# Eros and the Moon
**URL:** https://boards.straightdope.com/t/eros-and-the-moon/39495
**Category:** Cecil's Columns/Staff Reports
**Created:** [November 3, 2000, 2:05am UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495 "2000-11-03T02:05:22Z")
**Posts on this page:** 20
**Page:** 1
### Author:  [@DrPorkchop](https://boards.straightdope.com/u/DrPorkchop)
#### Post date: [November 3, 2000, 2:05am UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/1 "2000-11-03T02:05:22Z")
In reference to the “[What Happens if I Blow Up the Moon?](http://www.straightdope.com/columns/001027.html)” column, a friend of mine who is too shy (translates “too lazy”) to post himself asks, “If our ‘moon destroying’ friend were able to alter the course of Eros so it would collide with the moon,
would that impact be sufficient to destroy the moon?”
Any takers??
---
### Author:  [@Arnold\_Winkelried](https://boards.straightdope.com/u/Arnold_Winkelried)
#### Post date: [November 3, 2000, 5:51pm UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/2 "2000-11-03T17:51:08Z")
I’m no astronomer, and physics was never my strong suit, but I can provide more information that might help spark the discussion. Here is some information I found out (sources listed below)
**Eros**
elongated body about 33 x 13 x 13 km
density : 2.4 grams per cubic centimeter – about the same as the density of Earth’s crust.
shape: potato
mineral composition: particularly pyroxene and olivine, iron-bearing minerals commonly found in meteorites and in the Earth’s crust. It is a monolithic rock, not a “rubble pile” like other asteroids.
rotation period: 5 hours 17 minutes
period of revolution: 642 days.
closest approach to earth: 22 million km
**Earth’s Moon**
diameter at the equator : 3,476 km
density : 3.34 grams per cubic centimetre (earth: 5.52)
composition: crust (rich in plagioclase), mantle (denser materials such as pyroxene and olivine), asthenosphere (which I think is a weak or molten layer of the mantle), possible metallic core
E.B. also says that the most popular theory for the formation of the Moon is the giant impact hypothesis: In this scenario, set more than four billion years ago, the early Earth is struck a glancing blow by a body the size of Mars. The ejecta coalesce to form the moon.
**My conclusion**
If a collision with a body the size of Mars is insufficient to destory the Earth, a collision by such a small body as Eros with the Moon would seem to me to be insufficient to destroy the moon. Especially since Eros is less dense than the moon. Of course, it would also depend on how fast Eros is going.
Sources:
[Encylopædia Britannica article on Eros](http://www.britannica.com/bcom/eb/article/3/0,5716,33513+1+32939,00.html)
[Science @ NASA: Highlights from Asteroid Eros](http://www.spacescience.com/headlines/y2000/ast19feb_1.htm)
[Encylopædia Britannica article on Earth’s satelllite](http://www.britannica.com/bcom/eb/article/8/0,5716,118798+1,00.html)
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### Author:  [@Fiver](https://boards.straightdope.com/u/Fiver)
#### Post date: [November 3, 2000, 8:33pm UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/3 "2000-11-03T20:33:31Z")
**Arnold Winkelried** :
> [@](#):
>
> E.B. also says that the most popular theory for the formation of the Moon is the giant impact hypothesis: In this scenario, set more than four billion years ago, the early Earth is struck a _glancing blow_ by a body the size of Mars. The ejecta coalesce to form the moon.
(italics mine)
I’ve never heard it described as a “glancing blow” before. My understanding has always been that the Mars-sized planetesimal was absorbed by the infant Earth. That’s definitely the implication in Neil Comins’s book What If the Moon Didn’t Exist?
Is that not correct? Sources?
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### Author:  [@Irishman](https://boards.straightdope.com/u/Irishman)
#### Post date: [November 3, 2000, 9:37pm UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/4 "2000-11-03T21:37:14Z")
Here is some interesting information along these lines.
Cool site to calculate impact energy of various asteroid sizes and types.
Asteroid Impact Calculator
[http://www.badastronomy.com/bitesize/kaboom.html](http://www.badastronomy.com/bitesize/kaboom.html)
Description of the biggest crater in the Solar System.
Biggest Crater
[http://www.badastronomy.com/bitesize/big\_crater.html](http://www.badastronomy.com/bitesize/big_crater.html)
Excerpt from a review of the TV movie “Doomsday Rock”.
Doomsday Rock
[http://www.badastronomy.com/bad/tv/doomsday.html](http://www.badastronomy.com/bad/tv/doomsday.html)
> [@](#):
>
> [The asteroid itself] is about 10 by 5 by 3 kilometers in size, made almost entirely of iron and moving in at 50,000 miles per hour… That volume, given the density of iron, will have a mass of 10^18 grams, or about ten trillion tons (that’s ten million million tons)… Moving at 50,000 mph (or about 2 million centimeters per second) means it has a kinetic energy of 5 x 10^30 ergs… However, it is nowhere near enough to actually shatter the Earth. Actually, it’s only enough to leave a big crater, but even that crater would only be a few kilometers deep. Using gravitational arguments, it can be shown that it would take about 10^40 ergs to shatter the Earth.
Given that information, I think Eros would be far from enough to shatter, pulverize, or “vaporize” the moon. It will make a nice thump.
---
### Author:  [@Arnold\_Winkelried](https://boards.straightdope.com/u/Arnold_Winkelried)
#### Post date: [November 3, 2000, 10:09pm UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/5 "2000-11-03T22:09:15Z")
**Five** , I don’t know what to say to you except that Encyclopædia Britannica definitely uses the terms “glancing blow”.
See here: [Moon: Internal activity of the past and present / Origin and evolution](http://www.britannica.com/bcom/eb/article/8/0,5716,118798+10+110148,00.html)
It’s too bad no one caught this cosmic event on film. What are all those time-travellers thinking?
Irishman, I don’t suppose you know (off the top of your head) how to calculate the volume of a potato-shaped object? 😉
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### Author:  [@jrepka](https://boards.straightdope.com/u/jrepka)
#### Post date: [November 3, 2000, 11:04pm UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/6 "2000-11-03T23:04:37Z")
> [@](#):
>
> _I’ve never heard it described as a “glancing blow” before. My understanding has always been that the Mars-sized planetesimal was absorbed by the infant Earth._
“Glancing blow” is an unfortunate choice of words by EB. Most of the impact models I’ve seen require the Mars-size planetesimal to collide with the proto-Earth at a low angle. Apparently if it hits head-on the resultant impact doesn’t create the proper scattering of debris to form the Moon. “Glancing blow” implies an impactor hitting at a very low angle and ricocheting off into space (to me, at least).
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### Author:  [@jrepka](https://boards.straightdope.com/u/jrepka)
#### Post date: [November 3, 2000, 11:24pm UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/7 "2000-11-03T23:24:21Z")
> [@](#):
>
> _…I don’t suppose you know (off the top of your head) how to calculate the volume of a potato-shaped object?_
Treat it as a cylinder with a diameter of 13 km: V=pi\*(D/2)^2\*L ~ 4400 km^3, or 4.4E+12 m^3. With a density of 2.4 tonnes/m^3, we get a mass of ~10^13 tonnes (10^19 grams).
In the vicinity of Earth, 30 km/s (3E+06 cm) is a pretty good velocity assumption. KE=1/2\*MV^2, so the KE of Eros as it approaches Earth is ~4.5-5E+31 ergs.
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### Author:  [@Loren\_Pechtel](https://boards.straightdope.com/u/Loren_Pechtel)
#### Post date: [November 4, 2000, 1:51am UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/8 "2000-11-04T01:51:21Z")
> [@](#):
>
> _Originally posted by jrepka \*
> \*Treat it as a cylinder with a diameter of 13 km: V=pi(D/2)^2_L ~ 4400 km^3, or 4.4E+12 m^3. With a density of 2.4 tonnes/m^3, we get a mass of ~10^13 tonnes (10^19 grams).
>
> In the vicinity of Earth, 30 km/s (3E+06 cm) is a pretty good velocity assumption. KE=1/2\*MV^2, so the KE of Eros as it approaches Earth is ~4.5-5E+31 ergs. \*\*
Two points:
```
First, there's a lot of mass lost due to the fact it's round, not a cylinder. A sphere loses about 1/2 vs a cube, since you've already rounded in two dimensions I would guess about 1/6. A fairly small adjustment overall.
The big problem is with the energy you give it. Yes, 30km/sec is reasonable for something near the Earth. However, the target also has that velocity. The true impact velocity is going to be *MUCH* smaller, probably no more than 5km/sec + lunar escape velocity. Better knock a zero off that impact energy.
```
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### Author:  [@jab1](https://boards.straightdope.com/u/jab1)
#### Post date: [November 4, 2000, 5:50pm UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/9 "2000-11-04T17:50:50Z")
When I saw the title, I thought it was about the Moon’s effect on our love lives.
I am very glad no one has said that without a moon, women’s menstrual cycles would be affected.
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### Author:  [@jrepka](https://boards.straightdope.com/u/jrepka)
#### Post date: [November 5, 2000, 1:08am UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/10 "2000-11-05T01:08:37Z")
> [@](#):
>
> \*The big problem is with the energy you give it. Yes, 30km/sec is reasonable for something near the Earth. However, the target also has that velocity. The true impact velocity is going to be _MUCH_ smaller, probably no more than 5km/sec + lunar escape velocity. Better knock a zero off that impact energy. \*
You may be right; but I believe that meteorites striking the Earth approach with velocities _averaging_ 30km/sec. Some objects have higher velocities 'cus they hit us head-on (they approach from the morning side – orbital velocity + escape velocity), while others have lower velocities since they approach from the night side (orbital velocity - escape velocity)…
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### Author:  [@Chronos](https://boards.straightdope.com/u/Chronos)
#### Post date: [November 5, 2000, 6:56pm UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/11 "2000-11-05T18:56:19Z")
Assuming that all orbital directions are random, and all speeds are 30 km/s, then the average impact speed would be about 42 km/s. However, the orbital directions are not random: Most bodies in the solar system are orbiting generally in the same direction, counterclockwise as seen from the North.
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### Author:  [@dtilque](https://boards.straightdope.com/u/dtilque)
#### Post date: [November 5, 2000, 10:47pm UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/12 "2000-11-05T22:47:58Z")
> [@](#):
>
> \*Originally posted by Chronos \*
> However, the orbital directions are not random: Most bodies in the solar system are orbiting generally in the same direction, counterclockwise as seen from the North.
And to be more specific, the only bodies orbiting the sun that don’t go around this way are some of the long period comets. But even one of those hitting the moon probably wouldn’t do more than give us a lovely new crater to look at.
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### Author:  [@John\_Kentzel-Griffin](https://boards.straightdope.com/u/John_Kentzel-Griffin)
#### Post date: [November 5, 2000, 10:58pm UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/13 "2000-11-05T22:58:52Z")
> [@](#):
>
> \*Originally posted by \*\*\*Arnold Winkelried \*\*
> Irishman, I don’t suppose you know (off the top of your head) how to calculate the volume of a potato-shaped object? 😉
Easy:
1. Fill a large bathtub with water up to the top.
2. Place potato-shaped object in tub. Be sure to submerse completely.
3. Measure volume of water displaced.
4. Run through the streets naked yelling “Eureka!”
😃
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### Author:  [@whitetho](https://boards.straightdope.com/u/whitetho)
#### Post date: [November 5, 2000, 11:10pm UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/14 "2000-11-05T23:10:42Z")
Today’s _Parade Magazine_ (a supplement that appears in many newspapers) has an article entitled “Why We Have A Moon”, by David H. Levy. Levy endorses a 1974 theory by William K. Hartmann and Donald R. Davis that about 4.5 million years ago a Mars-sized planet on a highly elliptical orbit smashed into the Earth to create the moon. According to Levy, the planet “sideswiped the Earth, bounced off and seconds later, tore right back into the Earth”.
Remnants of the colliding planet plus parts of the Earth’s crust created two rings of debris, which in about a year consolidated into the moon. Levy notes the moon originally was about 1/24th of the distance it is now, as its orbit has over the years slowly moved out from Earth, currently at the rate of about a meter every century.
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### Author:  [@Arnold\_Winkelried](https://boards.straightdope.com/u/Arnold_Winkelried)
#### Post date: [November 6, 2000, 5:04pm UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/15 "2000-11-06T17:04:58Z")
> [@](#):
>
> \*Originally posted by DrMatrix \*
> Easy:
>
> 1. Fill a large bathtub with water up to the top.
> 2. Place potato-shaped object in tub. Be sure to submerse completely.
> 3. Measure volume of water displaced.
> 4. Run through the streets naked yelling “Eureka!”
Dr. Matrix, how about if I skip steps 1 to 3 and go directly to step 4? That will save me a lot of time and trouble.
---
### Author:  [@Arnold\_Winkelried](https://boards.straightdope.com/u/Arnold_Winkelried)
#### Post date: [November 6, 2000, 5:30pm UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/16 "2000-11-06T17:30:41Z")
> [@](#):
>
> _Originally posted by jrepka_
> Treat it as a cylinder with a diameter of 13 km: V=pi\*(D/2)^2\*L ~ 4400 km^3, or 4.4E+12 m^3. With a density of 2.4 tonnes/m^3, we get a mass of ~10^13 tonnes (10^19 grams).
When I tried the calculations I came up with a different number (a factor of 10 smaller). My calculations:
length of Eros = 33 km, diameter = 13
V = pi \* (6.5)^2 \* 33 = 4,380 km^3 = 4,380 \* 10^ **15** cm^3
mass = V \* density = 4,380 \* 10^15 cm^3 \* 2.4 g\*cm^-3 = 10,512 \* 10^15 g = 10.512 \* 10^12 metric tonnes.
> [@](#):
>
> _Originally posted by Loren Pechtel_
> First, there’s a lot of mass lost due to the fact it’s round, not a cylinder. A sphere loses about 1/2 vs a cube, since you’ve already rounded in two dimensions I would guess about 1/6. A fairly small adjustment overall.
I don’t understand this statement. If we are calculating the volume of Eros, from the description and dimensions I’ve read, it sounds like it would be more similar to a cylinder than a sphere (potato-shaped object 13 \* 13 \* 33 km).
Anyway, I would like to point out that **The Bad Astronomer** posted in [this thread](http://boards.straightdope.com/sdmb/showthread.php?threadid=44050) the following information:
> [@](#):
>
> Using numbers for the Moon (radius = 1738 km, mass=7.35x10^25 grams) I find that it takes about 30 trillion megatons (3x10^13) to blow the Moon up. Considering the largest bomb ever claimed to have been exploded was less than 100 megatons, we’re a bit short of the mark.
>
> A collision with Eros could do it, but Eros would have to be moving incredibly fast, about 5000 kilometers/second, to give it enough energy to wipe out the Moon. That’s a good fraction of the speed of light.
---
### Author:  [@Irishman](https://boards.straightdope.com/u/Irishman)
#### Post date: [November 6, 2000, 10:41pm UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/17 "2000-11-06T22:41:51Z")
Arnold, calculating it would be difficult, not knowing exactly the shapes of the curves on the ends, the elbow in the middle, the craters, etc. I could _approximate_ it like jrepka did, by assuming a cylinder. Of course you have to realize that’s probably on the high side. As Loren points out, a cylinder has straight wall sides and circular edges at the ends. A potato is more like a sphere cut in half and extended with a cylinder between the two halves. You have to round the ends as well as make the cross section circular.
jab1, I’m with you. Got a friend who brought up the moon affecting her mood the other night. I didn’t bother arguing - regardless of whether true or not, the mood was not conducive.
Source of Bad Astronomer quote:
[http://www.badastronomy.com/wwwboard/messages/327.html](http://www.badastronomy.com/wwwboard/messages/327.html) .
---
### Author:  [@jrepka](https://boards.straightdope.com/u/jrepka)
#### Post date: [November 7, 2000, 1:50am UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/18 "2000-11-07T01:50:27Z")
> [@](#):
>
> _Originally posted by Arnold Winkelried \*
> My calculations:
> length of Eros = 33 km, diameter = 13
> V = pi \* (6.5)^2 \* 33 = 4,380 km^3 = 4,380 \* 10^ **15** cm^3
> mass = V \* density = 4,380 \* 10^15 cm^3 \* 2.4 g_cm^-3 = 10,512 \* 10^15 g = 10.512 \* 10^12 metric tonnes.
OK, looking over my post i wasn’t as specific as I could’ve been with units, but the result is the same: 10\*10^12 tonnes = 10^13 tonnes.
If we round the edges (treat it as two halves of a 13 km sphere bounding a 20km cylinder) this reduces the volume, thus the mass, by ~13%. (change in volume between a “potato” and a cylinder = (3L - D)/(3L), where L = 33 km and D = 13 km).
---
### Author:  [@Arnold\_Winkelried](https://boards.straightdope.com/u/Arnold_Winkelried)
#### Post date: [November 7, 2000, 5:18pm UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/19 "2000-11-07T17:18:49Z")
> [@](#):
>
> _Originally posted by jrepka \*
> OK, looking over my post i wasn’t as specific as I could’ve been with units, but the result is the same: 10_10^12 tonnes = 10^13 tonnes.
You’re right and I was stupid and didn’t read your answer carefully enough.
Irishman, jrepka, thank you for clarifying for me what Loren Pechtel meant.
---
### Author:  [@wolfman](https://boards.straightdope.com/u/wolfman)
#### Post date: [November 10, 2000, 7:04am UTC](https://boards.straightdope.com/t/eros-and-the-moon/39495/20 "2000-11-10T07:04:56Z")
> [@](#):
>
> \*Originally posted by Arnold Winkelried \*
> \*\*
>
> > [@](#):
> >
> > \*Originally posted by DrMatrix \*
> > Easy:
> >
> > 1. Fill a large bathtub with water up to the top.
> > 2. Place potato-shaped object in tub. Be sure to submerse completely.
> > 3. Measure volume of water displaced.
> > 4. Run through the streets naked yelling “Eureka!”
>
> Dr. Matrix, how about if I skip steps 1 to 3 and go directly to step 4? That will save me a lot of time and trouble. \*\*
If that really would save you a lot of time and trouble then I suggest you consult with Jack Dean Tyler about various shape corective techniques. 🙂
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