Hi
Yes, there's not a simple equation. The table serves - however - as a basic reference of standards values in normal conditions. To know for sure, and accuratly, would involve many calculations and variables.
But it gives a general idea - which is good.
Clear skies and... merry Christmas.
Hi
Yes, you are right. I believe that the table used in Alcyone, that relates the magnitude with the arc is a good guess. For instance, for a body with an apparent magnitude of -1.5, the heliacal rising should happen with an arc of 8.4 degrees. But a body with an app. magnitude of 2.0, would...
Hi
Thank you. But I have that software (actually all the Alcyone software, free and paid - it's great). But it does not allow (I think) asteroids or comets.
Clear skies,
Good afternoon,
I was wondering if someone can point me out a study or perhaps a solution, to know with some assurance if a body with a certain magnitude (m) rising before the sun with a certain angular separation (a) can or cannot been seen with the naked eye.
It's the heliacal phenomenon I...
Ok. Using () in all the input I get:
(0.5/(1.6e-19))*(9.1e-31)
= 2.843750000000000e-12
Usually I don't have to do this. Also, in decimal model (not scientific/engineer) I don't get this error - I get the correct value. I'll download the latest version.
As for the positron, I knew that, but...
Ok - this is a mistery. If I make the math by hand, I get 2.84375e-12 Kg.
BUT. if I make using the calculator SpeedCrunch, I get:
0.5/(1.6e-19)*(9.1e-31)
= 2.843750000000000e19...?
Of course the hand made is right (the first). I don't understand this error in SpeedCrunch. It NEVER happened. :/
Hi
So the charge is 0.5 C. If I divide this by 1.6E10-19 C I get 3.125e18 electrons. Multiplying this by the mass of one electron 9.1E10-31 kg, I get 2.84375e29 Kg. And I must add 1.5 Kg from the ball...
Is this really right?...
Hi
Ok, then the ball seems simple enough to solve. I divide 0.5 by the module of the electron's charge, and multiply by the electron mass and sum it up to 1.5 - is this correct? But I get a value of 2.84375e29 + 1.5 Kg?...
Regarding the electron, I was thinking on that: the mass of the...
Good morning
I'm no expert in classical physics, so I have one doubt. If I have a body with mass M, and I charge it electrically with x C (Coulombs), will its mass remain the same, or will it change (in which amount)? Can you help me in the case of a ball, let's say, with 1.5 Kg charged with...
Greetings
Thanks for the compliments. The confirmation was made by the Azores base, to the local department, given a conic location section I presented. As for what and how, I don't know.
As for the information you said that it was not given, look better to the graphic, and to the published...
Good afternoon,
I forgot to mention the trail left by the plane, rocket or whatever - a jet trail. So it was not superman or a bird.
My apologies for the "ecliptic"- I meant, of course, "elliptic". And no it was not a plane. It was a rocket - I just confirmed.
I was not looking for an ufo...
Hi
I'm aware of that. But my question is simple: can this be a plane? In 2 minutes, it traveled 8.5º in the azimuth, and rised 33º. The innacuracy becomes from the fact that I had less than 2 minutes to figure this measures out.
I just wanted your opinion, that's all.
Clear skies,
Ptolemy
Hi
Thanks. Yes, I understand. The latitude is 41 degrees North, and the longitude 8.5º West.
So the rocket was around north-west, at 9h10m UT in the morning (sun at az. = 99.72º and alt. = 43.14º).
The sunset happened at azimuth = 299.26º.
Clear skies,
Ptolemy
Summary: orbital path of an object
Good morning
I would like to ask the members of this forum their opinion regarding the object, which path I present in the image. The x-axis is the azimuth (towards me, the observer), and the y-axis the altitude. So it goes up in the trajectory shown (from A...