<blockquote>
<p><strong>Was there ever a meteorite with measurable gravity?</strong></p>
</blockquote>
<p>At risk of stating the obvious:   Yes.  All mass, as far as every test I know about, has measured gravity.  No test has ever measured no gravity within the range of accuracy, so even low mass objects can have their gravity measured.</p>
<p><a href="https://en.wikipedia.org/wiki/Rubble_pile" rel="nofollow noreferrer">Rubble pile asteroids</a> couldn't form if they didn't exert gravity.  Asteroids have been observed in orbiting pairs, so there's little question that they exert measurable gravity, if you can get close enough and have good enough equiptment to one to measure it.</p>
<blockquote>
<p><strong>Did we ever find a meteorite (a meteoroid that has fallen down to Earth) whose mass, density and gravity were high enough to demonstrate
their gravity similar to the Cavendish experiment?</strong></p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Cavendish_experiment" rel="nofollow noreferrer">Cavendish measured gravity using 158 kg lead balls</a> and some sensitive equiptment.   These larger lead balls were stationary sources of perpendicular gravity on two smaller, 0.73 kg lead balls as the suspended objects on a wire upon which the gravity acted.   By the twisting of the suspension wire, and from a distance, so his mass wouldn't affect the experiment, Cavendish measured gravity.</p>
<p>A 158 kg lead ball is about 15 cm in radius and the smaller balls about 2.5 cm.  The distance between them (and I'm assuming this was center to center as that's standard in physical notations of this type, is listed as about 9 inches or 29.5 cm.    That means, surface to surface the distance was about 12 cm, which doesn't change the equation, but you wouldn't want to calculate gravity with two surfaces inside each other, even if the math lets you get away with that, so, I thought I'd mention it.</p>
<p>Some quick math and 29.5 cm, 158 KG, the gravitational force is easy to calculate. F=Gm/r^2, which comes to about 2.7/10,000,000 meters per second squared about 35 million times weaker than earth gravity.   A meteor could certainly have that force of gravity pretty easily.  While less dense than lead, greater overall could accomplish 1/35 million G forces easily enough.</p>
<p>What does that mean?  1/35 million G forces?   Lets call that amount of gravity one Cavendish - for obvious reasons.</p>
<p>Have you ever walked up a flight of stairs, one step at a time?  By just taking one step up a flight of stairs, you've reduced the gravitational force acting on you by one Cavendish.  A bit more than one, actually.   By taking an elevator up one floor works out to about 33 Cavendishes drop in gravity, for every 3 meters.    That's how gravity works.   Mass increases (roughly speaking) with the cube of the radius.  Gravity falls off with the square of the distance, so larger objects not denser objects, usually win in the gravity contest.   There are exceptions if there's enough variation in density, obviously.  Earth, for example has higher surface gravity than Uranus and Mercury has very nearly equal surface gravity to Mars, but as a general rule, when the overall mass is much different, density matters less.   The Sun and the Moon appear about the same size to our eyes and as a result have similar tidal forces, but the Sun exerts hundreds of times more gravity on the Earth than the Moon does.</p>
<p>The weakness of this force indicates how sensitive Cavendish' experiment was for it's time.</p>
<p>How did Cavendish know that there wasn't a mountain range in the distance that would exert equal pull on his test, or maybe some part of the Earth that was heavier on one side of where he lived than the other side?</p>
<p>Lets imagine that there was a mountain 12 km from Cavendish's laboratory, lets pretend this mountain was a 1.5 km high and weighs about 4 trillion kg.  That mountain would exert nearly 7 times the gravitational pull on Cavendish's test balls that the his 158 kg stationary lead balls did, so, did Cavendish need to find a flat region of equal gravitational pull in all directions?</p>
<p>Actually, the answer is no, because a mountain in the distance would have generated a gravitational field that would have been part of the initial conditions and he was measuring the change from the initial conditions.</p>
<p>And even if somebody had decided to mess up Cavendish's experiment by putting a mountain 12 miles from his house during his experiment, the effect would have been nearly equal on the two balls so it wouldn't have made much difference. It would just have been a tidal force, and a very small specific force on the torque of the wire that Cavendish was interested in.   Cavendish's experiment only works under very specific conditions, one 158 kg weight pulling both suspended balls on the wire in the same clockwise direction so the twist can be measured.  That doesn't work with a single large mass. It's the wrong kind of setup.</p>
<p>But to answer your question, forces greater than 1 Cavendish are easy for any relatively large body and a meteor could do that.  Bennu, which you mentioned in a comment exerts 6 micro-G, or about 22 Cavendishes.  It's radius is about 250 meters, so within about 1100 or 1200 meters, an object that size would exert 1 Cavendish of gravitational force.</p>
<p>Ofcourse, if Bennu was to actually strike the Earth, it would likely decrease, not increase gravity at the spot it hits because it would eject material over a wide area, leaving a crater.  Likewise, the effect would be very small, but on the edge of the crater, probably several Cavendish's worth of decreased gravity.</p>
<p>Perpendicular forces like this happen on Earth.  As land rises and falls.  As glaciers shift, the perpendicular gravitational force near those massive objects changes and this is most noticeable on the ocean.  The gravity of the Greenland ice sheet or the Antarctic ice sheet is measurable on sea level and calculating sea level rise if those ice sheets melt (which will take some time), but accurate calculation has to take into account the gravitational pull of the mass of glacial ice on the ocean's sea level.</p>
<p>The mass of a meteorite is generally quite small in the grand scheme of things and it's gravitational force on a nearby blade of grass, for example, while perhaps several Cavendishes, is negligible for all practical purposes, but measurable if you have sensitive enough equiptment.  You can now tell your friend when you drive past a mountain range that your car is experiencing a force greater than the force Cavendish first use to measure the gravitational constant and they probably won't know what you're talking about.</p>
<p>And if you're talking a much smaller, mostly Iron meteorite fragment, then you need to be quite close.  <a href="https://en.wikipedia.org/wiki/List_of_largest_meteorites_on_Earth#:%7E:text=Iron%20%5B1%5D%20%5B2%5D%20%20%20%20N%C2%B0%20,%EF%BB%BF%2027.609%20...%20%2010%20more%20rows%20" rel="nofollow noreferrer">The largest meteorite fragment or solid piece is the Hoba</a>, about 60,000 KG.  You'd need to be about 4 meters from it's center to experience 1 Cavendish of gravitational force.</p>
<p>It's fun to think that walking past a very dense, very heavy object, that you might feel it's gravity, but most things aren't nearly massive enough.  You don't feel the gravitational pull of a mountain range when you walk past it because the pull is negligible and mountain ranges are much more massive than most asteroids that are likely to hit the Earth.   Mountain ranges also are generally made up of lighter rock than the surrounding area, so there's also that, but even if they weren't, you still couldn't feel the gravitational pull of even the largest string of mountains.</p>
<p>Hope that wasn't too long.  I got answering and it came out longer than probably necessary.</p>

CC BY-SA 4.0
2020-12-02T12:01:06.697Z
I have some questions on your answer: First, you say the Sun and the Moon have similar tidal forces, but why are ocean tides caused by the Moon only then? 2nd, when riding an elevator the change of gravity are multiple hundredths of Gs. I once measured the gravity when riding an elevator, the change was about plus/minus 0.08 Gs. So when you start to ride down (or arrive up) in the lift you experience 0.92g while when starting to ride up (or arriving down) you experience 1.08g. I don't know an elevator where the difference would be several cavendishes. Third, you mention the mountains' mass...
Greenhorn
CC BY-SA 4.0
2020-12-02T12:02:52.990Z
...and this gravity difference is called a mascon (mass concentration) which also seems to be more than several cavendishes. Here's a map showing the differences in mascons: https://thumbs-prod.si-cdn.com/70gjnV4C2bY6VUgVj2OHyVLvGkg=/800x600/filters:no_upscale()/https://public-media.si-cdn.com/filer/20110520102356GOCE-Geoidkarte_mit_Schatten_H.jpg
Greenhorn
CC BY-SA 4.0
2020-12-02T12:04:19.067Z
And your mention of a decreased gravity concerning Bennu is pointless because I'm asking on the gravity of _Bennu_ itself, not how the _Earth's_ local gravity changes.
Greenhorn
CC BY-SA 4.0
2020-12-02T12:25:37.387Z
Tides are caused by both the Moon and the Sun.  But the Lunar force is about 3 times stronger.  I said "about the same" whether 3 times as much qualifies, I guess depends on ones flexibility on the use of the word about.   And I agree, Mascons probably have gravity variations of several Cavendishes, though I've not checked.  Equator to Pole the difference is well into the tens of thousands.
userLTK
CC BY-SA 4.0
2020-12-02T12:41:44.090Z
The equatorial surface gravity is 9.78 m/s² (0.997g) while the polar is 9.832 m/s² (1.003g), so the difference between pole and equator is plus/minus 0.003g in reference to average.
Greenhorn
CC BY-SA 4.0
2020-12-02T11:17:56.560Z
2020-12-02T11:17:56.560Z
userLTK
<p>That would be very difficult to do.  The meteor will pulverize when striking earth and it would be difficult to determine the original size of the meteor. Estimates of the initial mass can maybe be determined from the size of the crater but I doubt the accuracy of this approach. Fragments of the meteorite can be collected and used for the experiments you mentioned, but so could earth rocks!</p>

CC BY-SA 4.0
2020-12-02T07:53:30.970Z
My question is either about meteoroids that make it down to Earth's surface, or about the fragments you mention. Bennu's surface gravity is 6 micro-g. Would that gravity remain if Bennu impacted the Earth, and if so, would its attraction be noticeable to everyone nearby?
Greenhorn
CC BY-SA 4.0
2020-12-02T04:47:28.343Z
2020-12-02T04:47:28.343Z
Natsfan
<p>Did we ever find a meteorite (a meteoroid that has fallen down to Earth) whose mass, density and gravity were high enough to demonstrate their gravity similar to the Cavendish experiment, so that something was attracted to it?</p>

CC BY-SA 4.0
2020-12-01T08:32:07.423Z
You don't need large masses to do the Cavendish experiment. See [Bending Spacetime in the Basement](https://www.fourmilab.ch/gravitation/foobar/) by John Walker of Autodesk.
PM 2Ring
CC BY-SA 4.0
2020-12-02T03:37:26.527Z
Related for asteroids: https://astronomy.stackexchange.com/questions/12897/do-asteroids-have-a-gravitational-field
Nilay Ghosh
CC BY-SA 4.0
2020-12-02T07:55:46.623Z
@NilayGhosh Of course everything with mass has gravity. The question is whether this meteorite's gravity would be noticeable, such as attracting things nearby.
Greenhorn
CC BY-SA 4.0
2020-12-02T09:02:53.077Z
A scale does what you ask for. Meteorites are often (always?) weighed when found, especially those which are sold. Why would you do a measurement of the gravitational attraction via a MUCH more complicated method when the equivalence of the heavy mass and the interial mass is demonstrated in countless experiments?
planetmaker
CC BY-SA 4.0
2020-12-02T09:08:47.430Z
@planetmaker Because I wonder about the environmental influence. Imagine a meteorite comes down in a desert. I wonder whether it can have enough gravity to noticeably attract or shift sand corns.
Greenhorn
CC BY-SA 4.0
2020-12-02T09:13:47.807Z
That question then is better placed in Earth science - and without any reference to meteorites - they are not special in this respect than an equivalently big other piece of rock. The answer is 'no' ;) Ask yourself the question whether your house or your care affects how a grain of sand falls down to Earth.
planetmaker
CC BY-SA 4.0
2020-12-02T09:15:20.660Z
@planetmaker But aren't there many meteorites that are much more massive and denser than terrestrial rocks? Afaik meteorites are unusually heavy.
Greenhorn
CC BY-SA 4.0
2020-12-02T09:38:39.583Z
Your presumption is wrong. Most are indistinguishable from terrestrial rocks to the laymen as they're chemically mostly the same, too https://sites.wustl.edu/meteoritesite/items/chemical-composition-of-meteorites/
planetmaker
CC BY-SA 4.0
2020-12-02T09:45:24.973Z
@Greenhorn when meteors burn up entering the Earth's atmosphere, what can sometimes remain, if the meteor has enough iron is a dense iron fragment, so those meteorites are more dense, but as Planetmaker points out, those are less common.
userLTK
CC BY-SA 4.0
2020-12-02T10:48:37.500Z
@planetmaker I think you'd recognize a meteorite especially by the crater around it. :-)
Greenhorn
CC BY-SA 4.0
2020-12-02T10:49:41.880Z
@userLTK I'm indeed asking on the denser more metallic meteorites.
Greenhorn
CC BY-SA 4.0
2020-12-02T11:20:46.537Z
I posted an answer that, I think, addresses that.   The problem is that metallic meteorites, while dense are very small in a gravitational sense, so they exert very little gravity, more than Cavendish's experiment if you're right next to one but not a lot more.
userLTK
CC BY-SA 4.0
2020-12-02T14:13:21.177Z
@Greenhorn if there is any crater at all. A (large) crater is actually not to be expected unless you have an extremely rare giant meteor. See also one of the largest historic meteor fall https://en.wikipedia.org/wiki/Chelyabinsk_meteorite - no crater was formed at all as it fell into water or on snow. In any case: a meteor affects gravity identically to a similar normal rock, be that natural or artificial, like a sky scraper or waste deposit or military installation or the litteral "brick in the wall"
planetmaker
CC BY-SA 4.0
2020-12-02T15:48:25.533Z
@planetmaker I see.
Greenhorn
<gravity><astrophysics><meteorite>
Was there ever a meteorite with a measurable gravity?
Greenhorn
# Was there ever a meteorite with a measurable gravity?

(Asked by: username_0 on 2020-12-01)

Did we ever find a meteorite (a meteoroid that has fallen down to Earth) whose mass, density and gravity were high enough to demonstrate their gravity similar to the Cavendish experiment, so that something was attracted to it?

**Tags:** gravity, astrophysics, meteorite

> **Comments:**
> **username_1**: You don't need large masses to do the Cavendish experiment. See [Bending Spacetime in the Basement](https://www.fourmilab.ch/gravitation/foobar/) by John Walker of Autodesk.  
> **username_2**: Related for asteroids: https://astronomy.stackexchange.com/questions/12897/do-asteroids-have-a-gravitational-field  
> **username_0**: @NilayGhosh Of course everything with mass has gravity. The question is whether this meteorite's gravity would be noticeable, such as attracting things nearby.  
> **username_3**: A scale does what you ask for. Meteorites are often (always?) weighed when found, especially those which are sold. Why would you do a measurement of the gravitational attraction via a MUCH more complicated method when the equivalence of the heavy mass and the interial mass is demonstrated in countless experiments?  
> **username_0**: @username_3 Because I wonder about the environmental influence. Imagine a meteorite comes down in a desert. I wonder whether it can have enough gravity to noticeably attract or shift sand corns.  
> **username_3**: That question then is better placed in Earth science - and without any reference to meteorites - they are not special in this respect than an equivalently big other piece of rock. The answer is 'no' ;) Ask yourself the question whether your house or your care affects how a grain of sand falls down to Earth.  
> **username_0**: @username_3 But aren't there many meteorites that are much more massive and denser than terrestrial rocks? Afaik meteorites are unusually heavy.  
> **username_3**: Your presumption is wrong. Most are indistinguishable from terrestrial rocks to the laymen as they're chemically mostly the same, too https://sites.wustl.edu/meteoritesite/items/chemical-composition-of-meteorites/  
> **username_4**: @username_0 when meteors burn up entering the Earth's atmosphere, what can sometimes remain, if the meteor has enough iron is a dense iron fragment, so those meteorites are more dense, but as Planetmaker points out, those are less common.  
> **username_0**: @username_3 I think you'd recognize a meteorite especially by the crater around it. :-)  
> **username_0**: @username_4 I'm indeed asking on the denser more metallic meteorites.  
> **username_4**: I posted an answer that, I think, addresses that.   The problem is that metallic meteorites, while dense are very small in a gravitational sense, so they exert very little gravity, more than Cavendish's experiment if you're right next to one but not a lot more.  
> **username_3**: @username_0 if there is any crater at all. A (large) crater is actually not to be expected unless you have an extremely rare giant meteor. See also one of the largest historic meteor fall https://en.wikipedia.org/wiki/Chelyabinsk_meteorite - no crater was formed at all as it fell into water or on snow. In any case: a meteor affects gravity identically to a similar normal rock, be that natural or artificial, like a sky scraper or waste deposit or military installation or the litteral "brick in the wall"  
> **username_0**: @username_3 I see.  

---

## 2 answers

### Answer by username_4 on 2020-12-02. Score: 3 | Accepted answer

> **Was there ever a meteorite with measurable gravity?**

At risk of stating the obvious: Yes. All mass, as far as every test I know about, has measured gravity. No test has ever measured no gravity within the range of accuracy, so even low mass objects can have their gravity measured.

[Rubble pile asteroids](https://en.wikipedia.org/wiki/Rubble_pile) couldn't form if they didn't exert gravity. Asteroids have been observed in orbiting pairs, so there's little question that they exert measurable gravity, if you can get close enough and have good enough equiptment to one to measure it.

> **Did we ever find a meteorite (a meteoroid that has fallen down to Earth) whose mass, density and gravity were high enough to demonstrate
> their gravity similar to the Cavendish experiment?**

[Cavendish measured gravity using 158 kg lead balls](https://en.wikipedia.org/wiki/Cavendish_experiment) and some sensitive equiptment. These larger lead balls were stationary sources of perpendicular gravity on two smaller, 0.73 kg lead balls as the suspended objects on a wire upon which the gravity acted. By the twisting of the suspension wire, and from a distance, so his mass wouldn't affect the experiment, Cavendish measured gravity.

A 158 kg lead ball is about 15 cm in radius and the smaller balls about 2.5 cm. The distance between them (and I'm assuming this was center to center as that's standard in physical notations of this type, is listed as about 9 inches or 29.5 cm. That means, surface to surface the distance was about 12 cm, which doesn't change the equation, but you wouldn't want to calculate gravity with two surfaces inside each other, even if the math lets you get away with that, so, I thought I'd mention it.

Some quick math and 29.5 cm, 158 KG, the gravitational force is easy to calculate. F=Gm/r^2, which comes to about 2.7/10,000,000 meters per second squared about 35 million times weaker than earth gravity. A meteor could certainly have that force of gravity pretty easily. While less dense than lead, greater overall could accomplish 1/35 million G forces easily enough.

What does that mean? 1/35 million G forces? Lets call that amount of gravity one Cavendish - for obvious reasons.

Have you ever walked up a flight of stairs, one step at a time? By just taking one step up a flight of stairs, you've reduced the gravitational force acting on you by one Cavendish. A bit more than one, actually. By taking an elevator up one floor works out to about 33 Cavendishes drop in gravity, for every 3 meters. That's how gravity works. Mass increases (roughly speaking) with the cube of the radius. Gravity falls off with the square of the distance, so larger objects not denser objects, usually win in the gravity contest. There are exceptions if there's enough variation in density, obviously. Earth, for example has higher surface gravity than Uranus and Mercury has very nearly equal surface gravity to Mars, but as a general rule, when the overall mass is much different, density matters less. The Sun and the Moon appear about the same size to our eyes and as a result have similar tidal forces, but the Sun exerts hundreds of times more gravity on the Earth than the Moon does.

The weakness of this force indicates how sensitive Cavendish' experiment was for it's time.

How did Cavendish know that there wasn't a mountain range in the distance that would exert equal pull on his test, or maybe some part of the Earth that was heavier on one side of where he lived than the other side?

Lets imagine that there was a mountain 12 km from Cavendish's laboratory, lets pretend this mountain was a 1.5 km high and weighs about 4 trillion kg. That mountain would exert nearly 7 times the gravitational pull on Cavendish's test balls that the his 158 kg stationary lead balls did, so, did Cavendish need to find a flat region of equal gravitational pull in all directions?

Actually, the answer is no, because a mountain in the distance would have generated a gravitational field that would have been part of the initial conditions and he was measuring the change from the initial conditions.

And even if somebody had decided to mess up Cavendish's experiment by putting a mountain 12 miles from his house during his experiment, the effect would have been nearly equal on the two balls so it wouldn't have made much difference. It would just have been a tidal force, and a very small specific force on the torque of the wire that Cavendish was interested in. Cavendish's experiment only works under very specific conditions, one 158 kg weight pulling both suspended balls on the wire in the same clockwise direction so the twist can be measured. That doesn't work with a single large mass. It's the wrong kind of setup.

But to answer your question, forces greater than 1 Cavendish are easy for any relatively large body and a meteor could do that. Bennu, which you mentioned in a comment exerts 6 micro-G, or about 22 Cavendishes. It's radius is about 250 meters, so within about 1100 or 1200 meters, an object that size would exert 1 Cavendish of gravitational force.

Ofcourse, if Bennu was to actually strike the Earth, it would likely decrease, not increase gravity at the spot it hits because it would eject material over a wide area, leaving a crater. Likewise, the effect would be very small, but on the edge of the crater, probably several Cavendish's worth of decreased gravity.

Perpendicular forces like this happen on Earth. As land rises and falls. As glaciers shift, the perpendicular gravitational force near those massive objects changes and this is most noticeable on the ocean. The gravity of the Greenland ice sheet or the Antarctic ice sheet is measurable on sea level and calculating sea level rise if those ice sheets melt (which will take some time), but accurate calculation has to take into account the gravitational pull of the mass of glacial ice on the ocean's sea level.

The mass of a meteorite is generally quite small in the grand scheme of things and it's gravitational force on a nearby blade of grass, for example, while perhaps several Cavendishes, is negligible for all practical purposes, but measurable if you have sensitive enough equiptment. You can now tell your friend when you drive past a mountain range that your car is experiencing a force greater than the force Cavendish first use to measure the gravitational constant and they probably won't know what you're talking about.

And if you're talking a much smaller, mostly Iron meteorite fragment, then you need to be quite close. [The largest meteorite fragment or solid piece is the Hoba](https://en.wikipedia.org/wiki/List_of_largest_meteorites_on_Earth#:%7E:text=Iron%20%5B1%5D%20%5B2%5D%20%20%20%20N%C2%B0%20,%EF%BB%BF%2027.609%20...%20%2010%20more%20rows%20), about 60,000 KG. You'd need to be about 4 meters from it's center to experience 1 Cavendish of gravitational force.

It's fun to think that walking past a very dense, very heavy object, that you might feel it's gravity, but most things aren't nearly massive enough. You don't feel the gravitational pull of a mountain range when you walk past it because the pull is negligible and mountain ranges are much more massive than most asteroids that are likely to hit the Earth. Mountain ranges also are generally made up of lighter rock than the surrounding area, so there's also that, but even if they weren't, you still couldn't feel the gravitational pull of even the largest string of mountains.

Hope that wasn't too long. I got answering and it came out longer than probably necessary.

> **Comments:**
> **username_0**: I have some questions on your answer: First, you say the Sun and the Moon have similar tidal forces, but why are ocean tides caused by the Moon only then? 2nd, when riding an elevator the change of gravity are multiple hundredths of Gs. I once measured the gravity when riding an elevator, the change was about plus/minus 0.08 Gs. So when you start to ride down (or arrive up) in the lift you experience 0.92g while when starting to ride up (or arriving down) you experience 1.08g. I don't know an elevator where the difference would be several cavendishes. Third, you mention the mountains' mass...  
> **username_0**: ...and this gravity difference is called a mascon (mass concentration) which also seems to be more than several cavendishes. Here's a map showing the differences in mascons: https://thumbs-prod.si-cdn.com/70gjnV4C2bY6VUgVj2OHyVLvGkg=/800x600/filters:no_upscale()/https://public-media.si-cdn.com/filer/20110520102356GOCE-Geoidkarte_mit_Schatten_H.jpg  
> **username_0**: And your mention of a decreased gravity concerning Bennu is pointless because I'm asking on the gravity of _Bennu_ itself, not how the _Earth's_ local gravity changes.  
> **username_4**: Tides are caused by both the Moon and the Sun.  But the Lunar force is about 3 times stronger.  I said "about the same" whether 3 times as much qualifies, I guess depends on ones flexibility on the use of the word about.   And I agree, Mascons probably have gravity variations of several Cavendishes, though I've not checked.  Equator to Pole the difference is well into the tens of thousands.  
> **username_0**: The equatorial surface gravity is 9.78 m/s² (0.997g) while the polar is 9.832 m/s² (1.003g), so the difference between pole and equator is plus/minus 0.003g in reference to average.  

---

### Answer by username_5 on 2020-12-02. Score: 0

That would be very difficult to do. The meteor will pulverize when striking earth and it would be difficult to determine the original size of the meteor. Estimates of the initial mass can maybe be determined from the size of the crater but I doubt the accuracy of this approach. Fragments of the meteorite can be collected and used for the experiments you mentioned, but so could earth rocks!

> **Comments:**
> **username_0**: My question is either about meteoroids that make it down to Earth's surface, or about the fragments you mention. Bennu's surface gravity is 6 micro-g. Would that gravity remain if Bennu impacted the Earth, and if so, would its attraction be noticeable to everyone nearby?  

---

