Article (space.com) on fly-by of Asteroid 2004 MN4 in 2029.

[Spaceman Spiff]

Hi All,

I found via SlashDot this article on a naked-eye visible fly-by of asteroid 2004 MN4 on 13 Apr 2029 (yes, it's a Friday! ). Expected magnitude at brightest: a reasonable 3.3.

http://www.space.com/spacewatch/050204_2004_mn4.html

For the young 'uns, something to look forward to?

Spiff.

[Guest]

They just updated it in the last 48 hours. It was estimated to miss by 58,000 km. Now it's down to 30,000 km. Let's hope they don't do that again!

[Evil Dr Ganymede]

So I wonder when the apocalypse cultists will be starting their hijinx...

[symaski62]

They just updated it in the last 48 hours. It was estimated to miss by 58,000 km. Now it's down to 30,000 km. Let's hope they don't do that again!

?? image ?

[Bob Hegwood]

Interesting...

Any information on how close it will come to the Moon?

Looks like that will be a close call too.

Take care, Bob

[tony873004]

Interesting...

Any information on how close it will come to the Moon?

Looks like that will be a close call too.

Take care, Bob

I didn't realize I wasn't logged in last time. Hi, I'm guest!

It will pass 97,764 km from the Moon's surface about 18 hours after Earth's close approach. January's estimate had it closer to the Moon at 79,319 km.

[dirkpitt]

Dude, it actually cuts through Earth's geosynchronous orbit? If a satellite could take pictures of the asteroid as it zips by, that would be cool.

[andersa]

If a satellite could take pictures of the asteroid as it zips by, that would be cool.

In order to provide pictures better than those that can be obtained from the ground, that satellite ought to be placed in orbit specifically to study the asteroid. I suppose most current geostationary satellites (which should be orbiting the Earth at a geocentric distance slightly above 40,000 km) are used for telecommunications relay, not imaging. There is plenty of time to have something new in orbit until 2029, but perhaps it should have other purposes beyond this asteroid as well. Maybe in 25 years it will be a routine task to place a radio beacon on a passing asteroid, in order to simplify future tracking of it, just as we do with wild animals today?

2004 MN4 is available for ephemeris generation in the JPL Horizons system, but it's hard to tell how accurate that data is. According to it, the asteroid will cross Earth's equatorial plane from south to north at a geocentric distance of 50,000 km, almost two hours before perigeum (can you say so about a passing asteroid?) at 37,000 km and a speed of 7 km/s. What is the risk that it will take out a satellite or two in its way? I saw there is a cloud of geosynchronous/geostationary satellites available from Motherlode, but I haven't tried it out as I don't think it will tell me a lot.

The orbital elements of 2004 MN4 after April 2029 are of course even harder to predict, but JPL Horizons suggests a significant increase in its average solar distance, moving its perihelion well out of reach for Venus (though its aphelion is getting a little closer to Mars; I haven't checked whether it will actually be near Mars' perihelion). Its inclination will decrease from 3.3 to maybe 2 degrees, thereby increasing the risk of future Earth crossings somewhat. JPL Horizons refuses to predict close approaches after 2036-Apr-17, due to the accumulated uncertainties.

[symaski62]

http://neo.jpl.nasa.gov/risk/2004mn4.html

^^

[Spaceman Spiff]

http://neo.jpl.nasa.gov/risk/2004mn4.html

^^

Ah, very handy, symaski62! And a few clicks away, NEA orbital elements (http://neo.jpl.nasa.gov/cgi-bin/neo_elem?type=NEA;hmax=20;sort=name;sdir=ASC;max_rows=20;action=Display%20Table;show=1&from=1685).

From this list, I extract:

Code: Select all

Object 2004
a (AU)       0.92
e            0.191
i (deg)      3.3
w (deg)    126.2
Node (deg) 204.6
M (deg)    274.7
q (AU)       0.746
Q (AU)       1.10
P (yr)       0.89
H (mag)     19.23
MOID (AU)    0.000708
ref   63
class   ATE*


... which doesn't seem like a very precise set of data. Anyway, did we previously have an .scc for it? Otherwise time to make a new .scc upon precise data.

In order to provide pictures better than those that can be obtained from the ground, that satellite [snip]

What I'd want to do with Celestia is see if the asteroid will pass within the field of view of any of the meteorological satellites. The satellites there then will be different, but the geosync. longitudes are reserved per purpose. Piccies have a 17?° wide field to accommodate Earth. Chances of a satellite getting hit? Very small. Only 'dead' ones risk getting hit, 'live' ones will manoeuvre.

But yes, we should sent a spacecraft to intercept and land, a radio beacon would help track future risks much better. Solar System Wide GPS Now! .

So I wonder when the apocalypse cultists will be starting their hijinx...

But they already did with the first announcement!?

Spiff.

[selden]

The NEODyS site has more accurate values. See
http://newton.dm.unipi.it/cgi-bin/neody ... 04MN4;main

[Sky Pilot]

Wouldn't it be soooo cool if we could figure out a way to capture it into a permenant earth orbit so we can study it?

[Spaceman Spiff]

The NEODyS site has more accurate values. See

http://newton.dm.unipi.it/cgi-bin/neodys/neoibo?objects:2004MN4;main

Hi Selden,

after visiting the above, I tried this in an SSC:

Code: Select all

"2004 MN4" "Sol"
{
  EllipticalOrbit
  {
    Epoch 2451544.500  # 2000 Jan 01 00:00UT
    Period 0.88542969954186335186066338099042 # 1 year = 365.2622.
    SemiMajorAxis 0.92209
    Eccentricity 0.191253
    Inclination 3.333
    AscendingNode 204.563
    LongOfPericenter 126.223
    MeanLongitude 274.4
  }
  Radius 0.160
  Albedo 0.15
}


but both Earth and 2004 MN4 are far separated from the orbit intersect point come 13 Apr 2029. That is, not only is 200 MN4 in the wrong place in the orbit, but the orbit itself isn't in the right place. I'm sure specification of orbital elements is fairly standard. I'm scratchin' my head here.

Spiff.

[andersa]

What I'd want to do with Celestia is see if the asteroid will pass within the field of view of any of the meteorological satellites. The satellites there then will be different, but the geosync. longitudes are reserved per purpose. Piccies have a 17?° wide field to accommodate Earth.

Are those satellites fixed in the equatorial plane? If so, then I doubt they will catch 2004 MN4, which will cross the equatorial plane outside the geostationary orbit and then reach 30 degrees of declination (geocentric view). As it will still remain closer to the satellites than to the Earth, the geostationaries will have to bend over backwards to see it, way outside that 17-degree field. Picking satellite 96030B (from Thomas Guilpain's Motherlode submission) for a test, it shows 2004 MN4 at 50 degrees of declination, in Cygnus and Lacerta (using a sampled trajectory for 2029 obtained from JPL Horizons).

However, if you can find some geosynchronous satellites with at least 30 degrees of orbit inclination, then I suppose one of them may stand a chance to catch an image of this asteroid with the Earth as a background.

But yes, we should sent a spacecraft to intercept and land, a radio beacon would help track future risks much better. Solar System Wide GPS Now! .

Do we know the specifications for sufficiently durable radio transmitters today? I guess they should be solar-driven, and they may need some way of securing themselves in upright position on arbitrarily small asteroids. Throw a few of them at each asteroid encountered, for redundancy. We may not see that rock again for 50 years.

Note to self: Enable radio command channel before ejecting unit for non-return mission.

[andersa]

but both Earth and 2004 MN4 are far separated from the orbit intersect point come 13 Apr 2029. That is, not only is 200 MN4 in the wrong place in the orbit, but the orbit itself isn't in the right place. I'm sure specification of orbital elements is fairly standard. I'm scratchin' my head here.

Your elements look pretty close to mine, which I have obtained from JPL Horizons. This is my file (note wrong radius):

Code: Select all

"2004 MN4 (pre-2029)" "Sol"
{
   Class      "asteroid"
   Texture      "asteroid.jpg"
   Mesh      "asteroid.cms"
   Radius      0.240

   Ending   "2030 01 01 00:00"

   EllipticalOrbit {
      Period           0.885618060
      SemiMajorAxis        0.922187699
      Eccentricity        0.191481713
      Inclination        3.345885889
      AscendingNode      203.840602823
      ArgOfPericenter      126.721441081
      MeanAnomaly      238.040230403
      Epoch          2462227.5
   }
}

"2004 MN4 (y2029)" "Sol"
{
   Class      "asteroid"
   Texture      "asteroid.jpg"
   Mesh      "asteroid.cms"
   Radius      0.240

   Beginning      "2029 01 01 00:00"
   SampledOrbit   "2004mn4.xyz"
   Ending      "2030 01 01 00:00"
}

"2004 MN4 (post-2029)" "Sol"
{
   Class      "asteroid"
   Texture      "asteroid.jpg"
   Mesh      "asteroid.cms"
   Radius      0.240

   Beginning   "2029 01 01 00:00"

   EllipticalOrbit {
      Period           1.169087500
      SemiMajorAxis        1.109736033
      Eccentricity        0.191679714
      Inclination        2.161757800
      AscendingNode      203.598345230
      ArgOfPericenter       70.043551397
      MeanAnomaly      314.849145408
      Epoch          2462246.5
   }
}


If I recall correctly, the pre-2029 orbit is defined by the elements on April 1, and the post-2029 one on April 20 (rather than the beginning and the end of the year). At those respective dates, the two orbits align fairly well (within 500 km) with the sampled trajectory for 2029 (I added some overlap merely to do the comparison). I could send you the XYZ file if you want, but I guess you can obtain the same (or better) data from JPL Horizons yourself.

[Spaceman Spiff]

Are those satellites fixed in the equatorial plane?

Yes, almost all are. Especially meteorological GEO ones, which have to have the Earth properly centered as they scan it. Any satellites off the equator tend to be discarded or lost.

If so, then I doubt they will catch 2004 MN4, which will cross the equatorial plane outside the geostationary orbit [sni-i-p]

Just what I wanted to check with celestia, as the minimu approach does not have to be over the Equator. Sadly, I'm having trouble with orbital elements in the SSC file. Even so, I was expecting further inaccuracies because Celestia wouldn't model the perturbation as 2004 MN4 passes Earth... I expect meteorologists will not be happy with a streak of an asteroid in one of their pictures.

Do we know the specifications for sufficiently durable radio transmitters today? I guess they should be solar-driven, [snip]

Better is to shoot the asteroid with a long long harpoon with a radar reflector on the end. The reflector (as on ships and boats) is merely a passive device of three planes of metal forming an octahedral shape, causing three reflections that send the echo straight back to where it came from. No need for power, and the large radar cross-section will give an excellent point spread function allowing auto-focus of ISAR (inverse synthetic radar aperture) imagery.

Note to self: Enable radio command channel before ejecting unit for non-return mission.

No need for that oversight with a passive target, though are you talking Huygens here?

Spiff.

[Spaceman Spiff]

Your elements look pretty close to mine, [8<]

but my Epoch and Mean Anomalies are so different. However, that can only explain the asteroid being in the wrong part of the orbit, not the whole orbit being wrong. I'll try your data in an SSC and see what happens...

Spiff.

[tony873004]

Try this. These values are from Horizons. This is the first time I edited one of these for Celestia, so no guarantees!

I get a much different number for LongOfPericenter than the two of you. see explanation below.

"2004 MN4" "Sol"
{
EllipticalOrbit
{
Epoch 2451544.500 # 2000 Jan 01 00:00UT
Period 0.885791356216703 # 1 year = 365.2622.
SemiMajorAxis 0.922341665613605
Eccentricity 0.1913927843096595
Inclination 3.331234282195888
AscendingNode 204.6565676542412
LongOfPericenter 281.408562843944
MeanLongitude 202.345836245579
}
Radius 0.160
Albedo 0.15
}

Notes:
Argument of Perapsis is not the same as Longitude of Perapsis.
Longitude implies that is measured from the vernal equonix
Argument implies that it is measured from the ascending node.
Longitude of Pericenter = argument of pericenter - longitude of ascending node (add 360 if your value is negative)

I had to translate period and SMA to days and AU
Horizons number for period: PR= 2.795438299789701E+07
Horizons number for SMA: 1.379803492336867E+08

And I had to compute Mean Longitude because Horizons doesn't supply it, but it does supply the numbers needed to compute it:
ml = om + w + ma
mean longitude = Long of Ascending Node + Argument of Perifocus + Mean Anomoly (subtract 360 if value is over 360)

---------------------------------------------------------
2451544.500000000 = A.D. 2000-Jan-01 00:00:00.0000 (CT)
EC= 1.913927843096595E-01 QR= 1.115719060138322E+08 IN= 3.331234282195888E+00
OM= 2.046565676542412E+02 W = 1.260651304981848E+02 Tp= 2451659.876414976548
N = 1.287812362115388E-05 MA= 2.316241380931534E+02 TA= 2.168251672099402E+02
A = 1.379803492336867E+08 AD= 1.643887924535411E+08 PR= 2.795438299789701E+07
$$EOE
*******************************************************************************
Coordinate system description:

Ecliptic and Mean Equinox of Reference Epoch

Reference epoch: J2000.0
xy-plane: plane of the Earth's orbit at the reference epoch
x-axis : out along ascending node of instantaneous plane of the Earth's
orbit and the Earth's mean equator at the reference epoch
z-axis : perpendicular to the xy-plane in the directional (+ or -) sense
of Earth's north pole at the reference epoch.

Symbol meaning

JDCT Epoch Julian Date, Coordinate Time
EC Eccentricity, e
QR Periapsis distance, q (km)
IN Inclination w.r.t xy-plane, i (degrees)
OM Longitude of Ascending Node, OMEGA, (degrees)
W Argument of Perifocus, w (degrees)
Tp Time of periapsis (Julian day number)
N Mean motion, n (degrees/sec)
MA Mean anomaly, M (degrees)
TA True anomaly, nu (degrees)
A Semi-major axis, a (km)
AD Apoapsis distance (km)
PR Orbital period (sec)

[selden]

Don't forget that these Keplerian orbital parameters describe the asteroid's current orbit, not the orbit it will have 24 years from now. Every time the asteroid comes close to the Earth or Venus, its orbital parameters change significantly.

Horizons takes the gravitational effects into account and can provide orbital parameters closer to the times of closest approach, although they aren't yet using the most recent ephemeris.

Code: Select all

"2004 MN4 13Apr29" "Sol"
{
   Mesh "asteroid.cms"
   Texture "asteroid.*"

  EllipticalOrbit
  {
    Epoch     2462239.750000000 # = A.D. 2029-Apr-13 06:00:00.0000 (CT)
    Period          0.884
    SemiMajorAxis   9.213165631331153E-01
    Eccentricity    1.971975371275526E-01
    Inclination     3.451227590422083E+00
    AscendingNode   2.037799255760677E+02
    ArgOfPericenter 1.265176974665853E+02
    MeanAnomaly     2.526098901409764E+02
  }
  Radius 0.160
  Albedo 0.15
}


Using these parameters, Celestia shows the closest approach to be about 38,000 km at 22:12 UT on 13 Apr 2029.

[Spaceman Spiff]

OK, I tried andersa's data, and it works. After tony873004's posting (thanks!) I see that I placed an Arg of Periwotsit into a Long of Periwotsit. Celestia lets you specify whether one or other with keywords 'LongOfPericenter' or 'ArgOfPericenter'. andersa got it right.

After all that, the pre and post asteroids pass many 10,000km apart, because the orbit is so stronlgy perturbed by Earth, so the XYZ is rather vital. Thanks for the further data selden. I think the approach by andersa with two time-limited Keplerian orbits tied by an xyz segment might be ideal. How about an uplink to the Motherlode?

Also, andersa, your radius may not be so wrong, I noted in passing that the NEA site estimate of 320m diameter is a guestimate that could be out be a factor of two. Choose your pick!

Spiff.