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A question on escape velocity
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reply to post by -PLB-
An object in orbit is not at escape velocity. Orbit depends on a specific angular velocity...if there is any radial velocity, it ceases to be an orbit (a positive radial velocity sends the object away from the gravitating body, while a negative radial velocity is an impact trajectory).
An object in orbit is not at escape velocity. Orbit depends on a specific angular velocity...if there is any radial velocity, it ceases to be an orbit (a positive radial velocity sends the object away from the gravitating body, while a negative radial velocity is an impact trajectory).
reply to post by -PLB-
Look at that JUNO trajectory again, without that in-flight maneuver, JUNO would have remained in solar orbit entirely independent of the earth. The in-flight maneuver guided (will guide) JUNO back to a close flyby of earth, for a speed boost via the slingshot effect, since JUNO is catching up to earth it is already at earth escape velocity and wont get recaptured. We know earth is traveling about 66,600 mph around the sun so JUNO is already going a healthy clip as it reproaches earth.
Notice in the animation how slow JUNO is going beyond Mars orbit, and how it gains speed as it approaches the earth, that is partially powered by the sun, and its rocket, approaching earth faster than earth is going, which is already over twice earth escape velocity of around 25,000 mph if it left earth's surface, at 300 miles up, it would be a little less. I believe JUNO left earth escape velocity well under 23,000 mph, when its upper stage Centaur rocket fired.
Look at that JUNO trajectory again, without that in-flight maneuver, JUNO would have remained in solar orbit entirely independent of the earth. The in-flight maneuver guided (will guide) JUNO back to a close flyby of earth, for a speed boost via the slingshot effect, since JUNO is catching up to earth it is already at earth escape velocity and wont get recaptured. We know earth is traveling about 66,600 mph around the sun so JUNO is already going a healthy clip as it reproaches earth.
Notice in the animation how slow JUNO is going beyond Mars orbit, and how it gains speed as it approaches the earth, that is partially powered by the sun, and its rocket, approaching earth faster than earth is going, which is already over twice earth escape velocity of around 25,000 mph if it left earth's surface, at 300 miles up, it would be a little less. I believe JUNO left earth escape velocity well under 23,000 mph, when its upper stage Centaur rocket fired.
Originally posted by Illustronic
Gravity emanates from the center of a mass, and never becomes totally absent, yes gravity travels forever. If there were only 2 objects in the Universe, they would eventually collide, due to gravity, no matter how far away they might be.
If you ignore the expansion of space time that is. If the bodies orbit each other, I guess you can also prevent collision, theoretically.
Originally posted by Illustronic
reply to post by -PLB-
Look at that JUNO trajectory again, without that in-flight maneuver, JUNO would have remained in solar orbit entirely independent of the earth. The in-flight maneuver guided (will guide) JUNO back to a close flyby of earth, for a speed boost via the slingshot effect, since JUNO is catching up to earth it is already at earth escape velocity and wont get recaptured. We know earth is traveling about 66,600 mph around the sun so JUNO is already going a healthy clip as it reproaches earth.
I can't tell if it would have remained in orbit from that video, but I expect it to make larger and larger orbits, moving further and further away from the sun. Unless it crashes into the sun of course.
reply to post by CLPrime
Yes, thats what I am saying. I would expect the object to spiral away from the sun indefinitely. Not sure if that is called escape velocity though.
Yes, thats what I am saying. I would expect the object to spiral away from the sun indefinitely. Not sure if that is called escape velocity though.
reply to post by -PLB-
Yet that's not what I'm saying. At the distance of the Earth, the Sun's escape velocity is 42 km/s. If a projectile is launched from the Earth with an initial velocity of 11.2 km/s (the Earth's surface escape velocity), it will escape the Earth's gravity but will be captured by the Sun's gravity, and, consequently, will be pulled back into the Sun.
Yet that's not what I'm saying. At the distance of the Earth, the Sun's escape velocity is 42 km/s. If a projectile is launched from the Earth with an initial velocity of 11.2 km/s (the Earth's surface escape velocity), it will escape the Earth's gravity but will be captured by the Sun's gravity, and, consequently, will be pulled back into the Sun.
edit on 17-3-2012 by CLPrime because: (no reason given)
reply to post by CLPrime
I think this is partially correct. Since the projectile is already in orbit of the sun it already has a velocity of ~30km/s. So in order to reach escape velocity of the sun it requires only an additional ~10km/s. It seems to me that a stable orbit is the default result when escaping the earth gravity with exactly escape velocity with respect to earth. Whether the objects moves away or towards the sun after that will depend the exact direction. I can't really see a reason why the projectile would always crash into the sun. Yet it does seems to be the case that it requires at least an additional 10km/s acceleration in order to escape the suns gravity. From what I understand, when the increase in velocity is less, the result will be a elliptical orbit.
I think this is partially correct. Since the projectile is already in orbit of the sun it already has a velocity of ~30km/s. So in order to reach escape velocity of the sun it requires only an additional ~10km/s. It seems to me that a stable orbit is the default result when escaping the earth gravity with exactly escape velocity with respect to earth. Whether the objects moves away or towards the sun after that will depend the exact direction. I can't really see a reason why the projectile would always crash into the sun. Yet it does seems to be the case that it requires at least an additional 10km/s acceleration in order to escape the suns gravity. From what I understand, when the increase in velocity is less, the result will be a elliptical orbit.
reply to post by -PLB-
That 30 km/s is orbital angular velocity, perpendicular to escape velocity. The two are non-commutable.
That 30 km/s is orbital angular velocity, perpendicular to escape velocity. The two are non-commutable.
reply to post by CLPrime
I must say I am getting my info from Wiki, so I am not completely certain, but it does make sense to me. Wiki says:
Of course the same principle counts for the sun. Maybe my additional 10km/s figure is not completely accurate, but the required increase in velocity is "far less" than the ~42km/s escape velocity of the sun. And for an regular orbit only ~30km/s is required, which the projectile already has. So I still don't see why it should crash into the sun.
I must say I am getting my info from Wiki, so I am not completely certain, but it does make sense to me. Wiki says:
For an actual escape orbit a spacecraft is first placed in low Earth orbit (160–2,000 km) and then accelerated to the escape velocity at that altitude, which is a little less — about 10.9 km/s. The required change in speed, however, is far less because from a low Earth orbit the spacecraft already has a speed of approximately 8 km/s.
Of course the same principle counts for the sun. Maybe my additional 10km/s figure is not completely accurate, but the required increase in velocity is "far less" than the ~42km/s escape velocity of the sun. And for an regular orbit only ~30km/s is required, which the projectile already has. So I still don't see why it should crash into the sun.
edit on 17-3-2012 by -PLB- because: (no reason given)
reply to post by CLPrime
Let me expand on that:
The angular velocity will add to the projectile's initial velocity if the two are parallell. In that case, with an initial velocity of (11+30 = 41 km/s), the projectile will enter into a parabolic orbit around the Sun - it won't be pulled back in.
The angular velocity will subtract from the initial velocity if the two are anti-parallell. In that case, with an initial velocity of (11-30 = -19 km/s), the projectile will fall into the Sun.
In fact, for any direction other than the first case above, the projectile will fall into the Sun. I was technically wrong when I said (a whole 10 minutes ago) that the two are non-commutable. What I should have said is that there is a single specific case where the addition of the two velocities will add to give the Sun's escape velocity. Other than that, the projectile falls into the Sun.
Let me expand on that:
The angular velocity will add to the projectile's initial velocity if the two are parallell. In that case, with an initial velocity of (11+30 = 41 km/s), the projectile will enter into a parabolic orbit around the Sun - it won't be pulled back in.
The angular velocity will subtract from the initial velocity if the two are anti-parallell. In that case, with an initial velocity of (11-30 = -19 km/s), the projectile will fall into the Sun.
In fact, for any direction other than the first case above, the projectile will fall into the Sun. I was technically wrong when I said (a whole 10 minutes ago) that the two are non-commutable. What I should have said is that there is a single specific case where the addition of the two velocities will add to give the Sun's escape velocity. Other than that, the projectile falls into the Sun.
edit on 17-3-2012 by CLPrime because: (no
reason given)
reply to post by CLPrime
Ok this does compute, and was in fact my initial point. And when you increase the initial velocity of the projectile above the escape velocity of the earth (for example to 15km/s instead of 11km/s) the object will also be capable of escaping the sun, right?
Ok this does compute, and was in fact my initial point. And when you increase the initial velocity of the projectile above the escape velocity of the earth (for example to 15km/s instead of 11km/s) the object will also be capable of escaping the sun, right?
reply to post by -PLB-
Yep. It's all a mater of vector addition... the projectile's Earth-centered initial velocity plus the Earth's angular velocity. If the two are greater than the Sun's escape velocity at 1 AU, the projectile won't be pulled back into the Sun.
Yep. It's all a mater of vector addition... the projectile's Earth-centered initial velocity plus the Earth's angular velocity. If the two are greater than the Sun's escape velocity at 1 AU, the projectile won't be pulled back into the Sun.
reply to post by CLPrime
Ok I think I got it, though I am not sure about "In fact, for any direction other than the first case above, the projectile will fall into the Sun." yet.
To give the 2 extremes: you can move along the radial direction (towards or away from the sun), or along the cross-radial direction (advancing the earth or lagging the earth). It seems to me that two of these 4 situation will result in an orbit: away from the sun and advancing the earth. It seems logical that any direction between these two extremes would also result in an orbit. Or is there only one spot between these extremes where the projectile reaches orbit?
I must say that this isn't as intuitive as I first though it to be.
Ok I think I got it, though I am not sure about "In fact, for any direction other than the first case above, the projectile will fall into the Sun." yet.
To give the 2 extremes: you can move along the radial direction (towards or away from the sun), or along the cross-radial direction (advancing the earth or lagging the earth). It seems to me that two of these 4 situation will result in an orbit: away from the sun and advancing the earth. It seems logical that any direction between these two extremes would also result in an orbit. Or is there only one spot between these extremes where the projectile reaches orbit?
I must say that this isn't as intuitive as I first though it to be.
reply to post by -PLB-
Dealing with an initial velocity exactly equal to the Earth's surface escape velocity, the only case where the projectile will not fall into the Sun is when it is launched in the direction of Earth's orbit, because the sum of the two approximately equal the Sun's escape velocity at this distance.
If you point it directly away from the Sun, then the resultant velocity relative to the Sun is going to be sqrt[11^2 + 30^2] = 32 km/s, which, of course, is less than 42, so it will fall into the Sun.
Any angle between radial (90) and tangential (0) is going to give a velocity somewhere between 32 and 42 km/s. Anything less than 42 sends the projectile into the Sun, and the only case in which we get ~42 is when the projectile is pointed in the direction of Earth's angular velocity.
ETA: and, of course, if you increase the projectile's initial velocity, then you can make it as fast as you want in any direction (except directed at the Sun) in order to keep it from falling into the Sun.
Dealing with an initial velocity exactly equal to the Earth's surface escape velocity, the only case where the projectile will not fall into the Sun is when it is launched in the direction of Earth's orbit, because the sum of the two approximately equal the Sun's escape velocity at this distance.
If you point it directly away from the Sun, then the resultant velocity relative to the Sun is going to be sqrt[11^2 + 30^2] = 32 km/s, which, of course, is less than 42, so it will fall into the Sun.
Any angle between radial (90) and tangential (0) is going to give a velocity somewhere between 32 and 42 km/s. Anything less than 42 sends the projectile into the Sun, and the only case in which we get ~42 is when the projectile is pointed in the direction of Earth's angular velocity.
ETA: and, of course, if you increase the projectile's initial velocity, then you can make it as fast as you want in any direction (except directed at the Sun) in order to keep it from falling into the Sun.
edit on 17-3-2012 by CLPrime because: (no reason given)
Originally posted by CLPrime
Anything less than 42 sends the projectile into the Sun,
But why? If we regard the earth itself as a projectile, going at 30km/s around the sun, wouldn't the earth also be pulled into the sun? Why would the projectile behave differently than the earth itself? Or am I missing something here?
reply to post by -PLB-
The Earth is getting pulled into the Sun. Orbit is free-fall. It just happens to be moving sideways fast enough to keep it's inward fall from ever amounting to anything. Essentially, the Earth is continuously moving out of the way while the Sun is continuously pulling it inward.
An orbit entered into at escape velocity is called a parabolic orbit, and it only happens once. After that, the object heads off into space, coming to a stop at infinite distance.
The Earth is getting pulled into the Sun. Orbit is free-fall. It just happens to be moving sideways fast enough to keep it's inward fall from ever amounting to anything. Essentially, the Earth is continuously moving out of the way while the Sun is continuously pulling it inward.
An orbit entered into at escape velocity is called a parabolic orbit, and it only happens once. After that, the object heads off into space, coming to a stop at infinite distance.
reply to post by -PLB-
Here's what Wiki has to say about this:
From here: Orbital Mechanics - Escape Velocity.
Aaaaannnnnddd you should check out this: Orbits. Especially the 3rd, 4th, and 5th paragraphs.
Here's what Wiki has to say about this:
The escape velocity from the Earth's surface is about 11 km/s, but that is insufficient to send the body an infinite distance because of the gravitational pull of the Sun. To escape the solar system from a location at a distance from the Sun equal to the distance Sun-Earth, but not close to the Earth, requires around 42 km/s velocity, but there will be "part credit" for the Earth's orbital velocity for spacecraft launched from Earth, if their further acceleration (due to the propulsion system) carries them in the same direction as Earth travels in its orbit.
From here: Orbital Mechanics - Escape Velocity.
Aaaaannnnnddd you should check out this: Orbits. Especially the 3rd, 4th, and 5th paragraphs.
edit on 17-3-2012 by CLPrime because: (no reason given)
reply to post by CLPrime
Ok, I took your words too literally. I thought you meant that it is actually send into the sun, as in colliding with the sun. So what you are saying is that a stable orbit is also possible from a certain range of directions? And only in one specific direction (where the vectors are aligned) the projectile will also (approximately) escape the suns orbit?
After giving it some more thought, it seems to me that when an object escapes earths gravity at escape velocity, no matter what direction, the velocity soon approaches 0 with respect to the earth. I understand 0 is only reached in infinity, but I expect low speeds at pretty short distances, as the earths gravitational pull becomes negligible (though calculation is required to confirm).
So when we ignore this low left-over velocity then the object is only left with the 30km/s it already inherited from the earth solar orbit. So In that case indeed only a solar orbit is possible, and escaping or crashing into the sun won't happen without significant additional acceleration (at least not any time soon). And I guess Illustronic was right all along, although he didn't worded it that well (its not pulled into solar orbit, it already was in solar orbit to begin with)
I will give it more thought tomorrow as it still not completely clear. Its interesting stuff, gn.
Ok, I took your words too literally. I thought you meant that it is actually send into the sun, as in colliding with the sun. So what you are saying is that a stable orbit is also possible from a certain range of directions? And only in one specific direction (where the vectors are aligned) the projectile will also (approximately) escape the suns orbit?
After giving it some more thought, it seems to me that when an object escapes earths gravity at escape velocity, no matter what direction, the velocity soon approaches 0 with respect to the earth. I understand 0 is only reached in infinity, but I expect low speeds at pretty short distances, as the earths gravitational pull becomes negligible (though calculation is required to confirm).
So when we ignore this low left-over velocity then the object is only left with the 30km/s it already inherited from the earth solar orbit. So In that case indeed only a solar orbit is possible, and escaping or crashing into the sun won't happen without significant additional acceleration (at least not any time soon). And I guess Illustronic was right all along, although he didn't worded it that well (its not pulled into solar orbit, it already was in solar orbit to begin with)
I will give it more thought tomorrow as it still not completely clear. Its interesting stuff, gn.
reply to post by OZtracized
About 18,000 mph to escape earths atmosphere
Iv'e been through a ,white, rabbit, black, hole and found Nirvana
Cran
About 18,000 mph to escape earths atmosphere
Iv'e been through a ,white, rabbit, black, hole and found Nirvana
Cran
Just wanted to again thank everyone in this thread for keeping education ABOVE any pointless and silly snide comments/remarks. Gave every one of you
stars *mwah!*
This thread has got me thinking all last night, I couldn't sleep! Certainly is fascinating stuff, and its given me the basis for a thread I will hopefully start this week once I've made some pics up.
Keep it up ATS!
This thread has got me thinking all last night, I couldn't sleep! Certainly is fascinating stuff, and its given me the basis for a thread I will hopefully start this week once I've made some pics up.
Keep it up ATS!
Originally posted by CLPrime
reply to post by -PLB-
Here's what Wiki has to say about this:
The escape velocity from the Earth's surface is about 11 km/s, but that is insufficient to send the body an infinite distance because of the gravitational pull of the Sun. To escape the solar system from a location at a distance from the Sun equal to the distance Sun-Earth, but not close to the Earth, requires around 42 km/s velocity, but there will be "part credit" for the Earth's orbital velocity for spacecraft launched from Earth, if their further acceleration (due to the propulsion system) carries them in the same direction as Earth travels in its orbit.
From here: Orbital Mechanics - Escape Velocity.
Aaaaannnnnddd you should check out this: Orbits. Especially the 3rd, 4th, and 5th paragraphs.edit on 17-3-2012 by CLPrime because: (no reason given)
Your post got me thinking of the Voyager work horses.
As of February 08, 2012Important Topic Updates
, Voyager 1 is about 120.06973 astronomical units (1.7962176×1010 km) from the Earth and about 119.70479 astronomical units (1.7907582×1010 km) from the Sun.[24] The magnitude of the Sun from Voyager 1 is −16.4, or the dimmest as seen from any of the five space probes leaving the Solar System. Radio signals traveling at the speed of light between Voyager 1 and Earth take 16.5 hours to cross the distance between the two. (To compare, Proxima Centauri, the closest star to our Sun, is about 4.2 light-years distant or 2.65×105 AU.) Voyager 1's current relative velocity to the sun is 17,060 m/s (61,400 km/h; 38,200 mph). This calculates as 3.599 AU per year, about 10% faster than Voyager 2. At this velocity, 73,600 years would pass before reaching the nearest star, Proxima Centauri, were the spacecraft traveling in the direction of that star. Voyager 1 will need about 14,000 years at its current velocity to travel one light year, therefore 40,000 years will pass before coming anywhere near other stars or planets. Voyager 1 is predicted to enter the interstellar medium between 2012–15, though some scientists say it will be in 2014. Voyager 1 is still the farthest man made object in the universe from Earth.
Voyager 1 is not heading towards any particular star, but in about 40,000 years it will pass within 1.6 light years of the star AC+79 3888, which is at present in the constellation Camelopardalis. That star is generally moving towards our Solar System at about 119 km/s (430,000 km/h; 270,000 mph).[25].
work horse
They are'nt made like that anymore extra DIV
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