Tuesday, May 06, 2008

An Asteroid Beckons


From the UK Guardian May 7: NASA engineers have identified asteroid 2000SG344 as a potential landing site for astronauts. A study to be released next month suggests a 3-6 month mission with 2 weeks spent on the surface. When this 40 meter object was discovered in 2000, it was briefly considered a danger to Earth. 2000SG344's orbit close to Earth makes it a tempting target, one of hundreds of new worlds.

Bode's Law suggests that another planet should orbit between Mars and Jupiter. When Ceres, the largest asteroid, was discovered it was first thought to be the missing planet. In subsequent years hundreds more asteroids have been found. Ceres was long classified as an asteroid but recently has been promoted to minor planet. The DAWN spacecraft will rendezvous with asteroid Vesta in 2011 and Ceres in 2015.

We are fortunate to have found meteorites from Vesta. The Camel Donga meteorite seen below now sits in the Hall of Meteorites and New York's Museum of Natural History. Science fiction writers have long dreamed about mining the asteroids for metal. Some theories suggest that Ceres is largely made of water, and could contain even more water than Earth. More than just big rocks, the asteroids are new worlds that could even harbour life.

Observations suggest that Ceres is differentiated into core and mantle, which would mean that it was melted early in its history. Ultraviolet observations have found water vapour near Ceres' North Pole. How such a small body could be heated is a complete mystery. The asteroid's 10^{21} kg mass could easily hide a small Black Hole.

One can imagine the adventure of landing on this new world. An Orion spacecraft launched on an Ares V booster would carry along a small hab module. Since 2000SG344 very likely rotates, the Orion would have to line up with the asteroid's spin axis and match rotation. This maneuver would be similiar to the Orion docking with a space station in 2001: A SPACE ODYSSEY. Since the asteroid's gravitational field is negligible, the ship would fire pitons to contact the surface. The final touchdown might be accomplished by reeling the spacecraft in.

New presidents like to impose their own Vision, and an asteroid mission would be a lasting legacy. After a return to the Moon, an asteroid mission will be far simpler than reaching Mars. In addition to the adventure of landing on a new world, the mission could easily be justified. Much needs to be learned if we are to prevent another big impact. If we do not venture into the asteroids, one day they will venture into us.

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21 Comments:

Blogger Yoyo said...

Louise wrote: "Ceres was long classified as an asteroid but recently has been promoted to minor planet. (It's bigger than Pluto.)"

Not by a long shot - Pluto's mass is ~10^(22) kg, Ceres' mass is ~9x10^(20) kg.

Perhaps you were thinking of Eris?

8:27 PM  
Blogger L. Riofrio said...

Fixed that in proofreading, thanks. You earlier had a question about metrics, (-1, -1, -1, -1) vs. other metrics. Einstein used this metric to get (ds)^2 = - (dX1)^2 - (dX2)^2 - (dX3)^2 - (dX4)^2 when dX4 = ict. This "imaginary" factor ict allowed him to reach his conclusions about Relativity, but has been abandoned by modern books. You would enjoy Einstein's books, which are surprisingly easy to read.

4:42 AM  
Blogger Yoyo said...

I have no problem with using $x^4 = ict$ as a mathematical convenience to make the metric look Euclidean, but this doesn't change the signature to $(-1, -1, -1, -1)$. The signature is still Lorentzian. The notion of "signature" is only defined for metrics on real vector spaces. It's undefined for complex spaces, since there you can make the "signature" anything you like by inserting factors of $i$ in the basis vectors.

I can show that this metric is Lorentzian when restricted to real vectors by choosing the real basis ${\partial_k, i \partial_4 }$, then $g(\partial_k, \partial_k) = -1$ but $g(i \partial_4, i\partial_4) = +1$.

1:45 PM  
Blogger Yoyo said...

While we're on the subject, I'd still like to know what metric components $g_{ij}$ you used to get the RHS of eq (4) here (the metric you pointed me to here doesn't appear to be the right one, since it doesn't have a $z$ or an $A$ term).

Without knowing what the original $g_{ij}$ was, having $r(t)$ is not very useful. How can I know what role $r(t)$ is supposed to play in the metric? It's not even possible do simple things like evaluate it on a pair of vectors $X$ and $Y$ ie. $g_{ij}X^iY^j$, or raise and lower indices, or calculate a geodesic without knowing what $g_{ij}$ is and where $r(t)$ appears in it.

So, could you please write down the form of the metric that is used to get the RHS of eq (4). What I'm after is something like

$ds^2 = g_{ij} dx^i dx^j$

where the terms $r$, $z$ and $A$ from eq (4) appear in their proper places.

2:07 PM  
Blogger Kea said...

yoyo, in a cosmology that uses an hbar hierarchy, why would you expect classical GR to be applied universally?

2:26 PM  
Blogger Yoyo said...

Kea

I don't expect GR to apply universally. I'm quite happy to contemplate the idea that it's an approximation to the "true" theory of gravity, quantum or otherwise, even with a varying $c$. It's just that in Louise's article below there are lots of terms left undefined, and I'm trying to establish what they are.

My question relates to Louise's derivation eqs (3)-(9) of the Einstein-Friedman equations. This appears to be using totally standard GR, no modifications inserted. What one usually does to derive these is

1) Propose a simple form (ansatz) for the metric $g_{ij}dx^idx^j$ containing some unknown functions (such as $r$).

2) Propose a form for the energy-momentum tensor (Louise uses pressure-free dust, so $T^{ij} = \rho U^iU^j$).

3) Apply the Einstein equations $R_{ij} - 1/2Rg_{ij} = 8\pi GT_{ij}$, yielding an equation for the unknown functions that appear in $g_{ij}^dx^idx^j$.

For the FRW-type metrics, it's usual to start with an ansatz for the most general metric for a homogeneous, isotropic spacetime. Louise doesn't give an ansatz for the $g_{ij} dx^i dy^j$ she's starting from (this undefined metric appears in the LHS of eq (4) here), so I don't see how to obtain the RHS from the LHS, since the RHS contains terms like $r$, $z$ and $A$ which are not defined. Now, when I compare her derivation to "textbook" derivations I can make an educated guess as to what she means by $r$, $z$ and $A$, but rather than putting words in her mouth I'd rather she tell me exactly what $g_{ij} dx^i dy^j$ she started with. Then I can see where $r = ct$ needs to be inserted to get the metric of her proposed solution.

4:30 PM  
Blogger L. Riofrio said...

Thank you again, Kea.

Ansatz?! Since my German isn't that great, consult somebody named Einstein. Equations (4) through (8) come from his book "The Meaning of Relativity" pages 117-118. I added equation (9) myself.

Einstein would have said: those who insist on doing things the way they have been done before will never come up with anything original.

6:35 PM  
Blogger Yoyo said...

This comment has been removed by the author.

11:52 PM  
Blogger Yoyo said...

Thanks for that, I will look it up. Ansatz means something like "educated guess". Google Scholar returns 708,000 results for "ansatz". It's a perfectly cromulent word :)

Louise wrote:
"Einstein would have said: those who insist on doing things the way they have been done before will never come up with anything original."

He might have. Since I can't read the minds of dead people I will respond with a quote from Richard Feynman: "If you're doing an experiment, you should report everything that you think might make it invalid — not only what you think is right about it... Details that could throw doubt on your interpretation must be given, if you know them."

11:54 PM  
Blogger Stephen said...

Ceres was recently promoted to Dwarf Planet. It was demoted to Minor Planet in something like 1850. IMO, it's round, it orbits a star, it does not orbit some other planet, so it's a planet. But, i recognize that i'm not the IAU. I'm not even a memeber. But i don't like the IAU's definition much, and even less the process they used to get it. By the letter of the definition, we have no planets, since none have 'cleared their orbital region' (whatever that means). By the best explanation i've heard, we have one planet, Jupiter, which is the solar system's gravitational bully (other than the Sun). Everything is in orbital resonance with Jupiter. The Earth clearly needs a minor planet number. I suggest zero.

10:00 AM  
Blogger Stephen said...

Apophis briefly seemed to have a one in 45 chance of hitting Earth. 2000SG344 isn't Apophis. The Guardian article isn't clear about this. It's just sloppy.

The Guardian article also doesn't say if the Orion capsule will be radiation shielded, in case a CME comes their way. We had one come to Earth last week, and we're at solar minimum. And, the Sun has been doing practically NOTHING for months. Hardly any sunspots. Hardly any flares. No reason to buy an h-alpha solar scope comes to mind. If we can have CME induced aurora now, expect to encounter one on a 3-4 month long trip. It WILL happen. Even if i were 80, i would not sign up for an unshielded 3-4 month mission outside the Earth's magnetic bubble. I MIGHT sign up to go to the Moon, if after three days, i'd get into a good shelter there. If i were 80. I might get lucky.

10:16 AM  
Blogger Yoyo said...

This comment has been removed by the author.

3:12 AM  
Blogger Yoyo said...

This comment has been removed by the author.

3:15 AM  
Blogger Yoyo said...

Louise

I've looked up your page references for "The Meaning of Relativity" and would like to confirm the metric with you. The coordinates are $(x_1, x_2, x_3, x_4)$ where $x_4 = c t$ (note, not $ict$ this time - Einstein uses a real time coordinate in this section). In the form given in Eq (2) on pg 120

$ds^2 = d{x_4}^2 - G^2(x_4)A^2(r)(d{x_1}^2 + d{x_2}^2 + d{x_3}^2)$

where $A = (1+z r^2/4)^{-1}$ for $r^2 = {x_1}^2 + {x_2}^2 + {x_3}^2$. The $G(x_4)$ corresponds to the $r(t)$ or $R(t)$ you use in your article. I run into a bit of a problem here, because $G$ was stated to be a function of $x_4$ (pg 117 just below eq (2a)), not $t$, whereas you explicitly use $t$ in your formula $R = ct$ and $GM = tc^3$. I think the only sensible way to interpret $R(x_4)$ is to use the fact that it should have the same value regardless of whether the coordinates are $(x_k, x_4)$ or $(x_k, c t)$, so I am assuming that $R(x_4) = x_4$ since $x_4 = ct$ is the way you would interpret that - let me know if this is not what you mean. So, replacing $G$ by $R$ and putting in your condition $z = 0$ which gives $A = 1$, your metric would be:

$ds^2 = d{x_4}^2 - {x_4}^2 (d{x_1}^2 + d{x_2}^2 + d{x_3}^2)$

Is this what you mean? If not, please tell me what it should be.

Another alternative would be to start from

$ds^2 = c^2(t) d{t}^2 - R^2(t) (d{x}^2 + d{y}^2 + d{z}^2)$

and just plug in your formulae directly to get

$ds^2 = (GM)^{2/3}t^{-2/3}d{t}^2 - (GM)^{2/3}t^{4/3}(d{x}^2 + d{y}^2 + d{z}^2)$

Is that right instead?

3:16 AM  
Blogger L. Riofrio said...

For yy: Not what I mean. Note that the G on p. 117-118 is different from the Newtonian gravitaional constant in GM=tc^3. Einstein states elsewhere (like in p. 80 of the same book) that x4 = ict. I recommend reading p. 80 for discussion of the difference between (-1, -1, -1, 1) and (-1, -1, -1, -1) metrics.

9:18 AM  
Blogger Yoyo said...

This comment has been removed by the author.

1:14 PM  
Blogger Yoyo said...

Louise

I realise the $G(x_4)$ on p117 is not Newton's constant. It has exactly the same role in the Einstein-Friedmann equations as your $R(t)$ does, but it is stated as being a function of $x_4$ rather than $t$. I only mentioned $GM = tc^3$ to emphasise the fact that you are using $t$ as a coordinate while Einstein was using $x_4 = ct$ as a coordinate.

I also realise that the coordinate $x_4$ is used to mean $ict$ in most places, but in this section specifically Einstein switches to the real coordinate $x_4 = ct$. You can see this from the metric, since the minus sign is in there. The metric as given on pg120 is

$ds^2 = d{x_4}^2 - G^2(x_4)A^2(r)(d{x_1}^2 + d{x_2}^2 + d{x_3}^2)$

See the minus sign? This metric is given explicitly with signature $(-1,-1,-1,+1)$. If you want to insist that $x_4 = ict$ then fine, but the metric won't have Lorentzian signature and it won't be possible (for example) to have real null (lightlike) vectors.

Ok, so if what I said above is not the correct metric, then what is? Once more, could you please state it in whatever coordinates you prefer?

1:15 PM  
Blogger L. Riofrio said...

Office hour nearly over. Read Einstein and see me next week.

5:37 PM  
Blogger Yoyo said...

This comment has been removed by the author.

8:01 PM  
Blogger Yoyo said...

Louise

On your blog you've repeatedly complained about being ignored by the physics community. Other people have berated your critics like Ethan for not making an effort to understand your work.

I've put many hours of time into trying to follow your derivation and relate it to the references you've given. This is made very difficult by the fact that you use different symbols and coordinates than your sources, and you avoid answering simple questions about what those symbols mean. You've ignored multiple requests to simply state your metric, something which ought to be very straightforward.

I've made a couple of guesses at what your metric is and you tell me they're wrong. OK, for the fourth time, I'm asking you to write down your metric. It's one line of mathematics I'm asking for here. You've expressed a lot of confidence in your theory - put your money where your mouth is.

8:05 PM  
Blogger L. Riofrio said...

Einstein also said that the definition of insanity is trying the same thing repeatedly and expecting a different result.

7:11 PM  

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