Size of planet with 1/50 gravity of earth

[YummyFur]

Basically I'm wondering how big a rock in space would need to be to draw an average person to it with a gravitational force of a kilo or two. Say it was the average density of Earth and it was spherical, what would its diameter be.

 

[Drakkith]

I'm not sure I understand you. I believe mass is measured in kilograms, not force. How much gravitational force are you looking at getting?

 

[Ryan_m_b]

Well I'm pretty sure you would need 50 times less mass. Every time you half the diameter of a sphere you reduce it's volume by 8. So a sphere 1/4 the size of Earth would have 1/65th of the gravity.

 

[willem2]

Well I'm pretty sure you would need 50 times less mass. Every time you half the diameter of a sphere you reduce it's volume by 8. So a sphere 1/4 the size of Earth would have 1/65th of the gravity.

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.

 

[YummyFur]

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.

@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.

 

[YummyFur]

@willem2, ok so then that would be say a rock that was about 160 miles across. So an astronaut standing in the middle would see about 80 miles to his horizon.

 

[Ryan_m_b]

@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.

http://www.google.com/search?q=density+of+moon". Though I may have forgotten to take in other factors.

 

[DaveC426913]

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.

Yes. He would see a horizon that's very near. Mars' horizon is something like a couple of kilometers, enough to be quite noticeable.

 

[willem2]

@willem2, ok so then that would be say a rock that was about 160 miles across. So an astronaut standing in the middle would see about 80 miles to his horizon.

The distance to the horizon on a perfect sphere is

\sqrt { 2 R h }

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

 

[tmiddlet]

The universal law of gravitation is:
F_g = \frac{mMG}{r^2}
Where G = 6.674*10^{-11} \frac{m^3}{kg*s^2}

The average mass density of the Earth is 5.515 g/cm^3 (According to Wikipedia),
so the acceleration due to gravity based on a planet with Earth's density and radius r is a = \frac{\frac{4}{3}*\pi*r^3*5.515*10^6*6.674*10^{-11}}{r^2}

Using gravity on Earth to be 9.807 m/s, we write the equation
9.807/50 = 1.5418*10^{-6}*r

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

 

[DaveC426913]

There has got to be a formula that, for spherical objects of a fixed density, takes only mass and radius and returns the surface gravity.

 

[granpa]

for a fixed density
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

 

[BadBrain]

Check out this website:

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.)

 

[granpa]

about 1000 km

ceres is the smallest one

 

[nasu]

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.)

It should be dependent on the composition of the body. I doubt that there is a general answer.

 

[BadBrain]

about 1000 km

ceres is the smallest one

About 1000km given what density?

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

 

[BadBrain]

It should be dependent on the composition of the body. I doubt that there is a general answer.

NO KIDDING!

See my last two posts on this thread.

 

[Redbelly98]

This can be solved without actually knowing the density of Earth, or plugging in the value of G. We just need to know the radius of the Earth.

The surface acceleration, for a sphere of mass M and radius r is
a = \frac{GM}{r^2}
Since M = ρV, and V = 4πr ³/3, we have
a = \frac{4 \pi G \rho r^3}{3 r^2} = \frac{4 \pi G \rho}{3}r
In other words, a and r are proportional for a fixed density ρ. To get 1/50th the force or acceleration as experienced at the surface of the Earth, the radius must be 1/50th that of Earth:
r = \frac{6378 \text{ km}}{50} = 128 \text{ km}

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

 

[nasu]

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.

 

[Drakkith]

There has got to be a formula that, for spherical objects of a fixed density, takes only mass and radius and returns the surface gravity.

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

 

[DaveC426913]

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

Yes, 1. Had someone posted the formula? I did not see it. Or did not realize that was the general one.

 

[Redbelly98]

Yes, 1. Had someone posted the formula? I did not see it.

No, not before your original comment in Post #11.

But after it, in Post #18, I give a at the surface in terms of either (take your pick) (1) mass and radius or (2) average density and radius.

 

[BadBrain]

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.

 

[YummyFur]

I know it's probably buried in the formulas somewhere but I can't work this out... further to the original query, I am now wondering what the escape velocity would be for this 127km asteroid where the average person would weigh a kilo or two.

thx

 

[BadBrain]

In other words, a and r are proportional for a fixed density ρ.

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.

 

[BadBrain]

This actually brings to mind a certain idea I once had of an ice moon orbiting a massive planet.

The idea began with my observation of these little pointy projections from the top of some of the ice cubes in my ice cube tray, which I figured were the result of tiny droplets of water in the middle of the cubes that, although below the freezing temperature of water, couldn't freeze due to the lack of room available to expand and crystallize due to the fact of its being surrounded by water that had earlier frozen. As the surrounding ice crystallizing would place a pressure upon incompressible water, the water would have to find a way out of this situation by exploiting any sort of fissure within the surrounding ice. As upward is the only way out not blocked by the inside of the tray, and straight up is the shortest distance to the surface from the center, these little geysers always show up right in the middle of the cube, and of course they instantly freeze, as the water is already below the freezing point of water.

An ice moon orbiting a massive planet in a highly eccentric orbit would experience massive tidal forces which might generate internal frictional heating sufficient to melt some of the subsurface ice, which would then undergo a process similar to that which I observed in my ice cube tray, leaving a surface full of conical mountains, whose height, of course, would depend on the radius of the planet corrected for the density of ice.

 

[D H]

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.

More likely, such an object would be considerably less dense than the Earth. The Earth has a fairly high density because the high pressure inside the Earth compresses the material inside the Earth considerably and because the Earth has a fairly large iron core. The compression most certainly isn't going to happen on this tiny little planetoid, and a large iron core is rather dubious. If the density was half that of the Earth (the density of continental crustal rock), the planetoid would have to be over 500 km in diameter.

In other words, just about the size of Pallas. Note that Pallas' mean surface gravity is about 1/54 that of the Earth.

Another point of interest: Pallas is not spherical. A rocky object that small just isn't massive enough to pull itself into hydrostatic equilibrium. The planetoid is going to be rather lumpy, and thus so will the gravity field.

Addendum
Hubble image of Pallas:

 

[BadBrain]

Here are the extensions for my last post:

Let's say my ice moon is the same size as Earth's moon.

Mean density of Earth: 5.515g/cm^3

Density of ice1 at 0 C (OK, I know I'm pushing things here): 0.9167g/cm^3

Relative density ice: Earth: 0.9167/5.515 = 0.16622

***

Mean Earth radius: 6371km

Mean Lunar radius: 1737.1km

Relative Lunar: Earth radius: 0.2717

***

(0.2717)(0.16622) = 0.045g

Assuming proportional amount of water at proportional depth of ice, and ignoring effects of atmospheric resistance, my observed "ice volcanoes" in my ice cubes were about 3mm high, so the volcanoes on my moon would be 3mm/0.045 = 66 2/3mm.

(OK, so they're not exactly the Himalayas!)

 

[BadBrain]

D-H:

Which brings me back to my original 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.)

Density seems, to me, to be the only factor that can explain the reason what there's no such thing as a gasbag planet with the radius or mass of earth, in that any gasbag smaller than a Jupiter-class planet wouldn't be able to cohere, and the gas would simply bleed away into interplanetary space. (In other words, why is Saturn, the smallest gasbag in our solar system, 95 times the mass of Earth (as I accept that Uranus and Neptune are more properly described as icebags)?) In yet other words, why does the lesser density of gasbags appear to require greater mass than mineral planets by orders of magnitude in order to form a coherent planet (assuming our solar system to be typical)?