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Old 01-13-2014, 07:05 PM
 
Location: Wasilla, Alaska
17,823 posts, read 23,348,841 times
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Old 01-13-2014, 07:16 PM
 
Location: Sarasota, FL
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Quote:
Originally Posted by geos View Post
I don't understand escape velocity. Why do we need a rocket to travel 25,000 mph to get into space? Why not a plane with wings and rocket engines that gradually climbs higher and higher till it leaves the atmosphere like this: https://www.youtube.com/watch?v=_OMx8Im16Sw

Speaking of gravity, if the earth's gravity is strong enough to hold the moon in orbit why do astronauts feel zero G just a few miles above earth. And I assume gravity doesn't just go from 1 to 0 at some threshold but decreases gradually. So, If I'm in an airplane at 40k' I weight less than at sea level?
The escape velocity (roughly 11.2Km/second, or roughly 25,000 mph) is what we need to get past gravity. Sure, it could gradually climb higher and higher, but it would take the same amount of energy if you did it all at once or in stages. If you throw an apple in the air, it'll fall back to you. Gravity pulls it back. You have the entire mass of the Earth pulling on it, versus the mass of the apple. So in order to combat that gravity, that's the velocity it takes to get into orbit.

But, once your there, there IS still gravity. This is commonly misunderstood. To escape our Milky Way Galaxy, that also requires an escape velocity!

Once your in space, you need to go into orbit. Once you are traveling at 17,500 mph, then your falling around the Earth. Not TO Earth, AROUND the Earth. As gravity pulls you to Earth, the Earth moves and curves around you, and you never land. And you get to fall around the Earth once every hour and a half, its a fast trip!

Gravity decreases the further your away from it. It's called the Inverse Square Law. That's shown pictorially by the image Glitch posted above my post. Feeling gravity is directly proportional to the product of their masses, and inversely proportional to the square of their separation distance. So, in English, that means the further away you are the less it interacts with you. Same thing applies to lights, too. The further you are from a light, the dimmer it appears.
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Old 01-13-2014, 07:53 PM
 
Location: Wasilla, Alaska
17,823 posts, read 23,348,841 times
Reputation: 6541
It should be noted that mass is a measure of how much material is in an object, and weight is a measure of the gravitational force exerted on that material in a gravitational field. Therefore, mass and weight are proportional to each other, with the acceleration due to gravity as the determining factor.
Weight = mg = G(Mm/r^2)
Where
M = Mass of Earth (5,972,000,000,000,000,000,000,000 = 5.972E+24 kilograms)
m = Mass of the object (in kilograms)
r = Distance from the center of the Earth to the center of the object (Earth Radius = 6,371,000 meters)
G = Gravitational Constant (0.00000000006673848 = 6.673848E-11 cubic meters per kilogram per second per second)

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Old 01-15-2014, 11:49 PM
 
Location: Sarasota, FL
1,713 posts, read 2,335,186 times
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a.) 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, ..........

b.) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ........

Take these numbers and continue with their pattern. Here's the question:

Q: If you continue the patterns of these numbers, and take them to infinity, which sequence has the largest amount of numbers in them? a? b?

(PS Glitch dont spoil the fun too early )
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Old 01-16-2014, 01:09 AM
 
Location: Seattle, Washington
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^ Wouldn't they both have an infinite amount of numbers in them?
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Old 01-17-2014, 10:46 PM
 
Location: Wasilla, Alaska
17,823 posts, read 23,348,841 times
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I have a question, but I have to provide you with some information first.

The Spanish recently discovered a binary system consisting of a rapidly rotating (~278 km/s) spectral type B5:ne C star of ~10 M☉ with a companion black hole with 3.8 to 6.9 M☉.


Weirdly Passive Black Hole Discovered Orbiting Fast Spinning Star Animation - YouTube

One of the interesting things about Herbig B(e) type stars is that they are all very young (<10 Myrs). A 10 M☉ star has a life expectancy of ~31.6 million years.

Assuming both stars originally formed from the same stellar nebula, in order for one of those stars to go supernova and become a black hole within 10 million years it would have to have a mass of 16 M☉ or more. The supernova would have spewed out more than 9 M☉ of material, and completely change the barycenter between the two stars. Before the supernova the B5 star would have been in orbit around the more massive (16+ M☉) star, but after the supernova sheds 9+ M☉ the black hole ends up in orbit around the now more massive B5 star.

Why did the resulting supernova that created the black hole not completely destroy its B(e) companion, or at the very least destroy their binary orbit?

Sources:
Spanish researchers discover the first black hole orbiting a
SIMBAD query result
[1201.1726] On the binary nature of the gamma-ray sources AGL J2241+4454 (=MWC 656) and HESS J0632+057 (=MWC 148)
http://www.nature.com/nature/journal...ture12916.html
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Old 01-20-2014, 10:58 PM
 
Location: Wasilla, Alaska
17,823 posts, read 23,348,841 times
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I created a simulation of a binary system with a 16 M☉ star and a 10 M☉ star companion. I put them in perfectly circular orbits around a common barycenter. Then, without changing the orbit, I changed the mass of the 16 M☉ star to a 6.9 M☉ black hole.

The result was the 10 M☉ star companion began pulling away from the black hole immediately, and the now less massive black hole followed the 10 M☉ star. They continued to orbit each other, only in a highly eccentric orbit, and both the 10 M☉ star and the 6.9 M☉ black hole were now moving in completely different spiral trajectory.

So that explains why they are still a binary pair, but it does not explain why the 9+ M☉ of ejecta from the resulting supernovae moving at almost the speed of light did not completely disrupt the 10 M☉ companion star.
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Old 01-25-2014, 03:28 PM
 
Location: Sarasota, FL
1,713 posts, read 2,335,186 times
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Damn, I'm sorry I didnt see your questions Glitch. Somehow I managed to null my subscription to this thread.

I dont know the answer to the question, though, as it's still a "Why is this happening?" question in the scientific community. Nobody really knows..... yet........

On your simulation, why start out in a perfect circular orbit? Most binary star systems have an elliptical orbit around the common center-of-mass. (pic1)

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Old 01-25-2014, 07:51 PM
 
Location: Wasilla, Alaska
17,823 posts, read 23,348,841 times
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Quote:
Originally Posted by beninfl View Post
Damn, I'm sorry I didnt see your questions Glitch. Somehow I managed to null my subscription to this thread.

I dont know the answer to the question, though, as it's still a "Why is this happening?" question in the scientific community. Nobody really knows..... yet........

On your simulation, why start out in a perfect circular orbit? Most binary star systems have an elliptical orbit around the common center-of-mass. (pic1)
The reason I put the two stars in a perfectly circular orbit was because it is the most stable. I had no doubt that in a highly elliptical orbit a sudden change in mass by one star would send both stars in opposite directions, destroying the binary orbit. However, in a perfectly circular orbit the sudden change of mass causes the orbit to change, but both the star and the now black hole remain in orbit about a common barycenter. Only instead of the barycenter being stationary as before the supernovae, it is now moving at a high rate of speed.

So it is possible for a star in a binary orbit to supernovae and still remain in a binary orbit. Which answers the first question. The only question that remains is how could a star survive its companion going supernovae? Large stars in particular have densities of less than 1.5 g/cm^3. I also understand that only about 2% or 3% of the supernovae blast will actually hit the companion star, but we are talking some of the densest material (nickel, iron, gold, platinum, etc.) impacting the star at almost the speed of light. That has to do some serious damage to the companion star, I would think.
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Old 01-25-2014, 08:10 PM
 
Location: Wasilla, Alaska
17,823 posts, read 23,348,841 times
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Here is way out there theoretical physics question. According to M-theory, gravitons are not confined to a single brane, their gravitational effects can be felt in other branes. If gravitons are capable of effecting other branes, then could gravitons from another brane be "bleeding" into our brane and be the cause of "Dark Matter?" Conservely, could gravitons in our brane be the source of "Dark Matter" in other branes?
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