Why is higher density causing higher surface gravity?

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SUMMARY

The discussion centers on the relationship between density and surface gravity, specifically how compressing Earth's mass to a smaller diameter significantly increases surface gravity. The participants reference the gravitational force equation F=Gm1m2/r², emphasizing that as the radius (r) decreases, the gravitational force increases due to the inverse square law. Despite the overall mass remaining constant, the proximity to the mass source results in a higher gravitational pull. A key takeaway is that the perceived increase in gravity is due to the reduced distance from the mass, not an increase in the Earth's gravitational force itself.

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SpiderET
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Normal density of Earth is 5,5 tons/m3 and surface gravity is 9,8 m/s and. But if we would compress mass of Earth to 50 km diameter, we would get density 1,4 mil tons/m3 and surface gravity would be 158 860 m/s.

I know the math, in basic gravity equotation F=Gm1m2/r2 with decreasing r you get higher gravity, but in this case the overall mass of Earth would be still the same, why is surface gravity so much higher? Is there some good physical explanation for this?
 
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I'm not sure I understand. Are you asking why the force of gravity follows the inverse square law? I.e., why it's stronger the closer you get to the source?
 
Bandersnatch said:
I'm not sure I understand. Are you asking why the force of gravity follows the inverse square law? I.e., why it's stronger the closer you get to the source?
The explanation that gravity follows inverse square law seems to me like not sufficient for explanation of increasing surface gravity as described in example and I am looking for some extended or better explanation.
 
What's not sufficient about it w/r to your example? You're closer to the source of gravity (mass), and the force is greater as per the inverse square law.

Can you illuminate the issue a bit more? Perhaps with a different example?
 
SpiderET said:
I know the math, in basic gravity equotation F=Gm1m2/r2 with decreasing r you get higher gravity, but in this case the overall mass of Earth would be still the same
So what happens to F, if you keep the masses constant, and decrease r?
 
SpiderET said:
in basic gravity equotation F=Gm1m2/r2 with decreasing r you get higher gravity
That is the explanation. The objection you raise is irrelevant.
 
The total strength of the Earth's gravitational force hasn't gotten any stronger you've just moved closer to source. Maybe it would help if you imagined yourself standing on the Earth when suddenly it shrank to 1/4th it's size. If you were able to (stay alive) and maintain the exact same distance (to the center of the Earth, not the surface) the "pull" of the Earth's gravitational force you would experience would be exactly the same.
 
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J.J.T. said:
Gravity doesn't get any stronger you've just moved closer to source. Maybe it would help if you imagined yourself standing on the Earth when suddenly it shrank to 1/4th it's size. If you were able to (stay alive) and maintain the exact same distance (to the center of the Earth, not the surface) the force of the Earth's gravity you would experience would be exactly the same.
I don't know what you mean by "gravity". Apparently you do NOT mean "gravitational force" since that does change with distance. So what do you mean?
 
HallsofIvy said:
I don't know what you mean by "gravity". Apparently you do NOT mean "gravitational force" since that does change with distance. So what do you mean?

Sorry that was poor wording. I meant to say that the gravitational force of the Earth didn't get any stronger in the sense that if you maintained your distance as the density increased there would be no effect.
 
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  • #10
You are still using "gravity" as a noun, apparently not as "gravitational force". What do you mean by that?
 
  • #11
SpiderET said:
I know the math, in basic gravity equotation F=Gm1m2/r2 with decreasing r you get higher gravity, but in this case the overall mass of Earth would be still the same, why is surface gravity so much higher? Is there some good physical explanation for this?

Remember that the Earth is not a point-like object. You are closer to the ground you are standing on than the opposite side of the Earth, so the force from the former is greater than that from the latter. Decreasing the radius of the Earth brings all that matter you aren't standing on closer, so the force of gravity from them increases via the inverse square law.
 
  • #12
HallsofIvy said:
You are still using "gravity" as a noun, apparently not as "gravitational force". What do you mean by that?

C'mon man this is a message board not an academic paper >.>, but if it really bothers you i can fix it.
And before you say anything I put "pull" in quotation marks for you so no one would take it literally.
 
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