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The Earth and its orbit around the sun

  1. Apr 10, 2008 #1
    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.
     
  2. jcsd
  3. Apr 10, 2008 #2

    mathman

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    The earth's gravity is much stronger than the sun's for things near the earth, such as satellites, astronauts, and the moon.
     
  4. Apr 10, 2008 #3

    chroot

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    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
     
  5. Apr 10, 2008 #4

    tiny-tim

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    Welcome to PF!

    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:
     
  6. Apr 10, 2008 #5

    russ_watters

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    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).
     
  7. Apr 10, 2008 #6
    if the suns gravitional attraction is so small on the moon how can it still be powerful enough to attract lets say uranius millions and millions of light years away
     
  8. Apr 10, 2008 #7

    D H

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    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.
     
  9. Apr 10, 2008 #8

    Danger

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    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.
     
  10. Apr 11, 2008 #9

    mathman

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    Not so! The best evidence is the fact that the moon's effect on tides is about three times that of the sun.
     
  11. Apr 11, 2008 #10

    Danger

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    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.
     
  12. Apr 11, 2008 #11

    D H

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    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.
     
    Last edited: Apr 11, 2008
  13. Apr 11, 2008 #12

    tiny-tim

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    … a fruity thought …

    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:
     
  14. Apr 11, 2008 #13

    Danger

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    I'm not disputing the math. I was merely pointing out that Mathman was arguing against a statement that hadn't been made.
     
  15. Apr 11, 2008 #14

    D H

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    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).
     
  16. Apr 11, 2008 #15

    Danger

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    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.)
     
  17. Apr 11, 2008 #16

    russ_watters

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    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...
     
    Last edited: Apr 11, 2008
  18. Apr 11, 2008 #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 cant '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.
     
  19. Apr 11, 2008 #18

    Danger

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    That's a very cool approach to the problem, Robert. Nice summation.
     
  20. Apr 12, 2008 #19

    mathman

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    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.
     
  21. Apr 12, 2008 #20

    D H

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    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 ...
    ... 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.
     
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