R: Re: Asteroid echos on 2380 MHz

• Subject: R: [amsat-bb] Re: Asteroid echos on 2380 MHz
• From: "i8cvs" <domenico.i8cvs@xxxxxx>
• Date: Sat, 17 Aug 2002 04:54:15 +0200

```Hi  Dan N8FGV,

The attenuation of a signal  sent to the asteroid and reflected back to the
earth,lets call it EAE  Earth-Asteroid-Earth must be computed using the
RADAR equation as normally we do for an EME signal.

Distance Earth-Asteroid = 527000 Km = 527000000  meters

Diameter of Asteroid= 500 meters

Reflectivity coefficient of the Asteroid :we estimate 7% of the incident RF
power as for the surface of the moon but it can be more than 7%
depending on the material by wich the Asteroid is composite but we don't
know this data.

The asteroid as seen as a RADAR target is a disk with a radius of 250 meters
wich surface Sa is:
2
Sa = 250    x  3.14 = 196350  square meters

Suppose now that we are located at the center of a sphere wich radius is the
distance from the earth to the asteroid or R=527000 Km and suppose that we
radiate all around the power P = 1 W using an isotropic antenna for 2380 MHz

The surface S of the above sphere can be computed with:
2
S= 4 x 3.14 x R     and

6   2                  18
S = 4 x 3.14 x ( 527 x 10   )  = 3.49 x 10     square meters

Transmitting isotropically 1 W from the center of the sphere the power
density D received on every square meter of the above sphere is

1                                 - 19
D=-------------------- = 2.86 x 10        W / square meter
18
3.49 x 10

The power density P1 collected by the asteroid surface as transmitted by
the earth using an isotropic antenna with the power of 1 W is than given
by D x Sa

- 19                                    -14
P1= 2.86 x 10       x  196350 = 5.62 x 10         watt

The reflectivity capability of the asteroid is 7% of the above power and so
the isotropic radiated power echo P2 from the asteroid is:

-15
P2= 3.9 x 10        watt

The above P2 power is now reflected back "isotropically" all around by the
asteroid and is totally collected again by the surface S of the sphere and
so the power density P3 collected  on  every square meter on the earth
surface is P2/S or:

-15
3.9 x 10                           -33
P3= ----------------- = 1.12 x 10       W/square meter
18
3.49 x 10

Now it is necessary to know how much of the P3 power is collected by an
isotropic antenna for 2380 MHz on the earth.

The aperture area A of an isotropic antenna  can be computed using the
following formula:
2
A= lambda    / 4 x 3.14  and for 2380 MHz  A= 0.00126 square meters

The power Pr received by the isotropic antenna and available on its
connector is than P3 x A or

-33                                     -36
Pr =  1.12 x 10       x  0.00126 = 1.41 x 10      watt

The isotropic path attenuation  Earth-Asteroid-Earth is than computed
dividing the power P=1 W isotropically radiated from the earth by the power
Pr received back from the asteroid  on the earth using the same isotropic
antenna.

1
Isotropic attenuation = 10 log       ------------------ = 358,5 dB
10                   -36
1.41 x 10

Following the ARRL Newington CT,informations on day 15, 2002 the Arecibo
Radiotelescope will transmit as well CW using a power of 900 Kw to its 73
dBi gain antenna so that having already computed the EAE attenuation of
358,5 dB at 2380 MHz we can now compute the signal to noise ratio S/N at
Arecibo considering that their receiver equivalent noise temperature is 4 K
and the cold sky is 3 K and that they receive on CW with a band width of 500
Hz as normally we do on EME

Under this conditions the Arecibo receiver noise floor Pn with the antenna
connected is:

-23                                            -20
Pn  = 1.38 x 10      x ( 4 + 3 ) x 500 = 4.83 x 10      watt = - 193.16 dBW
= - 163 .16 dBm

Arecibo power = 900 KW = ..................+ 59.54 dBW
Arecibo antenna gain =..........................+ 73.00 dBi
--------

Arecibo EIRP............................................+132.54 dBW
Isotropic attenuation AEA.................... - 358.5 0 dB
---------

Power received at earth on isotropic
antenna ......................................................- 225.96 dBW

Arecibo antenna gain............................... +  73 .00 dBi
---------

Available power at Arecibo antenna      - 152.96 dBW
Arecibo receiver noise floor on CW       - 193.16 dBW
---------
-
Arecibo  (S+N)/N  ratio on CW...............+  40.20 dB

CONCLUSION:

The antenna gain at Arecibo is 73 dBi and a (S+N)/N ratio of 40.2 dB

In order to receive the same signal but with a (S+N)/N = 0 dB  is
necessary to use an antenna with a gain of   73- 40.2 = 32.8 dBi

that assuming to use a state of the art preamplifier with NF= 0,35 dB the
receiver noise floor under the same condition rises to  -186,65 dBW so that
and antenna extra gain of  -186,65 -(-193,16) = 6,51 dB is necessary.

The total necessary antenna gain is than 32,8 + 6,51 = 39.31 dBi

At 2380 MHz a dish of  minimum 5,5 meter in diameter is necessary to achive
such gain but many amateur EME users are using even biger dishes so that the
experiment seams to be succesfully possible.

In addition using software type that of AF9Y is possible to detect CW
signals almost 30 dB belove the noise.

In addition many EME operators are trained to receive CW signal well down
the noise by ears

I hope this help a bit

73" de i8CVS Domenico

----- Original Message -----
From: Dan Schultz <n8fgv@AMSAT.Org>
To: <amsat-bb@AMSAT.Org>
Sent: Friday, August 16, 2002 11:39 PM
Subject: [amsat-bb] Re: Asteroid echos on 2380 MHz

> A good source of information on the asteroid flyby is
> http://skyandtelescope.com. The site includes finder charts.
>
> Sky and Telescope reports the diameter as 500 meters (1/3 mile) and a
> flyby distance of  527,000 kilometers (327,000 miles). Given that the
> moon is 3475 kilometers (2159 miles) diameter at an average distance of
> 384,000 kilometers (239,000 miles), we can calculate as follows:
>
> Size difference:
> (500 meters) ^ 2 / (3475 km) ^ 2 = 48,302,500 times smaller projected
> area than the moon
> 10*LOG (48,302,500) = 76.8 dB worse.
>
> Range difference:
> (527,000 kilometers) ^ 4 / (384,000 kilometers) ^ 4 = 3.55 times less
> signal
> 10*LOG(3.5) = 5.5 dB worse.
>
> Total: 82.3 dB worse.
>
> Using your figures for Arecibo uplink power and antenna gain, the 80 dB
> gain will just about offset the 82.3 dB loss due to the smaller size and
> greater distance of the asteroid. It should be only 2.3 dB worse than a
> 100 watt amateur EME station. However I think most EME stations run more
> than 100 watts, the asteroid reflection would be 12.3 dB worse than an
> amateur EME station running 1 KW. So let's give it a listen and see if
> you can hear it.
>
> Dan Schultz N8FGV
>
>
> >The asteroid is purportedly about 1/2 mile in diameter and passing at a
>
> >distance of 300,000 miles (just beyond the orbit of the moon). So how
> >strong an echo will this tiny rock produce compared to a moonbounce
> signal
> >(off a much bigger rock!)? Of course Arecibo will be running a signal
> 9000
> >times as powerful as is used on 2304 MHz eme (avg 100w), and with 40 dB
>
> >more antenna gain! That's about 80 dB stronger! the moon is about a
> >quarter the size of the earth (guessing 2000 miles diameter) so that is
>
> >(very) roughly (Drock)^2/ (Dmoon)^2 = 1/16,000,000 weaker reflection!
>
> >So whatdya say? Is this possible?
>
> >Ed, AL7EB
>
> ----
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