the sun and the same kind of inverse cube relation will hold.

Far be it from me to criticize but when the OP stated he was in grade 9 and needed a simple explanation I would hardly credit the above as useful!

Write for your target audience!

 

[D H]

Far be it from me to criticize but when the OP stated he was in grade 9 and needed a simple explanation I would hardly credit the above as useful!

Write for your target audience!

I thought I was writing for the target audience! I intentionally avoided calculus. Others talked about gradients, which is calculus (and not even freshman level calculus). Algebra is taught in the 8th or 9th grade. My post was algebraic; it should be accessible to a bright eighth or ninth grade student.

 

[Cosmo Novice]

I thought I was writing for the target audience! I intentionally avoided calculus. Others talked about gradients, which is calculus (and not even freshman level calculus). Algebra is taught in the 8th or 9th grade. My post was algebraic; it should be accessible to a bright eighth or ninth grade student.

Well I was in one of the higher tier maths classes during 8th/9th grade and I don't think it would have been accessable to me. Maybe high school mathematics taught in the UK is not upto standard - which may be more likely.

Anyways this was in no way intended to be rude or as a criticism! I am sure your efforts are appreciated and I have seen a number of very good posts from you which have helped me.

Cosmo

 

[Natko]

Hi Natko!

(just got up … :zzz:)

Imagine a spaceship in space, pointing away from the Sun (it can be orbiting, or falling, it doesn't matter).

The gravity on the bottom of the spaceship is stronger than the gravity on the top.

If the spaceship was stretchy, this difference in force would stretch it.

The closer the spaceship is to the Sun, the bigger the difference, and so the bigger the stretching force.

Near a black hole, the difference is very strong, and the extreme stretching there (of spaceships or of astronauts) is popularly known as "spaghettification".

Of course, physicists don't call it "stretching force", they call it "tidal force".

Let's calculate how much the stretching force is.

Suppose the spaceship has length L, and suppose its centre is at distance R from the centre of the Sun.

Then the gravity on the bottom of the spaceship is proportional to 1/(R - L/2)²,

and the gravity on the top of the spaceship is proportional to 1/(R + L/2)² (slightly less).

The difference is 1/(R - L/2)² - 1/(R + L/2)²

= ((R + L/2)² - (R - L/2)²)/((R + L/2)²(R - L/2)²)

= 2RL/(R² - L²/4)²

which if L is very much less than R, is very nearly 2RL/R⁴, or 2L/R³.

So the difference in force on a spaceship of length L is 2L/R³ …

ie the stretching force per length is proportional to 1/R³.

Water is stretchy (well, it flows, which has the same effect ), and so the water round the Earth is stretched away from the centre of the Earth, in line with the Sun.

This stretching ("tidal") force is proportional to 1/R³.

There is of course also a stretching force cause by the Moon, proportional to 1/R_Moon³, where R_Moon is the distance to the Moon.

The Sun is about 400 times further from us than the Moon is (R/R_Moon ~ 400),

so (Sun's tidal force)/(Moon's tidal force) is about 400 times smaller than (Sun's gravity)/(Moon's gravity). ​

So what you are saying is that gradient and tidal force are the same thing.
And since the Sun is farther away from Earth than the moon is, it has a smaller gradient/tidal force. Thus, this causes the Moon to pull on one side of the Earth, creating a tide on each side, and the Sun has an effect on the entire Earth, pulling it into orbit. But the Sun still has a stronger gravitational pull on the Earth than the Moon does, even the gradient/tidal force is smaller.

 

[tiny-tim]

So what you are saying is that gradient and tidal force are the same thing.

Yes.

And since the Sun is farther away from Earth than the moon is, it has a smaller gradient/tidal force.…But the Sun still has a stronger gravitational pull on the Earth than the Moon does, even the gradient/tidal force is smaller.

Yes.

Thus, this causes the Moon to pull on one side of the Earth, creating a tide on each side, and the Sun has an effect on the entire Earth, pulling it into orbit.

No, gravitational forces pull (in one direction), but tidal forces don't pull, they stretch, and stretching is both ways …

that's why there's a tidal bulge on both sides of the Earth (facing the Moon and opposite), ie two high tides every day!

The Moon stretches the water on both sides of the Earth,

the Sun does the same, but less so (much smaller tidal bulges),

and both the Moon and the Sun affect the Earth's path through space.

 

[Natko]

No, gravitational forces pull (in one direction), but tidal forces don't pull, they stretch, and stretching is both ways …

that's why there's a tidal bulge on both sides of the Earth (facing the Moon and opposite), ie two high tides every day!

The Moon stretches the water on both sides of the Earth,

the Sun does the same, but less so (much smaller tidal bulges),

So you're saying that gravity both pushes and pulls at the same time? I don't get this "stretching" concept.

and both the Moon and the Sun affect the Earth's path through space.

How does the Moon affect Earth's orbit around the Sun?

 

[tiny-tim]

So you're saying that gravity both pushes and pulls at the same time? I don't get this "stretching" concept.

No, tidal forces stretch.

I explained that earlier, and I though you agreed.

Well, stretching is a both-ways effect …

just try stretching something by pulling on only one end!

How does the Moon affect Earth's orbit around the Sun?

It makes it wobble.

If you watch the Earth and Moon from "above", you'll see that they sort-of dance round each other! :tongue2:

 

[Natko]

If you watch the Earth and Moon from "above", you'll see that they sort-of dance round each other! :tongue2:

Although the Moon has an effect on Earth's movement, it's the Sun's gravitational pull that Earth to orbit it, right?

 

[Natko]

Although the Moon has an effect on Earth's movement, it's the Sun's gravitational pull that Earth to orbit it, right?

 

[Drakkith]

Although the Moon has an effect on Earth's movement, it's the Sun's gravitational pull that Earth to orbit it, right?

Both the Earth and the Moon are orbiting the Sun. They also both orbit each other around a point called a Barycenter: http://en.wikipedia.org/wiki/Barycentric_coordinates_(astronomy [Broken])

The Sun pulls BOTH the Earth and the Moon, keeping us in orbit around it as we both orbit around each other.

 

[D H]

So you're saying that gravity both pushes and pulls at the same time? I don't get this "stretching" concept.

The tidal forces do indeed push and pull at the same time. The following figure (source: wikipedia article on tidal forces) shows the tidal forces exerted by some other object on a planet. Notice that the tidal forces stretch the planet along the line between the center of the planet and the center of the satellite, but squeezes the planet inward on the plane normal to this line.

How does the Moon affect Earth's orbit around the Sun?

It makes it wobble.

If you watch the Earth and Moon from "above", you'll see that they sort-of dance round each other! :tongue2:

That depends on what you mean by "wobble". The curvature of the Earth's orbit about the Sun is always directed sunward. In fact, even the curvature of the Moon's orbit about the Sun is always directed sunward.

This leads to a devil's advocate question: Does the Moon orbit the Earth or the Sun? Since the curvature is always sunward (this is not the case for a vehicle in low Earth orbit) and the gravitational force exerted by the Sun on the Moon is more than twice that exerted by the Earth on the Moon, why do we say that the Moon orbits the Earth?

 

[Drakkith]

This leads to a devil's advocate question: Does the Moon orbit the Earth or the Sun? Since the curvature is always sunward (this is not the case for a vehicle in low Earth orbit) and the gravitational force exerted by the Sun on the Moon is more than twice that exerted by the Earth on the Moon, why do we say that the Moon orbits the Earth?

I like to think that an object can have more than 1 orbit at once.

 

[D H]

I like to think that an object can have more than 1 orbit at once.

Bingo!

 

[tony873004]

...This leads to a devil's advocate question: Does the Moon orbit the Earth or the Sun? Since the curvature is always sunward (this is not the case for a vehicle in low Earth orbit) and the gravitational force exerted by the Sun on the Moon is more than twice that exerted by the Earth on the Moon, why do we say that the Moon orbits the Earth?

Another way to look at it is to consider the Moon another solar-orbiting object that is in a 1:1 resonance with Earth. The set of 1:1 orbits include horseshoe orbits (L4 to L3 to L5 and back), tadpole orbits (L4 or L5), quasi orbits (circles the Earth in a rotating frame of reference), and stuff that actually does orbit the Earth, regardless of what direction its sunward curvature points.

In the case of the Moon, you can picture it orbiting the Sun. Anytime it gets too far ahead of the Earth, the Earth pulls it into a slower orbit where it ultimately falls behind the Earth. The Earth then pulls it forward until its solar orbit is faster than the Earth's and it gets ahead of the Earth. Repeat indefinately.

If the Sun were to suddenly disappear (yes I know its hypothetical, but I can ponder that anyway), horseshoe, tadpole and quasi orbits would cease to exist, while an object in a direct orbit would fly away from the solar system, bound to its planet.

Unlike the horseshoe, tadpole and quasi orbits, a direct orbit has an energy PE+KE < 0, an eccentricity < 1, and a positive semi-major axis.

 

[Cosmo Novice]

This leads to a devil's advocate question: Does the Moon orbit the Earth or the Sun? Since the curvature is always sunward (this is not the case for a vehicle in low Earth orbit) and the gravitational force exerted by the Sun on the Moon is more than twice that exerted by the Earth on the Moon, why do we say that the Moon orbits the Earth?

We could take it even further and argue that actually the moon orbits the galactic centre as it rotates on the spiral arm! I don't know how far we could take these orbital frames but ultimately they are all orbits.

So the moon does orbit the erath, but it also orbits the sun and the galactic centre. This is all quite interesting!

Cosmo

 

[tony873004]

We could take it even further and argue that actually the moon orbits the galactic centre as it rotates on the spiral arm! I don't know how far we could take these orbital frames but ultimately they are all orbits.

So the moon does orbit the earth, but it also orbits the sun and the galactic centre. This is all quite interesting!

Cosmo

An interesting side note is that Jupiter's large moons do not always simultaneously orbit both Jupiter and the Sun. There are portions of their orbits where their velocity with respect to the Sun is greater than solar escape velocity. It's as if during each orbit, Jupiter captures Io, Europa, Ganymede, and Callisto from interstellar trajectories, then ejects them back onto an interstellar trajectories.

 

[Natko]

So the Moon orbits the Earth, which orbits the Sun, so the Moon ultimately orbits the Sun as well.

 

[Drakkith]

So the Moon orbits the Earth, which orbits the Sun, so the Moon ultimately orbits the Sun as well.

Sure.