Gravity moving toward the center of the Sun?

  1. Gravity moving toward the center of the Sun??

    Ok can someone tell me if I am wrong in my understanding here (in a heated debate with a physics professor) : As you move from the surface of the sun to the center, does the force of gravity increase or decrease and why? The point I am trying to make is that it decreases because once you are at the center you are being pulled equally in all directions by the mass of the sun that is surrounding you and if you were at the surface, you essentially have the entire mass of the sun "beneath" you. So just looking at it simply mathematically you are going from maximum gravity at the surface to minimum at the core and therefore the force of gravity must be decreasing. HOW AM I WRONG????
     
  2. jcsd
  3. D H

    Staff: Mentor

    You are assuming that because gravitation acceleration is non-zero at the surface but zero at the center necessarily means a monotonically decreasing function. That assumption is not valid. You are ignoring density variations.

    By way of analogy, look at the Earth. Surface gravity is about 9.81 m/s2, and it is of course zero at the Earth's core. This does not mean that gravitational acceleration decreases monotonically from the surface to the core. In fact, it doesn't. The greatest gravitational acceleration is not at the surface but at the core/mantle boundary, almost halfway down toward the center of the Earth.

    Density variations within the Sun are much greater than those within the Earth. Gravitational acceleration will first increase with increasing depth inside the Sun. It's only going to start dropping toward zero well within the Sun's core.
     
  4. What causes it to not decrease monotonically? Why would the gravitational acceleration be greater within the earth than it would be on the surface? Isn't the mass of the earth now above you pulling on you and therefore decreasing the gravitational attraction toward the center of the earth?
     
  5. D H

    Staff: Mentor

    Increased density is what causes this. The Earth's core is significantly denser than are the Earth's mantle and crust.

    Define [itex]\rho(r)[/itex] to be the density of the material inside the Earth whose distance to the center of the Earth is exactly r and define [itex]\bar{\rho}(r)[/itex] to be the average density of all of the material inside the Earth whose distance to the center of the Earth is r or less. The gravitational force at r will increase if [itex]\rho(r) < \frac 2 3 \bar{\rho(r)}[/itex].

    No. Suppose the Earth was hollow inside, a spherical shell of matter in which density is a function of distance from the center of the Earth. The gravitational acceleration toward the Earth at any point inside the hollow part would be zero. This is Newton's shell theorem.
     
  6. phinds

    phinds 8,755
    Gold Member

    brains3234, I often find that when thinking about such things, it can be helpful to take things to an unlikely extreme to help see what's going on.

    Suppose the first 3,000 miles of the Earth's surface were Styrofoam, with almost negligible weight and the full mass of the earth was concentrated uniformly from 3,000 miles down from the surface all the way to the center.

    Now, on the surface you would be 3,000 miles further away from the center of mass, so as you moved down to the 3,000 mile point, you'd be attracted by the same mass but you'd be significantly closer. You do the math.
     
  7. Right, ok so let's say you get through that 3000 miles of styrofoam and are now at the styrofoam/core boundary. . . once you start digging into the core, doesn't the force of gravity begin to decrease then? In this case sure it is still increasing while you go from the surface into the styrofoam and then hit the core, but once you start digging into the core and approach the center, shouldn't it be decreasing?
     
  8. But the earth isn't a hollow sphere, it's a solid sphere. Look at these: Figure 14.22 (http://www.kgarbageij-iitjee.com/gravitational-force-between-particle-and-spherical-mass)
    or second page on (http://courses.physics.northwestern.edu/Phyx125/Gravity Inside A Spherical Shell.pdf)
     
  9. phinds

    phinds 8,755
    Gold Member

    Absolutely, as was implied in D H's post where he said " It's only going to start dropping toward zero well within the Sun's core" --- he was also making the point that the sun is non-uniform in density (as is the earth).
     
  10. D H

    Staff: Mentor

    Both of those references discussed a simplified version of the shell theorem. The gravitational acceleration inside a uniform density spherical shell is zero. The uniform density qualifier is not needed. What is needed is a spherical density distribution: Density is a function of radial distance only.

    Look at the last reference, read the start of the last paragraph. "For a uniform sphere (which the Earth isn’t) ..." That parenthetical remark is important. The Earth is not a uniform sphere. The Earth's core comprises but about 1/6 of the Earth's volume but contains about 1/3 of the Earth's mass. The core is significantly more dense than the mantle or the crust.

    Both of those references said exactly what I did about a spherical shell. One way to look at a point inside the Earth is to think of the mass below and the mass above as two separate objects. The total gravitational force is the sum of the gravitational forces of those separate objects. The second object contributes *nothing*. It's only the mass below that contributes. Both of your references said exactly that.

    The core/mantle boundary is at a distance of about 3480 km from the center of the Earth, or 54.6% of the Earth's radius. The mass of the core is 32.4% of the total mass of the Earth. Do the math. Gravitation at the core/mantle boundary is about 9% more than surface gravity.
     
  11. D H

    Staff: Mentor

    To bring this back to the question raised in the opening post, the conditions inside the Sun are even more extreme than those inside the Earth. The Earth's core occupies about 16.3% of the Earth's total volume and contains 32.4% of the Earth's total mass. Compare that to the Sun, where the core is but 2.3% of the total volume but contains 40% of the total mass. Gravitational force inside the Sun rises all the way down to, and even well into, the Sun's core.
     
  12. Thanks you guys I really appreciate you explaining this to me. It makes a lot more sense now.
     
  13. In this context I am just wondering how one can work out expected gravitational redshift of photons emanating from the sun. For example if Lyman Alpha photons (1216 A) are generated half way down below the 'surface' of the sun, they would presumably experience much greater redshift than if they were generated at the surface.
     
  14. Drakkith

    Staff: Mentor

    Essentially all of the photons sent out from the Sun are created in the photosphere. Past that the solar material is opaque and absorbs them, so they can't escape.
     
    neilparker62 likes this.
  15. Ok - thanks for the clarification. Will have to look elsewhere (ie outside of gravity) for possible causes of redshift in the Lyman lines.
     
  16. Drakkith

    Staff: Mentor

    What redshift are you referring to?
     
  17. davenn

    davenn 3,675
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    Gold Member
    2014 Award

    Drakkith likes this.
  18. davenn

    davenn 3,675
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    2014 Award

  19. It relates to an "interesting co-incidence" I found and which is dealt with in this thread. For my (debunked) hypothesis that the solar Lyman Alpha line might be located at 'peak left' rather than 'dip central' (see Lyman Alpha profile in the above thread), there would have to be an appreciable red shift between terrestrial Lyman Alpha and solar Lyman Alpha lines.

    Apologies to the OP if I have somewhat "red shifted" this thread !
     
  20. Yes - that's where my query originates from. For my (debunked) hypothesis that the Lyman Alpha line is at "peak left" rather than "dip central" in the profile, there would have to be an appreciable reshift between terrestrial and solar Lyman Alpha lines.

    Apologies to the OP if I have somewhat 'red shifted' this particular thread!
     
  21. ' ... appreciable redshift' that is
     
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