Earth, Sun, Moon Orbit


by Natko
Tags: earth, gravity, horseshoe orbit, moon, orbit, sun, tadpole orbit
Natko
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#1
Feb1-12, 03:30 PM
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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.
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tiny-tim
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Feb1-12, 03:55 PM
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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
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Feb1-12, 05:06 PM
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Quote Quote by tiny-tim View Post
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
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Feb1-12, 05:29 PM
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Earth, Sun, Moon Orbit


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 … would anyone else like to carry on from there? )
Drakkith
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Feb1-12, 06:56 PM
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Quote Quote by Natko View Post
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
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Feb1-12, 07:54 PM
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Quote Quote by Natko View Post
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

[tex]a_{\text{earth}} = \frac{GM_{\text{moon}}}{R^2}[/tex]

where G is the universal gravitational constant, Mmoon, 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:

[tex]a_{\text{drop}} = \frac{GM_{\text{moon}}}{(R-r)^2}[/tex]

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

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

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

[tex]\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}[/tex]

The first term in parentheses can be rewritten as

[tex]\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[/tex]

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

[tex]\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}[/tex]
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

[tex]
a_{\text{rel}} \approx \frac{GM_{\text{moon}}}{R^2} 2\frac r R
= \frac{2GM_{\text{moon}}r}{R^3}
[/tex]

So, 1/R3. Substitute the moon for the sun and the same kind of inverse cube relation will hold.
Natko
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#7
Feb1-12, 08:02 PM
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Quote Quote by Drakkith View Post
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
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#8
Feb1-12, 08:03 PM
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Quote Quote by D H View Post
The earth as a whole is accelerating toward the moon. This acceleration is given by

[tex]a_{\text{earth}} = \frac{GM_{\text{moon}}}{R^2}[/tex]

where G is the universal gravitational constant, Mmoon, 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:

[tex]a_{\text{drop}} = \frac{GM_{\text{moon}}}{(R-r)^2}[/tex]

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

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

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

[tex]\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}[/tex]

The first term in parentheses can be rewritten as

[tex]\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[/tex]

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

[tex]\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}[/tex]
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

[tex]
a_{\text{rel}} \approx \frac{GM_{\text{moon}}}{R^2} 2\frac r R
= \frac{2GM_{\text{moon}}r}{R^3}
[/tex]

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
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Feb2-12, 07:00 AM
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Hi Natko!

(just got up … )

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/RMoon3, where RMoon is the distance to the Moon.
The Sun is about 400 times further from us than the Moon is (R/RMoon ~ 400),

so (Sun's tidal force)/(Moon's tidal force) is about 400 times smaller than (Sun's gravity)/(Moon's gravity).
Cosmo Novice
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#10
Feb2-12, 07:24 AM
P: 366
Quote Quote by D H View Post
The earth as a whole is accelerating toward the moon. This acceleration is given by

[tex]a_{\text{earth}} = \frac{GM_{\text{moon}}}{R^2}[/tex]

where G is the universal gravitational constant, Mmoon, 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:

[tex]a_{\text{drop}} = \frac{GM_{\text{moon}}}{(R-r)^2}[/tex]

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

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

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

[tex]\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}[/tex]

The first term in parentheses can be rewritten as

[tex]\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[/tex]

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

[tex]\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}[/tex]
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

[tex]
a_{\text{rel}} \approx \frac{GM_{\text{moon}}}{R^2} 2\frac r R
= \frac{2GM_{\text{moon}}r}{R^3}
[/tex]

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 criticise 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
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#11
Feb2-12, 07:49 AM
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Quote Quote by Cosmo Novice View Post
Far be it from me to criticise 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
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Feb2-12, 07:56 AM
P: 366
Quote Quote by D H View Post
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 dont 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
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#13
Feb2-12, 03:30 PM
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Quote Quote by tiny-tim View Post
Hi Natko!

(just got up … )

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/RMoon3, where RMoon is the distance to the Moon.
The Sun is about 400 times further from us than the Moon is (R/RMoon ~ 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
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Feb2-12, 04:22 PM
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Quote Quote by Natko View Post
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
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#15
Feb2-12, 04:47 PM
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Quote Quote by tiny-tim View Post

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.
Quote Quote by tiny-tim View Post
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?
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Feb2-12, 05:00 PM
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Quote Quote by Natko View Post
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!
Natko
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Feb2-12, 05:38 PM
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Quote Quote by tiny-tim View Post
If you watch the Earth and Moon from "above", you'll see that they sort-of dance round each other!
Although the Moon has an effect on Earth's movement, it's the Sun's gravitational pull that Earth to orbit it, right?
Natko
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#18
Feb2-12, 05:40 PM
P: 40
Although the Moon has an effect on Earth's movement, it's the Sun's gravitational pull that Earth to orbit it, right?


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