What is the total size of the Asteroid Belt in kilometers?

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    Asteroid Belt
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SUMMARY

The total size of the Asteroid Belt cannot be accurately calculated by simply adding the diameters of its largest asteroids. The four largest asteroids—Ceres (950 km), Vesta (530 km), Pallas (550 km), and Hygiea (500 km by 350 km)—combine for a total diameter of 2,380 km, which represents only half the mass of the belt. The total estimated mass of the Asteroid Belt is approximately 2.8–3.2×10^21 kg, or about 4% of the Moon's mass. To determine the size of a hypothetical planet formed from the Asteroid Belt, one must calculate volumes based on the average density of the asteroids rather than simply summing their diameters.

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  • Understanding of basic geometry, specifically volume calculations of spheres.
  • Familiarity with asteroid classification and characteristics.
  • Knowledge of mass and density relationships in astrophysics.
  • Ability to interpret scientific data and references, such as those from NASA or Wikipedia.
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  • Research the volume calculation formula for spheres and apply it to irregular shapes.
  • Investigate the average density of asteroids and how it affects mass and volume calculations.
  • Explore the mass distribution of the Asteroid Belt and its largest bodies, particularly Ceres.
  • Read scientific articles on the formation and composition of the Asteroid Belt for deeper insights.
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Astronomy enthusiasts, astrophysicists, educators, and students interested in planetary science and the dynamics of the Asteroid Belt.

eurus
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Can someone please help me in calculating the total size in km of the Asteroid Belt? I have read the Asteroid Belt if lumped together would create a planet 4 times the size of Earth, however, I am trying to validate these numbers and am having difficulty in doing so.

Here are the facts I have been able to locate:

7,000 known asteroids
billions of estimated asteroids - sizes range from specs of dust to large hand sized objects
the 4 largest were once considered planets and account for 1/2 the total mass of the Asteroid Belt
- Ceres, with a diameter of 950 kilometers
- Vesta has a diameter of 530 kilometers
- Pallas has a diameter of 550 kilometers but is less massive than Vesta
- Hygiea is 500 by 350 kilometers

These 4 combine for a total diameter of 2,380 kilometers. If accounts for 1/2 the mass of the Asteroid Belt then when I add in the other 1/2 the total km would be 4,760 which is 1/4th the size of Earth's equatorial diameter (12,756 km). Which does not match to the original estimates where I have read the Asteroid Belt (combined) would be 4 times the size of Earth.

Where am I miscalculating? Thanks.
 
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eurus said:
I have read the Asteroid Belt if lumped together would create a planet 4 times the size of Earth
Where have you read this? It's patently wrong by many orders of magnitude.
There is barely enough material in the belt to make up a small fraction of the Moon, let alone more than Earth.

eurus said:
Where am I miscalculating?
You're miscalculating all the same, because you can't add diameters. You need to add volumes. This will net you a body with density equal to avg. of the asteroids.
Or add masses, select the density as equal to earth, and get volume from there.
 
It's readily available via a simple search regarding the size of the Asteroid Belt. And, yes, the Asteroid belt contains very large objects such as Ceres that are part of the Asteroid Belt so I'm not sure where you get there barely being enough material. You can e-mail NASA and tell them how wrong they are with their calculations as this is where I gathered the data.
 
If it's so simple, you should be able to link to an example article. Here on PF we require claims to be substantiated.
 
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Bandersnatch is right. The asteroid belt contains much less material than the moon. This Wikipedia article is a good start, and it says,

"The mass of all the objects of the asteroid belt, lying between the orbits of Mars and Jupiter, is estimated to be about 2.8–3.2×10^21 kg, or about 4% of the mass of the Moon. Of this, Ceres comprises 0.95×10^21 kg, a third of the total.[38] "
 
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Bandersnatch said:
If it's so simple, you should be able to link to an example article. Here on PF we require claims to be substantiated.

If you can't contribute then stop responding. The question was simple yet you don't seem to have an appropriate response on how to solve the equation.
 
But I have provided you with the response on how to calculate the size of your planet made of asteroids. You just need to use volumes instead of diameters. What you did by adding diameters was string a number of rocks one next to another. What you need to do is mash them all together to make one large rock - that is, add volumes. Just assume they're all spherical with given diameters.

When I asked you to look up your sources, I wasn't trying to be nasty - I was trying to make you look again at what you read, because it most certainly either doesn't say what you wrote, or is not a reputable source. We can point out the mistake if you provide the source - that's why the forum has this policy on giving references in questions. Otherwise we can't address it.

In any case, you won't get the number you want from any sort of properly done calculations, because that number is wrong. Just look at the Wiki article, or do what you said and google for 'asteroid belt size'.
 
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OP - Your source of data is wrong, if you have one, and your attitude is worse. I suggest working on the latter if you wish to be successful in any field.
 
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Great points. So please tell me if the below logic is the math I need to perform.

Assumption: I understand each individual asteroid is not a perfect sphere but for the sake of making the math easier I am going to assume perfect spheres.

Let's assume the data of the original statistics I initially posted is true. I have the 4 known diameters of the 4 largest asteroids in the belt. If those 4 comprise 1/2 the mass of the entire Asteroid Belt then tell me if this is what I need to calculate:

First I have to determine the radius based on diameter. Once I have the answer to r then I need to calculate the volume of a sphere. Since I need radius in order to calculate the volume right?
 
  • #10
eurus:
Calculate (or locate) the masses of the largest bodies.
Add them together.
Then calculate the volume of a single body of that same mass using an average density.

As a rough math-check, consider that Ceres has 1/3 the mass of the whole belt. A spherical body with the times the mass of Ceres will have an approximate radius equal to (the cube root of 3 about 1.44x) larger than Ceres. So, your final answer should be in the neighborhood of (950x1.44) km.
 
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  • #11
DaveC426913 said:
eurus:
Calculate (or locate) the masses of the largest bodies.
Add them together.
Then calculate the volume of a single body of that same mass using an average density.

Thank you. Once I do this is it as simple as multiplying my result by 2 to get the other half that is not known?
 
  • #12
eurus said:
That aside, how can you make an absolute statement that my "source of data is wrong" without actually knowing the answer to begin with?
Because you said this:

I have read the Asteroid Belt (combined) would be 4 times the size of Earth

eurus said:
If you cannot contribute then please don't respond. If you don't like my attitude then grow some thicker skin. Either way I could care less.
Eurus, this is not the PF way.

You're asking for help. It behooves you to be polite. And yes, some of what you get may not be want you want to hear, but it is nonetheless helpful in your search for the answer.
 
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  • #13
eurus said:
Thank you. Once I do this is it as simple as multiplying my result by 2 to get the other half that is not known?
Multiply the mass by 2, yes, Before calculating the final volume.

Note the inherent ambiguities in any of these figures:
- the estimate that the belt's minor components is estimated to be "half" its total mass
- the "average density" of all the objects in the asteroid is very sketchy.
 
  • #14
The OP has failed to provide a reference for the stated size in the OP, rendering the question moot. Several good answers are already in the thread.

The thread will remain closed.
 
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