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| | :: OTOH, from the table above i'm thinking that the 5.4 might be the Venus figure, and it was wrongly placed besides Earth... | | :: OTOH, from the table above i'm thinking that the 5.4 might be the Venus figure, and it was wrongly placed besides Earth... |
| | :: Secondly, what i found interesting was that the Earth's 6.4 looks so much like its radius! I wonder if it's merely a coincidence, or there's a connection between the two... -- [[Special:Contributions/141.101.99.233|141.101.99.233]] 21:25, 30 October 2013 (UTC) | | :: Secondly, what i found interesting was that the Earth's 6.4 looks so much like its radius! I wonder if it's merely a coincidence, or there's a connection between the two... -- [[Special:Contributions/141.101.99.233|141.101.99.233]] 21:25, 30 October 2013 (UTC) |
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| | :::The table is great, it must be included in the article; layout and time is just my problem right now. PRO TIP: Do not care about the x-axis.--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 22:18, 30 October 2013 (UTC) | | :::The table is great, it must be included in the article; layout and time is just my problem right now. PRO TIP: Do not care about the x-axis.--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 22:18, 30 October 2013 (UTC) |
| − | : The fact that the density of the Earth is 5478 kilograms per cubic kilometer makes me pretty sure it is a typo. [[User:Fewmet|Fewmet]] ([[User talk:Fewmet|talk]]) 03:04, 4 July 2014 (UTC)
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| − | :: Hehe, you might be right. That's the best explanation. It would be a strange coincidence otherwise. But your units are wrong: a cubic kilometer of water, ice-cream or Natalie Portmans would be already something like a billion kilograms. Or a trillion, if you are American. Oh, you might be American. In this case: happy 4th of July! -[[Special:Contributions/188.114.102.35|188.114.102.35]] 12:39, 4 July 2014 (UTC)
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| − | :::Thanks for catching that (and for the July 4 wishes). It should be kilograms per cubic meter. Looking into that, though, leaves me less sure that is the origin of the problem. I thought I had multiple sources for Earth having a density of 5478 kg/m3, but can find only [//atharvatutorials.com/doc/physics_paper.docx one] (and not a very compelling one at that). I have sounder sources for [//www.universetoday.com/26771/density-of-the-earth/ 5513 kg/m3]. [//nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html 5514 kg/m3], [//www.wolframalpha.com 5515 kg/m3], [//www.esa.int/Our_Activities/Space_Science/Venus_Express/Venus_compared_to_Earth 5520 kg/m3] and [//principles.ou.edu/earth_planet/ 5540 kg/m3]. It may be trivial in that all round to 5500 kg/m3.
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| − | ::It was corrected on the poster version. Earth's well in the main graphic is marked as 6379km, just like the inset.[[Special:Contributions/108.162.216.86|108.162.216.86]] 00:19, 21 August 2014 (UTC)
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| − | :::I solved for the wells on Earth, Moon and Mars using the equation Randall gave and masses and equatorial radii from NASA, getting 6371 km, 287 km and 1286 km, respectively. [[User:Fewmet|Fewmet]] ([[User talk:Fewmet|talk]]) 23:07, 5 July 2014 (UTC)
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| | The Oberth Effect mentioned in the title text is [//www.askamathematician.com/2013/01/q-how-does-the-oberth-effect-work-and-where-does-the-extra-energy-come-from-why-is-it-better-for-a-rocket-to-fire-at-the-lowest-point-in-its-orbit/ well-explained here] (assuming you are not intimidated by the algebra in squaring a binomial). The gist of it is that using a bit of fuel in a rocket thrust will increase the rocket’s kinetic energy . The higher the kinetic energy at the time of the thrust, the greater the increase in kinetic energy. It works because the energy of the fuel goes into increasing the kinetic energy of the ship and the kinetic energy of the spent fuel. The faster you go, the greater the portion of the energy the ship gets. | | The Oberth Effect mentioned in the title text is [//www.askamathematician.com/2013/01/q-how-does-the-oberth-effect-work-and-where-does-the-extra-energy-come-from-why-is-it-better-for-a-rocket-to-fire-at-the-lowest-point-in-its-orbit/ well-explained here] (assuming you are not intimidated by the algebra in squaring a binomial). The gist of it is that using a bit of fuel in a rocket thrust will increase the rocket’s kinetic energy . The higher the kinetic energy at the time of the thrust, the greater the increase in kinetic energy. It works because the energy of the fuel goes into increasing the kinetic energy of the ship and the kinetic energy of the spent fuel. The faster you go, the greater the portion of the energy the ship gets. |
| − | :The image at the top is out of date and should be fixed as well as the earth's comment about how it's different--[[Special:Contributions/172.70.178.107|172.70.178.107]] 10:52, 7 September 2022 (UTC)
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| − | ::I, for one, don t understand at all what you think is out-of-date/different. There's a new planet, or some rescaling of gravity? If serious, do elaborate. (If not serious, elaborate to add to the humour.) [[Special:Contributions/141.101.99.154|141.101.99.154]] 12:27, 7 September 2022 (UTC)
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| − | The “gravity assist” is also known as the slingshot effect. The [//en.wikipedia.org/wiki/Gravity_assist#Explanation Wikipedia explanation] is good, especially with its diagram. In it a spaceship (or other body) accelerates toward a planet (or moon, star, etc.) in the same direction that body was going. The ship picks up a little of the body’s momentum and so goes faster, although only according to an external reference frame. An observer at rest with respect to that other body would actually see the ship approach and depart with the same speed. | + | The “gravity assist” is also known as the slingshot effect. The [//en.wikipedia.org/wiki/Gravity_assist#Explanation Wikipedia explanation] is good, especially with its diagram. In is a spaceship (or other body) accelerates toward a planet (or moon, star, etc.) in the same direction that body was going. The ship picks up a little of the body’s momentum and so goes faster, although only according to an external reference frame. An observer at rest with respect to that other body would actually see the ship approach and depart with the same speed. |
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| | The title text reference to orbital speed is unclear to me. I suppose it just means that the given gravity wells assume you are at rest on the surface of the planet. Then being in orbit (and necessarily having an orbital speed) would mean you are part way out of the well already. [[User:Fewmet|Fewmet]] ([[User talk:Fewmet|talk]]) 02:57, 4 July 2014 (UTC) | | The title text reference to orbital speed is unclear to me. I suppose it just means that the given gravity wells assume you are at rest on the surface of the planet. Then being in orbit (and necessarily having an orbital speed) would mean you are part way out of the well already. [[User:Fewmet|Fewmet]] ([[User talk:Fewmet|talk]]) 02:57, 4 July 2014 (UTC) |
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| − | If the first stage of a rocket is still supplying lift for a while after its fuel is used up and the stage is cut adrift, would there be any saving in waiting for the next phase to cut in when forward motion is almost ended rather than continuing the burn immediately from the second stage?
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| − | The higher the vehicle gets the more productive the fuel becomes.Or is it preferable to continue the journey as fast as possible? {{unsigned|Weatherlawyer}}
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| − | My first instinct would be to say burn as continuously as possible. If you wait until your speed is almost zero, you have to use a whole load of energy (fuel) to get back to the speed you were going in the first place. --[[User:Pudder|Pudder]] ([[User talk:Pudder|talk]]) 17:12, 27 January 2015 (UTC)
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| − | :Hence the need to use ultra light containers in the first stage?{{unsigned|Weatherlawyer}}
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| − | Maybe the typo is based on Randall's days at NASA? It might already incorporate gravity assists and the Oberth effect. That number might even be what NASA was using as the minimum potential with known cost-effective techniques. [[User:Flewk|flewk]] ([[User talk:Flewk|talk]]) 21:22, 8 January 2016 (UTC)
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| − | == Earth's Geosynchronous Orbital Altitude ==
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| − | In the XKCD strip, the artist states above Earth in the lower right popout that the geosynchronous altitude is well below top of Earth's gravity well. While the rest of his strip is a wonderful representation of the science behind gravity wells, this one bit is not accurate. A geosynchronous altitude for Earth is nearly 36,000 km, not under 6000 km. Kudos for the rest of the strip, though.
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| − | : The strip scales the heights of the corresponding wells based on the assumption of constant Earth surface gravity; in other words, it takes the same amount of energy to climb such a well as it does to escape the real gravity well. By contrast, as one ascends from the Earth's surface, gravity decreases, so it requires less energy to climb to an orbital altitude than it does to reach the same height in the hypothetical well. The amount of energy required to put a geostationary satellite in orbit, for example, is equivalent to that used in raising it 5413 km in Earth surface gravity, and thus it is located 5413 km from the bottom of the well. [[User:Arcorann|Arcorann]] ([[User talk:Arcorann|talk]]) 03:42, 8 February 2019 (UTC)
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| − | I have a question relating to this topic. I've learnt how to calculate well depth, but how did Randall Munroe calculate the position of things inside the gravity well (moons of planets, for example, or Saturn's rings)?
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| − | : I'm also confused by this. Anyone have the answer? [[Special:Contributions/162.158.78.78|162.158.78.78]] 17:50, 5 May 2020 (UTC)
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| − | == How much gravity can be overcome? ==
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| − | How much of the gravity well can be overcome by launching from a high mountain?
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| − | If we constructed a spaceport on a mountaintop that was, say, 14,505 ft (Mt. Whitney, CA), or even 20,310 ft (Denali, Alaska), or slightly less after clearing a flat surface, would it significantly reduce the amount of the gravity well a rocket had to climb, and hence the amount of fuel needed to reach LEO? Would thinner air reduce drag and increase efficiency significantly as well?
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| − | This would be tough to build and maintain, but would it be worth delivering spaceships to mountaintops by truck to reduce the need for fuel to escape Earth's Gravity well?
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| − | :I am not an expert, but to my knowledge, the effect of latitude on gravity is much stronger than those of a mountain. Especially one as far north as Denali. That is why Nasa uses Florida and Texas, Russia uses {{w|Baikonur_Cosmodrome}} in Southern Kazakhstan , Europe uses {{w|Guiana_Space_Centre}} in south america (French Guiana, so technically Europe...). --[[User:Lupo|Lupo]] ([[User talk:Lupo|talk]]) 08:52, 29 May 2019 (UTC)
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| − | :FWIW in the early stages of planning the Space Shuttle, the Air Force was looking at launching from the top of the Rockies, maybe in Colorado or such. The reason being exactly the savings in propellant and consequent gain in payload. I was made to believe the savings was quite significant.
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| − | :What killed it was the issue of what happens when one malfunctions or crashes on launch. You've got Space Shuttle & parts raining down on the Denver metro area; not good.
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| − | :Source: My dad worked on the project for the Air Force back in the 70s or thereabouts. [[User:Flug|Flug]] ([[User talk:Flug|talk]]) 21:52, 22 September 2019 (UTC)
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