Talk:681: Gravity Wells - Revision history

92.23.2.228 at 20:33, 8 August 2025

2025-08-08T20:33:52Z

185.70.55.130 at 15:40, 8 August 2025

2025-08-08T15:40:06Z

185.115.7.100 at 11:29, 8 August 2025

2025-08-08T11:29:24Z

185.115.7.100 at 11:28, 8 August 2025

2025-08-08T11:28:46Z

185.115.7.100 at 11:22, 8 August 2025

2025-08-08T11:22:54Z

FaviFake: Shortened title for {{toc}}

2023-05-23T10:34:14Z

Shortened title for {{toc}}

141.101.99.154 at 12:27, 7 September 2022

2022-09-07T12:27:01Z

172.70.178.107: Relocated comment

2022-09-07T10:54:04Z

Relocated comment

172.70.178.107: Commented

2022-09-07T10:52:01Z

Commented

162.158.78.78 at 17:50, 5 May 2020

2020-05-05T17:50:29Z

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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)
 : You can use the same formula as in the explanation: D = GM/(gd), where D is the depth in the gravity well, G is the gravitational constant, M is the mass of the planet, g is acceleration due to gravity on Earth's surface (9.81 m/s²) and d is the distance from the center of mass of the planet. Note that D is the '''depth''' in the well, while d is the '''distance''' from the center of mass, so they are reversed (one counts "from the top", the other "from the bottom"). In the explanation, d is substituted for the radius of the planet (i.e. the distance from the center of the planet to its surface), which is useful because that's where you would consider yourself "on the planet", and also where you would launch your rockets (or baseballs) from. That said, you can also use other values. E.g. for Saturn's rings: Saturn has a gravity well depth of ~66000 km at its surface; rings are 7000 to 80000 km above its equator, or 67000 - 140000 km from its center of mass; that gives them the '''depths''' in the gravity well of ~57000 to ~28000 km, or 90% to 40% (once again, counting from the top). This about lines up with the comic. (also note how things at the very top of the gravity well are infinitely far away from the actual planet). [[Special:Contributions/185.70.55.130|185.70.55.130]] 15:40, 8 August 2025 (UTC)
 == How much gravity can be overcome? ==

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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)?
 : 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)
 == How much gravity can be overcome? ==

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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)
-:: The reason we build spaceports closer to the equator is not any effect on gravity, but rather the fact that Earth "moves" (in a linear sense) faster at the equator, therefore you don't need to spend as much fuel to accelerate to orbital velocity, assuming you are going eastwards. For polar orbits (where you go around the earth north-to-south or vice versa), you can launch from anywhere you like (hence Plesetsk, Kodiak, Vandenburg, etc). For (rare) westward launches, you want to be as far away from the equator as possible.
+:: The reason we build spaceports closer to the equator is not any effect on gravity, but rather the fact that Earth "moves" (in a linear sense) faster at the equator, therefore you don't need to spend as much fuel to accelerate to orbital velocity, assuming you are going eastwards. For polar orbits (where you go around the earth north-to-south or vice versa), you can launch from anywhere you like (hence Plesetsk, Kodiak, Vandenburg, etc). For (rare) westward launches, you want to be as far away from the equator as possible. [[Special:Contributions/185.115.7.100|185.115.7.100]] 11:29, 8 August 2025 (UTC)
 :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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 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?
 :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)
 :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.

← Older revision		Revision as of 11:22, 8 August 2025
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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)		: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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		+	I think the sentence "This is why it took a huge rocket to get to the moon but only a small one to get back" is wrong (or at least very misleading). The main reason for this is that the "huge" rocket has to carry the "small" rocket all the way to the moon, while the "small" rocket only has to carry back its useful payload, so of course the "huge" rocket will have to be much bigger as per the rocket equation. The secondary (and more interesting) reason is that the "big" rocket has to (1) ascend through the Earth's atmosphere (which takes a lot of energy due to drag), (2) escape most of the Earth's gravity well, (3) slow down enough to land at the Moon; while the "small" rocket doesn't have to deal with an atmosphere on its way up, and doesn't need to slow down at Earth using its own engines because it can aerobrake (use the drag of the atmosphere to slow down). If neither Earth nor Moon had an atmosphere, and your rockets only had to go one way (i.e. didn't have to carry a return vehicle on them), then the rockets would be roughly the same size.
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		+	If noone minds within a couple days, I will note this in the explanation. [[Special:Contributions/185.115.7.100\|185.115.7.100]] 11:22, 8 August 2025 (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 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)
+::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)
 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.

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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)
 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 “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.
@@ Line 52: / Line 54: @@
 :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)

← Older revision		Revision as of 10:52, 7 September 2022
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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)		: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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		+	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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 : 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)
 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)?
 == How much of the gravity well can be overcome by launching from a high mountain? ==