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Traveling through the asteriod belt?

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Traveling through the asteriod belt?


What's the (particle) density of the asteroid belt?How frequently do asteroids collide with each other?What would be the delta-v of rendezvousing with temporarily captured asteroids in Sun-Earth L-points?How useful is the Interplanetary Transport Network?How much of the asteroid belt is discovered?How feasible will be to inject an object into the asteroid orbitHow many asteroids could a retrograde probe in the Asteroid Belt flyby?Timing shadows from the Kuiper belt! Any news? Did it work?Can we destroy an asteroid by spinning it?Why paint only one-half of Bennu?Can small asteroids in the asteroid belt be detected on the fly and how much of a threat do they represent for a human manned space mission there?













2












$begingroup$


When we send spacecraft through the asteroid belt how much finger crossing is going on? Is it better to travel over the solar system rather than through it?
enter image description here



sajri.astronomy.cz










share|improve this question









$endgroup$








  • 1




    $begingroup$
    Possible duplicate of What's the (particle) density of the asteroid belt?
    $endgroup$
    – Hobbes
    2 hours ago






  • 1




    $begingroup$
    voting to leave open since the other question is only about density and does not offer much in terms of the OP's question on collision probability.
    $endgroup$
    – uhoh
    51 mins ago


















2












$begingroup$


When we send spacecraft through the asteroid belt how much finger crossing is going on? Is it better to travel over the solar system rather than through it?
enter image description here



sajri.astronomy.cz










share|improve this question









$endgroup$








  • 1




    $begingroup$
    Possible duplicate of What's the (particle) density of the asteroid belt?
    $endgroup$
    – Hobbes
    2 hours ago






  • 1




    $begingroup$
    voting to leave open since the other question is only about density and does not offer much in terms of the OP's question on collision probability.
    $endgroup$
    – uhoh
    51 mins ago
















2












2








2





$begingroup$


When we send spacecraft through the asteroid belt how much finger crossing is going on? Is it better to travel over the solar system rather than through it?
enter image description here



sajri.astronomy.cz










share|improve this question









$endgroup$




When we send spacecraft through the asteroid belt how much finger crossing is going on? Is it better to travel over the solar system rather than through it?
enter image description here



sajri.astronomy.cz







probe trajectory asteroid interplanetary






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 5 hours ago









MuzeMuze

1,3311160




1,3311160








  • 1




    $begingroup$
    Possible duplicate of What's the (particle) density of the asteroid belt?
    $endgroup$
    – Hobbes
    2 hours ago






  • 1




    $begingroup$
    voting to leave open since the other question is only about density and does not offer much in terms of the OP's question on collision probability.
    $endgroup$
    – uhoh
    51 mins ago
















  • 1




    $begingroup$
    Possible duplicate of What's the (particle) density of the asteroid belt?
    $endgroup$
    – Hobbes
    2 hours ago






  • 1




    $begingroup$
    voting to leave open since the other question is only about density and does not offer much in terms of the OP's question on collision probability.
    $endgroup$
    – uhoh
    51 mins ago










1




1




$begingroup$
Possible duplicate of What's the (particle) density of the asteroid belt?
$endgroup$
– Hobbes
2 hours ago




$begingroup$
Possible duplicate of What's the (particle) density of the asteroid belt?
$endgroup$
– Hobbes
2 hours ago




1




1




$begingroup$
voting to leave open since the other question is only about density and does not offer much in terms of the OP's question on collision probability.
$endgroup$
– uhoh
51 mins ago






$begingroup$
voting to leave open since the other question is only about density and does not offer much in terms of the OP's question on collision probability.
$endgroup$
– uhoh
51 mins ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

The asteroid belt isn't nearly as dense as popular movies make it out to be.



An answer from the Dawn Mission's FAQ, specifically "What is the average distance between individual asteroids? (6/13/10)", helps here.




Asteroids are not distributed uniformly in the asteroid belt, but
could be approximated to be evenly spaced in a region from 2.2 AU (1
AU is 93 million miles, or the average distance between Earth and the
Sun) to 3.2 AU from the Sun and extending 0.5 AU above and below the
ecliptic (the plane of Earth's orbit, which is a convenient reference
for the solar system). That yields a volume of roughly 16 cubic AU, or
about 13 trillion trillion cubic miles. (Note: space is big!)



If there were 2 million asteroids 1 mile or larger in that volume,
each asteroid would have 6.7 million trillion cubic miles to itself,
so the average distance between individual asteroids 1 mile in
diameter or larger would be about 1.9 million miles. That is nearly 8
times the distance between Earth and the Moon.




Just for something interesting, it looks like your gif only covers the Jupiter trojans, which are the asteroids that Jupiter shepherds around the solar system at its Lagrange points. A more accurate image is available on Wikipedia.



Asteroid map from Wikipedia






share|improve this answer









$endgroup$





















    1












    $begingroup$

    The asteroid belt is roughly 6 Astronomical Units wide, and so when it is drawn only 600 pixels wide with each asteroid a handful of pixels wide, *you end up with each asteroid being five times bigger than the Sun!**



    I've borrowed a small piece of the image used in @jos' excellent answer to show what I mean.



    enter image description here





    As pointed out in this answer, Wikipedia says:




    Contrary to popular imagery, the asteroid belt is mostly empty. The asteroids are spread over such a large volume that it would be improbable to reach an asteroid without aiming carefully.






    The Exposing PseudoAstronomy blogpost Asteroid Belts – Proximity of Rocks and Why Navigation Is Not Dangerous (Sorry, Han Solo) says:




    How Many Asteroids of What Size?



    In terms of what is known, there are about 20,000 asteroids between 2-3 km, which is about the smallest that we likely have a complete sampling of. What that statement means is that, while we have identified asteroids that are smaller, our detection technology is not good enough to have found all of the asteroids that are smaller.



    If we extrapolate, assuming a -3 power low, down to, say, 100-meter asteroids, there are probably ~82 million asteroids that are ~100-200-meters across. If we extrapolate further, down to 1-meter asteroids, then we really have a gargantuan number of objects – about 1014 (100 quadrillion) objects of that size. That’s quite a lot.



    What Does this Mean for Navigation?



    If we add up all of those objects, we have about 1.2×1014 asteroids larger than 1 meter. Now, let’s look at the asteroid belt. It stretches from 1.8 to 3.3 A.U., which is a distance of 1.5 A.U., or about 225,000,000 km. That’s a fairly large distance (that’s actually about the distance between the Sun and Mars).



    The area of a disk that size, however, is gargantuan: A = π · r2 = π · ((3.3 A.U.)2 – (1.8 A.U.)2) = π · (1.7·1017 km2) = 5.4·1017 km2. That is a huge area. Simple division shows that each asteroid, regardless of its own size, could have 4,500 km2 all to itself – a little bit more than the entire U.S. state of Rhode Island.



    And that’s if they were all just in one plane. In reality, they occupy a volume of space, some orbiting “above” or “below” others (where those terms are relative to the plane that the Earth’s orbit makes).



    Even if we cut the size of asteroids in half again, and were interested in all asteroids larger than half a meter (1.5 ft) in size, then we have 8 times as many asteroids, but each one still has over 500 km2 all to itself, and even more space if we consider the vertical component.



    What does this mean for navigation? It’s easy! In fact, you really have to try to hit an asteroid, at least in our own belt. And so, the next time you see a tiny ship careening through an asteroid field in a TV show or movie, remember that in real life, asteroid belts really aren’t that dangerous for navigation.




    @AlanSE's answer to the related but different question What's the (particle) density of the asteroid belt? shows an illustration of the power law behavior of size at large, observable scale. Unfortunately the link to the source (dated 2001) is now broken.



    enter image description here






    share|improve this answer











    $endgroup$













    • $begingroup$
      1017 should be 10E17 or 10^17
      $endgroup$
      – Uwe
      2 mins ago











    Your Answer





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    2 Answers
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    2 Answers
    2






    active

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    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    The asteroid belt isn't nearly as dense as popular movies make it out to be.



    An answer from the Dawn Mission's FAQ, specifically "What is the average distance between individual asteroids? (6/13/10)", helps here.




    Asteroids are not distributed uniformly in the asteroid belt, but
    could be approximated to be evenly spaced in a region from 2.2 AU (1
    AU is 93 million miles, or the average distance between Earth and the
    Sun) to 3.2 AU from the Sun and extending 0.5 AU above and below the
    ecliptic (the plane of Earth's orbit, which is a convenient reference
    for the solar system). That yields a volume of roughly 16 cubic AU, or
    about 13 trillion trillion cubic miles. (Note: space is big!)



    If there were 2 million asteroids 1 mile or larger in that volume,
    each asteroid would have 6.7 million trillion cubic miles to itself,
    so the average distance between individual asteroids 1 mile in
    diameter or larger would be about 1.9 million miles. That is nearly 8
    times the distance between Earth and the Moon.




    Just for something interesting, it looks like your gif only covers the Jupiter trojans, which are the asteroids that Jupiter shepherds around the solar system at its Lagrange points. A more accurate image is available on Wikipedia.



    Asteroid map from Wikipedia






    share|improve this answer









    $endgroup$


















      3












      $begingroup$

      The asteroid belt isn't nearly as dense as popular movies make it out to be.



      An answer from the Dawn Mission's FAQ, specifically "What is the average distance between individual asteroids? (6/13/10)", helps here.




      Asteroids are not distributed uniformly in the asteroid belt, but
      could be approximated to be evenly spaced in a region from 2.2 AU (1
      AU is 93 million miles, or the average distance between Earth and the
      Sun) to 3.2 AU from the Sun and extending 0.5 AU above and below the
      ecliptic (the plane of Earth's orbit, which is a convenient reference
      for the solar system). That yields a volume of roughly 16 cubic AU, or
      about 13 trillion trillion cubic miles. (Note: space is big!)



      If there were 2 million asteroids 1 mile or larger in that volume,
      each asteroid would have 6.7 million trillion cubic miles to itself,
      so the average distance between individual asteroids 1 mile in
      diameter or larger would be about 1.9 million miles. That is nearly 8
      times the distance between Earth and the Moon.




      Just for something interesting, it looks like your gif only covers the Jupiter trojans, which are the asteroids that Jupiter shepherds around the solar system at its Lagrange points. A more accurate image is available on Wikipedia.



      Asteroid map from Wikipedia






      share|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        The asteroid belt isn't nearly as dense as popular movies make it out to be.



        An answer from the Dawn Mission's FAQ, specifically "What is the average distance between individual asteroids? (6/13/10)", helps here.




        Asteroids are not distributed uniformly in the asteroid belt, but
        could be approximated to be evenly spaced in a region from 2.2 AU (1
        AU is 93 million miles, or the average distance between Earth and the
        Sun) to 3.2 AU from the Sun and extending 0.5 AU above and below the
        ecliptic (the plane of Earth's orbit, which is a convenient reference
        for the solar system). That yields a volume of roughly 16 cubic AU, or
        about 13 trillion trillion cubic miles. (Note: space is big!)



        If there were 2 million asteroids 1 mile or larger in that volume,
        each asteroid would have 6.7 million trillion cubic miles to itself,
        so the average distance between individual asteroids 1 mile in
        diameter or larger would be about 1.9 million miles. That is nearly 8
        times the distance between Earth and the Moon.




        Just for something interesting, it looks like your gif only covers the Jupiter trojans, which are the asteroids that Jupiter shepherds around the solar system at its Lagrange points. A more accurate image is available on Wikipedia.



        Asteroid map from Wikipedia






        share|improve this answer









        $endgroup$



        The asteroid belt isn't nearly as dense as popular movies make it out to be.



        An answer from the Dawn Mission's FAQ, specifically "What is the average distance between individual asteroids? (6/13/10)", helps here.




        Asteroids are not distributed uniformly in the asteroid belt, but
        could be approximated to be evenly spaced in a region from 2.2 AU (1
        AU is 93 million miles, or the average distance between Earth and the
        Sun) to 3.2 AU from the Sun and extending 0.5 AU above and below the
        ecliptic (the plane of Earth's orbit, which is a convenient reference
        for the solar system). That yields a volume of roughly 16 cubic AU, or
        about 13 trillion trillion cubic miles. (Note: space is big!)



        If there were 2 million asteroids 1 mile or larger in that volume,
        each asteroid would have 6.7 million trillion cubic miles to itself,
        so the average distance between individual asteroids 1 mile in
        diameter or larger would be about 1.9 million miles. That is nearly 8
        times the distance between Earth and the Moon.




        Just for something interesting, it looks like your gif only covers the Jupiter trojans, which are the asteroids that Jupiter shepherds around the solar system at its Lagrange points. A more accurate image is available on Wikipedia.



        Asteroid map from Wikipedia







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 5 hours ago









        josjos

        173110




        173110























            1












            $begingroup$

            The asteroid belt is roughly 6 Astronomical Units wide, and so when it is drawn only 600 pixels wide with each asteroid a handful of pixels wide, *you end up with each asteroid being five times bigger than the Sun!**



            I've borrowed a small piece of the image used in @jos' excellent answer to show what I mean.



            enter image description here





            As pointed out in this answer, Wikipedia says:




            Contrary to popular imagery, the asteroid belt is mostly empty. The asteroids are spread over such a large volume that it would be improbable to reach an asteroid without aiming carefully.






            The Exposing PseudoAstronomy blogpost Asteroid Belts – Proximity of Rocks and Why Navigation Is Not Dangerous (Sorry, Han Solo) says:




            How Many Asteroids of What Size?



            In terms of what is known, there are about 20,000 asteroids between 2-3 km, which is about the smallest that we likely have a complete sampling of. What that statement means is that, while we have identified asteroids that are smaller, our detection technology is not good enough to have found all of the asteroids that are smaller.



            If we extrapolate, assuming a -3 power low, down to, say, 100-meter asteroids, there are probably ~82 million asteroids that are ~100-200-meters across. If we extrapolate further, down to 1-meter asteroids, then we really have a gargantuan number of objects – about 1014 (100 quadrillion) objects of that size. That’s quite a lot.



            What Does this Mean for Navigation?



            If we add up all of those objects, we have about 1.2×1014 asteroids larger than 1 meter. Now, let’s look at the asteroid belt. It stretches from 1.8 to 3.3 A.U., which is a distance of 1.5 A.U., or about 225,000,000 km. That’s a fairly large distance (that’s actually about the distance between the Sun and Mars).



            The area of a disk that size, however, is gargantuan: A = π · r2 = π · ((3.3 A.U.)2 – (1.8 A.U.)2) = π · (1.7·1017 km2) = 5.4·1017 km2. That is a huge area. Simple division shows that each asteroid, regardless of its own size, could have 4,500 km2 all to itself – a little bit more than the entire U.S. state of Rhode Island.



            And that’s if they were all just in one plane. In reality, they occupy a volume of space, some orbiting “above” or “below” others (where those terms are relative to the plane that the Earth’s orbit makes).



            Even if we cut the size of asteroids in half again, and were interested in all asteroids larger than half a meter (1.5 ft) in size, then we have 8 times as many asteroids, but each one still has over 500 km2 all to itself, and even more space if we consider the vertical component.



            What does this mean for navigation? It’s easy! In fact, you really have to try to hit an asteroid, at least in our own belt. And so, the next time you see a tiny ship careening through an asteroid field in a TV show or movie, remember that in real life, asteroid belts really aren’t that dangerous for navigation.




            @AlanSE's answer to the related but different question What's the (particle) density of the asteroid belt? shows an illustration of the power law behavior of size at large, observable scale. Unfortunately the link to the source (dated 2001) is now broken.



            enter image description here






            share|improve this answer











            $endgroup$













            • $begingroup$
              1017 should be 10E17 or 10^17
              $endgroup$
              – Uwe
              2 mins ago
















            1












            $begingroup$

            The asteroid belt is roughly 6 Astronomical Units wide, and so when it is drawn only 600 pixels wide with each asteroid a handful of pixels wide, *you end up with each asteroid being five times bigger than the Sun!**



            I've borrowed a small piece of the image used in @jos' excellent answer to show what I mean.



            enter image description here





            As pointed out in this answer, Wikipedia says:




            Contrary to popular imagery, the asteroid belt is mostly empty. The asteroids are spread over such a large volume that it would be improbable to reach an asteroid without aiming carefully.






            The Exposing PseudoAstronomy blogpost Asteroid Belts – Proximity of Rocks and Why Navigation Is Not Dangerous (Sorry, Han Solo) says:




            How Many Asteroids of What Size?



            In terms of what is known, there are about 20,000 asteroids between 2-3 km, which is about the smallest that we likely have a complete sampling of. What that statement means is that, while we have identified asteroids that are smaller, our detection technology is not good enough to have found all of the asteroids that are smaller.



            If we extrapolate, assuming a -3 power low, down to, say, 100-meter asteroids, there are probably ~82 million asteroids that are ~100-200-meters across. If we extrapolate further, down to 1-meter asteroids, then we really have a gargantuan number of objects – about 1014 (100 quadrillion) objects of that size. That’s quite a lot.



            What Does this Mean for Navigation?



            If we add up all of those objects, we have about 1.2×1014 asteroids larger than 1 meter. Now, let’s look at the asteroid belt. It stretches from 1.8 to 3.3 A.U., which is a distance of 1.5 A.U., or about 225,000,000 km. That’s a fairly large distance (that’s actually about the distance between the Sun and Mars).



            The area of a disk that size, however, is gargantuan: A = π · r2 = π · ((3.3 A.U.)2 – (1.8 A.U.)2) = π · (1.7·1017 km2) = 5.4·1017 km2. That is a huge area. Simple division shows that each asteroid, regardless of its own size, could have 4,500 km2 all to itself – a little bit more than the entire U.S. state of Rhode Island.



            And that’s if they were all just in one plane. In reality, they occupy a volume of space, some orbiting “above” or “below” others (where those terms are relative to the plane that the Earth’s orbit makes).



            Even if we cut the size of asteroids in half again, and were interested in all asteroids larger than half a meter (1.5 ft) in size, then we have 8 times as many asteroids, but each one still has over 500 km2 all to itself, and even more space if we consider the vertical component.



            What does this mean for navigation? It’s easy! In fact, you really have to try to hit an asteroid, at least in our own belt. And so, the next time you see a tiny ship careening through an asteroid field in a TV show or movie, remember that in real life, asteroid belts really aren’t that dangerous for navigation.




            @AlanSE's answer to the related but different question What's the (particle) density of the asteroid belt? shows an illustration of the power law behavior of size at large, observable scale. Unfortunately the link to the source (dated 2001) is now broken.



            enter image description here






            share|improve this answer











            $endgroup$













            • $begingroup$
              1017 should be 10E17 or 10^17
              $endgroup$
              – Uwe
              2 mins ago














            1












            1








            1





            $begingroup$

            The asteroid belt is roughly 6 Astronomical Units wide, and so when it is drawn only 600 pixels wide with each asteroid a handful of pixels wide, *you end up with each asteroid being five times bigger than the Sun!**



            I've borrowed a small piece of the image used in @jos' excellent answer to show what I mean.



            enter image description here





            As pointed out in this answer, Wikipedia says:




            Contrary to popular imagery, the asteroid belt is mostly empty. The asteroids are spread over such a large volume that it would be improbable to reach an asteroid without aiming carefully.






            The Exposing PseudoAstronomy blogpost Asteroid Belts – Proximity of Rocks and Why Navigation Is Not Dangerous (Sorry, Han Solo) says:




            How Many Asteroids of What Size?



            In terms of what is known, there are about 20,000 asteroids between 2-3 km, which is about the smallest that we likely have a complete sampling of. What that statement means is that, while we have identified asteroids that are smaller, our detection technology is not good enough to have found all of the asteroids that are smaller.



            If we extrapolate, assuming a -3 power low, down to, say, 100-meter asteroids, there are probably ~82 million asteroids that are ~100-200-meters across. If we extrapolate further, down to 1-meter asteroids, then we really have a gargantuan number of objects – about 1014 (100 quadrillion) objects of that size. That’s quite a lot.



            What Does this Mean for Navigation?



            If we add up all of those objects, we have about 1.2×1014 asteroids larger than 1 meter. Now, let’s look at the asteroid belt. It stretches from 1.8 to 3.3 A.U., which is a distance of 1.5 A.U., or about 225,000,000 km. That’s a fairly large distance (that’s actually about the distance between the Sun and Mars).



            The area of a disk that size, however, is gargantuan: A = π · r2 = π · ((3.3 A.U.)2 – (1.8 A.U.)2) = π · (1.7·1017 km2) = 5.4·1017 km2. That is a huge area. Simple division shows that each asteroid, regardless of its own size, could have 4,500 km2 all to itself – a little bit more than the entire U.S. state of Rhode Island.



            And that’s if they were all just in one plane. In reality, they occupy a volume of space, some orbiting “above” or “below” others (where those terms are relative to the plane that the Earth’s orbit makes).



            Even if we cut the size of asteroids in half again, and were interested in all asteroids larger than half a meter (1.5 ft) in size, then we have 8 times as many asteroids, but each one still has over 500 km2 all to itself, and even more space if we consider the vertical component.



            What does this mean for navigation? It’s easy! In fact, you really have to try to hit an asteroid, at least in our own belt. And so, the next time you see a tiny ship careening through an asteroid field in a TV show or movie, remember that in real life, asteroid belts really aren’t that dangerous for navigation.




            @AlanSE's answer to the related but different question What's the (particle) density of the asteroid belt? shows an illustration of the power law behavior of size at large, observable scale. Unfortunately the link to the source (dated 2001) is now broken.



            enter image description here






            share|improve this answer











            $endgroup$



            The asteroid belt is roughly 6 Astronomical Units wide, and so when it is drawn only 600 pixels wide with each asteroid a handful of pixels wide, *you end up with each asteroid being five times bigger than the Sun!**



            I've borrowed a small piece of the image used in @jos' excellent answer to show what I mean.



            enter image description here





            As pointed out in this answer, Wikipedia says:




            Contrary to popular imagery, the asteroid belt is mostly empty. The asteroids are spread over such a large volume that it would be improbable to reach an asteroid without aiming carefully.






            The Exposing PseudoAstronomy blogpost Asteroid Belts – Proximity of Rocks and Why Navigation Is Not Dangerous (Sorry, Han Solo) says:




            How Many Asteroids of What Size?



            In terms of what is known, there are about 20,000 asteroids between 2-3 km, which is about the smallest that we likely have a complete sampling of. What that statement means is that, while we have identified asteroids that are smaller, our detection technology is not good enough to have found all of the asteroids that are smaller.



            If we extrapolate, assuming a -3 power low, down to, say, 100-meter asteroids, there are probably ~82 million asteroids that are ~100-200-meters across. If we extrapolate further, down to 1-meter asteroids, then we really have a gargantuan number of objects – about 1014 (100 quadrillion) objects of that size. That’s quite a lot.



            What Does this Mean for Navigation?



            If we add up all of those objects, we have about 1.2×1014 asteroids larger than 1 meter. Now, let’s look at the asteroid belt. It stretches from 1.8 to 3.3 A.U., which is a distance of 1.5 A.U., or about 225,000,000 km. That’s a fairly large distance (that’s actually about the distance between the Sun and Mars).



            The area of a disk that size, however, is gargantuan: A = π · r2 = π · ((3.3 A.U.)2 – (1.8 A.U.)2) = π · (1.7·1017 km2) = 5.4·1017 km2. That is a huge area. Simple division shows that each asteroid, regardless of its own size, could have 4,500 km2 all to itself – a little bit more than the entire U.S. state of Rhode Island.



            And that’s if they were all just in one plane. In reality, they occupy a volume of space, some orbiting “above” or “below” others (where those terms are relative to the plane that the Earth’s orbit makes).



            Even if we cut the size of asteroids in half again, and were interested in all asteroids larger than half a meter (1.5 ft) in size, then we have 8 times as many asteroids, but each one still has over 500 km2 all to itself, and even more space if we consider the vertical component.



            What does this mean for navigation? It’s easy! In fact, you really have to try to hit an asteroid, at least in our own belt. And so, the next time you see a tiny ship careening through an asteroid field in a TV show or movie, remember that in real life, asteroid belts really aren’t that dangerous for navigation.




            @AlanSE's answer to the related but different question What's the (particle) density of the asteroid belt? shows an illustration of the power law behavior of size at large, observable scale. Unfortunately the link to the source (dated 2001) is now broken.



            enter image description here







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            edited 50 mins ago

























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            • $begingroup$
              1017 should be 10E17 or 10^17
              $endgroup$
              – Uwe
              2 mins ago


















            • $begingroup$
              1017 should be 10E17 or 10^17
              $endgroup$
              – Uwe
              2 mins ago
















            $begingroup$
            1017 should be 10E17 or 10^17
            $endgroup$
            – Uwe
            2 mins ago




            $begingroup$
            1017 should be 10E17 or 10^17
            $endgroup$
            – Uwe
            2 mins ago


















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