The Earth and its orbit around the sun

In summary: Its an easy matter to compute the gravitational acceleration of the Moon toward the Sun and Earth:\begin{aligned}\mathbf a_{M\to S} &= \frac {GM_S}{r_{M \to S}^2} \approx 5.9 \,\text{mm}/\text{s}^2 \\[6pt]\mathbf a_{M\to E} &= \frac {GM_E}{r_{M \to E}^2} =\approx 2.7 \,\text{mm}/\text{s}^2\end{
  • #1
njay22
2
0
If the Earth orbits the sun how can astronauts fell no attraction to the sun in space. Even pluto which is 4500 million years away, still orbits the sun.
 
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  • #2
The Earth's gravity is much stronger than the sun's for things near the earth, such as satellites, astronauts, and the moon.
 
  • #3
They do feel gravitational attraction to the Sun! If they did not, they wouldn't stay near the Earth -- they'd fly out of the solar system.

- Warren
 
  • #4
Welcome to PF!

njay22 said:
If the Earth orbits the sun how can astronauts fell no attraction to the sun in space. Even pluto which is 4500 million years away, still orbits the sun.

Hi njay22! Welcome to PF! :smile:

Astronauts haven't gone any further than the moon yet.

So the sun's attraction is felt by the astronauts, but they don't notice it because it's so small compared with the attraction of the Earth.

But when astronauts go to Mars or Venus … then they'll be so far from Earth that the sun will be the main source of attraction! :smile:
 
  • #5
The question is functionally equivalent to asking why astronauts in orbit of Earth don't feel the force of gravity from the earth. The answer in both cases is that they are in orbit. You may not exactly "feel" it, but that force is what keeps you in orbit. Since objects around you are also in orbit, you feel no force between you and those objects, which is how gravitational force manifests on Earth (you'd fall through it and feel no force if it weren't for the floor getting in your way).
 
  • #6
if the suns gravitational attraction is so small on the moon how can it still be powerful enough to attract let's say uranius millions and millions of light years away
 
  • #7
What makes you think it is small? The gravitational force exerted on the Moon by the Sun exceeds the force exerted on the Moon by the Earth.
 
  • #8
njay22 said:
how can it still be powerful enough to attract let's say uranius millions and millions of light years away

Hmmm... the last time that I checked, Uranus wasn't millions and millions of light-years away. Even at that range, though, gravity has some infinitesimal influence.
 
  • #9
D H said:
What makes you think it is small? The gravitational force exerted on the Moon by the Sun exceeds the force exerted on the Moon by the Earth.

Not so! The best evidence is the fact that the moon's effect on tides is about three times that of the sun.
 
  • #10
mathman said:
Not so! The best evidence is the fact that the moon's effect on tides is about three times that of the sun.

Apples and oranges? DH was referring to the gravitational force from the sun, as opposed to from the Earth, upon the moon. You're talking about the difference between the moon's and the sun's effect upon Earth.
 
  • #11
Danger said:
mathman said:
Not so! The best evidence is the fact that the moon's effect on tides is about three times that of the sun.
Apples and oranges? DH was referring to the gravitational force from the sun, as opposed to from the Earth, upon the moon. You're talking about the difference between the moon's and the sun's effect upon Earth.
Its an easy matter to compute the gravitational acceleration of the Moon toward the Sun and Earth:

[tex]
\begin{aligned}
\mathbf a_{M\to S} &= \frac {GM_S}{r_{M \to S}^2} \approx 5.9 \,\text{mm}/\text{s}^2 \\[6pt]
\mathbf a_{M\to E} &= \frac {GM_E}{r_{M \to E}^2} =\approx 2.7 \,\text{mm}/\text{s}^2
\end{aligned}
[/tex]

A little thought implies that the gravitational acceleration of the Earth toward the Moon is a whole lot smaller than of the Earth toward the Sun. So what's the deal with tides? Simple. The tides result from the gradient in the gravitational acceleration. This is roughly an inverse-cube relationship compared to the gravitational force, which of course has an inverse square relationship. Apples and oranges.
 
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  • #12
… a fruity thought …

D H said:
… Apples and oranges.

hmm … so if Isaac Newton had been hit on the head by a cube-shaped orange …

:smile: he would have worked out the tides as well! :smile:
 
  • #13
I'm not disputing the math. I was merely pointing out that Mathman was arguing against a statement that hadn't been made.
 
  • #14
Danger, I was agreeing with you that the comparing the strength of the tides with the strength of the gravitational force is a matter of apples and oranges (or more precisely, cubes and squares).
 
  • #15
I realize that, DH, and appreciate it. My comment was actually directed to Mathman, not to your mathematical post.
(Although I must admit that I rather like the idea of cubic apples; it would save me a lot of space in the fridge.)
 
  • #16
mathman said:
Not so! The best evidence is the fact that the moon's effect on tides is about three times that of the sun.
DH gave the answer to the previous issue, but the answer to the tides issue is that the tides are more dependent on distance than the actual force of gravity is because the tidal force is a measure of the difference in gravitational force between the near and far sides of the earth. It's an inverse cube relationship instead of an inverse square relationship.

Danger was right - they are two different issues with different answers. And Tim's joke was a little weak, but I'll let that one go...
 
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  • #17
njay22, simply ask yourself what gravity is. if the entire universe was still its current dimensions but only consisted of two atoms located as "far" away from each other as spacialy possible they would still eventually 'fall' into each other just as you are 'falling' into the Earth and the Earth is 'falling' into the sun and the sun is 'falling' into the milky way's super-massive black hole. Essential every single elementary particle of matter is "falling" into every other particle of matter. Simply by existing, matter is gravatationally 'attracted' to every other piece of matter.

Now for why me or your astronaut friend can't 'feel' the gravational force of the sun, well even though we know about all this stuff our lowly human sences are designed to keep us alive here on earth, not to show us the astronomical force of gravatiy applied to us by the matter surrounding us. Simply put we are small and far from the sun, but also we are small and much much closer to the earth, therefore (since gravatiy is based on mass AND distance) we can 'feel' with our various biological senses the effects of the mass of the Earth but not that of the sun. Survival for creatures on Earth has so far had very little to do with being sensitve to the Fg from the sun, so though we know its there from our technology we have no clue its there from our natural senses.
 
  • #18
That's a very cool approach to the problem, Robert. Nice summation.
 
  • #19
Just to get it in context, I did a quick calculation of the various gravitational forces involved (omitting G).
sun on Earth or sun on moon ~10^14
moon on Earth ~5 x 10^11
earth on moon ~4 x 10^13

I hope this helps, although it doesn't address the orignal question.
 
  • #20
robertm said:
Now for why me or your astronaut friend can't 'feel' the gravational force of the sun,
The answer is extremely simple: we can't 'feel' the gravitational force from almost everything, including the Earth. The one exception is a future astronaut who ventures a bit to close to a black hole or neutron star. The difference in the gravitational acceleration between the astronaut's head and feet might literally tear the astronaut apart. With the exception of these extreme tidal forces, we cannot feel gravity, including ...
We can 'feel' with our various biological senses the effects of the mass of the Earth but not that of the sun.
... the earth. We do not feel the gravitational force exerted on us by the Earth. What we feel instead is the upward normal force exerted by the Earth that keeps us from sinking into the Earth. Gravity itself is undetectable (directly). The accelerometers in airplanes, in spacecraft , and in your inner ear measure all forces acting on the plane, the spacecraft , or your head except for gravitational force.
 
  • #21
Thanks for clearing that up D H. I think i understand, we can detect
the resultant forces from the Fg but not the Fg directly. Hmmm... very interesting, given no other reference point could you detect, with any mechanism, pure gravatational force say in a perfect vacuum with no other forces involved?
 

What is the Earth's orbit around the sun?

The Earth's orbit around the sun is the path that the Earth takes as it revolves around the sun. It is an elliptical orbit, meaning that it is shaped like an oval rather than a perfect circle.

What is the duration of the Earth's orbit around the sun?

The Earth's orbit around the sun takes approximately 365.24 days, or one year, to complete. This is why we have leap years every four years to account for the extra quarter day.

What causes the Earth's orbit around the sun?

The Earth's orbit around the sun is caused by the gravitational pull of the sun. The sun's massive size and gravitational force keeps the Earth in its orbit.

Why does the Earth's distance from the sun vary during its orbit?

The Earth's distance from the sun varies during its orbit due to its elliptical shape. This means that at certain points in its orbit, the Earth is closer to the sun (perihelion) and at other points, it is farther away (aphelion).

How does the Earth's orbit around the sun affect the seasons?

The Earth's orbit around the sun is responsible for the changing of the seasons. As the Earth revolves around the sun, its tilt causes different parts of the Earth to receive more or less direct sunlight, resulting in the different seasons.

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