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Drakkith

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Ryan_m_b

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But you would als get closer to the center. Every time you halve the diameter, there will be 8 times less mass, the 1/r^2 in Newton's law. The end result is that an object with 1/4 times the diameter of Earth, will have 1/4 times the gravity, so you'll need 1/50 of the diameter.

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@ryan, that doesn't seem right because the moon is about 1/4 the diameter of the earth and is about 1/6th the gravity.

@drakkith, my bad. I meant how much would he weigh.

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Ryan_m_b

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http://www.google.com/search?q=density+of+moon". Though I may have forgotten to take in other factors.@ryan, that doesn't seem right because the moon is about 1/4 the diameter of the earth and is about 1/6th the gravity.

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DaveC426913

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Yes. He would see a horizon that's very near. Mars' horizon is something like a couple of kilometers, enough to be quite noticeable.basically I was wondering if a future astronaut landing on an asteroid that just had enough gravity to keep him on the surface would be able to tell he was on a small object by it's appearance.

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The distance to the horizon on a perfect sphere is

[tex] \sqrt { 2 R h } [/tex]

where R is the radius of the sphere and h the height above the surface. This will be somewhat less than 1 km.

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[tex]F_g = \frac{mMG}{r^2}[/tex]

Where [tex]G = 6.674*10^{-11} \frac{m^3}{kg*s^2}[/tex]

The average mass density of the earth is 5.515 g/cm^3 (According to Wikipedia),

so the

Using gravity on Earth to be 9.807 m/s, we write the equation

[tex] 9.807/50 = 1.5418*10^{-6}*r[/tex]

This can easily be solved to find a radius of 127 kilometers.

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DaveC426913

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m = r^3

force = m/r^2 = r^3/r^2 = r

1/50 = force = r

http://www.wolframalpha.com/input/?i=radius+of+earth

6367.5 kilometers

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http://www.world-builders.org/lessons/less/les1/gravity.html

***

By the way, what's the limit for mass and/or radius of a body to be gravitationally compacted into a sphere? (I used to know it, but I forgot.)

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about 1000 km

ceres is the smallest one

ceres is the smallest one

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It should be dependent on the composition of the body. I doubt that there is a general answer.Check out this website:

By the way, what's the limit for mass and/or radius of a body to be gravitationally compacted into a sphere? (I used to know it, but I forgot.)

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About 1000km given what density?about 1000 km

ceres is the smallest one

The gravitational attraction of a planet or a moon is dependent on it's diameter and its density!

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NO KIDDING!!!It should be dependent on the composition of the body. I doubt that there is a general answer.

See my last two posts on this thread.

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The surface acceleration, for a sphere of mass

[tex]a = \frac{GM}{r^2}[/tex]

Since

[tex]a = \frac{4 \pi G \rho r^3}{3 r^2} = \frac{4 \pi G \rho}{3}r[/tex]

In other words,

[tex]r = \frac{6378 \text{ km}}{50} = 128 \text{ km}[/tex]

This agrees with tmiddlet's calculation posted earlier, and with what others have been saying without showing a derivation.

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NO KIDDING!!!

See my last two posts on this thread.

Sorry, I cannot see any other posts between your claim that you used to know the answer and my remark. Still don't see why would you yell at me.

Your question:

"By the way, what's the limit for mass and/or radius of a body to be gravitationally compacted into a sphere? (I used to know it, but I forgot.)"

is not well defined as it depends on the initial state of the "body" to be (or not to be) compacted (whatever that means). I don't dispute that you may have had an answer but it should have been to a more specifically formulated question.

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Drakkith

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Wouldn't it be the standard formula, but with the variable density replaced with just a number?

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DaveC426913

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Yes, 1. Had someone posted the formula? I did not see it. Or did not realize that was the general one.Wouldn't it be the standard formula, but with the variable density replaced with just a number?

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No, not before your original comment in Post #11.Yes, 1. Had someone posted the formula? I did not see it.

But after it, in Post #18, I give

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nasu:

Sorry about the shouting. I should have put a smiley after that. I tai care to be more polite in the future.

And you're correct that the post to which you responded wasn't very clear, but when I used the phrase:

"...mass and/or radius of a body..."

I intended to refer to density, as the radius plugged into the sphere equation gives you the volume, and mass divided by volume gives density.

Sorry about the shouting. I should have put a smiley after that. I tai care to be more polite in the future.

And you're correct that the post to which you responded wasn't very clear, but when I used the phrase:

"...mass and/or radius of a body..."

I intended to refer to density, as the radius plugged into the sphere equation gives you the volume, and mass divided by volume gives density.

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thx

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And for a different density, you could complete the equation by plugging in the absolute density, or by ignoring the gravitational constant and dividing your final result by the relative density. (In other words, if your density is 0.8 that of earth, your final answer would be 128km/0.8 = 160km.In other words,aandrare proportional for a fixed densityρ.

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