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(The following is in reply to User talk:Vegaswikian#SpaceX CRS-7 not seventh overall flight for Falcon 9 v1.1. -- ToE 19:17, 5 August 2014 (UTC))[reply]

I just copied the format of the previous flight when I created that article. I guess my question is how important is that information after the first flight? Maybe that level of detail should only be in the first flight. Given that the booster will be used for many different named flights, trying to keep track of this across multiple articles with constantly changing launch dates is too much. So if you want to remove this from this article and some others, I'm OK. Maybe you could also check at WikiProject Spaceflight for their opinion. Vegaswikian (talk) 18:47, 5 August 2014 (UTC)[reply]

Done. I removed the v1.1 sequence number from SpaceX CRS-5, SpaceX CRS-6, and SpaceX CRS-7, primarily because they were factually incorrect, but it is also premature to include our guesses. -- ToE 19:17, 5 August 2014 (UTC)[reply]

Typing error in Project Morpheus corrected. Thank you for pointing it out.

In the article whose link added: A final line to say, the Morpheus Lander has successfully flown, automatically found a landing space and landed without crashing, may be appropriate. Without the line some people will suspect failure is being covered up. Good layout for your article. Andrew Swallow (talk) 19:47, 28 September 2014 (UTC)[reply]

p.s. The project has just issued a conference paper that says the latest version's main engine has a thrust of 5,400 lbf (24,000 N). Andrew Swallow (talk) 11:01, 29 September 2014 (UTC)[reply]

Okay. Robert McClenon (talk) 03:13, 30 December 2014 (UTC)[reply]

The sockpuppet tag remains valid as far as I'm concerned, however if you wish to remove it yourself then I won't re-add it. GiantSnowman 19:46, 7 February 2015 (UTC)[reply]

The above is a reply to User talk:GiantSnowman#WoolSalesman. -- ToE 00:47, 8 February 2015 (UTC) [reply]

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You're wasting your time and energy, you're now suffering from that tag-team effect. I'm sorry about that, but there's no known solution, the community seems to encourage it. Please accept my condolences and try to move onto something completely different. The Rambling Man (talk) 20:27, 15 May 2015 (UTC)[reply]

I do have the sense that BB is coming to the defense of a perceived attack on M, which is a shame since I intended no such attack and quite a few times recently I've interacted with M concerning the hatting of questions, and each time she has been very reasonable, standing by her initial decision but offering no objection to unhatting, as happened in this case via her talk page. -- ToE 21:05, 15 May 2015 (UTC)[reply]

Yes, I did go back and I did see your post. Thanks for the information as this is exactly what I wanted to find. Gurumaister (talk) 15:46, 16 October 2015 (UTC)[reply]

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A few weeks ago, I asked for help in the reference desk. I needed the temperature in a certian location in space and you helped me a lot.The information was crucial for a project and I'm asking to know if I can thank you in it by mentioning your wikipedia username. 84.228.177.168 (talk) 10:08, 18 December 2015 (UTC)[reply]

Yes, if you wish. I do appreciate the direct thanks here. (And thanks for that thanks!) But I don't particularly care one way or another about mention elsewhere, so do whatever you feel best about it. Do feel free to use what I showed you without attribution if that works better. The section has been archived here. As explained in the derivation in the section I linked, the equation I gave determines the temperature at which the spherical object is at equilibrium with respect to the incoming insolation and its own outgoing thermal radiation. In practice, the geometry and imperfect thermal conduction of a probe will mean that side facing toward the sun will be warmer and the side facing away will be cooler. -- ToE 14:16, 18 December 2015 (UTC)[reply]

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Thanks Katie! -- ToE 14:46, 29 January 2016 (UTC)[reply]

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Thank you BracketBot, I've fixed it. -- ToE 12:24, 25 June 2016 (UTC)[reply]

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People like you keep my existential contemplation skewed to the bright side. Massive thanks. Splićanin (talk) 00:56, 22 April 2017 (UTC)[reply]

Thanks for your note on August 3rd! HappyCamper 23:41, 15 August 2017 (UTC)[reply]

I still don't understand where is a mistake in Solutions explanations. They added the potential energies. Feynman did the same in this part of 13-3:

...

We can understand why it should be the energy of every pair this way: Suppose that we want to find the total amount of work that must be done to bring the objects to certain distances from each other. We may do this in several steps, bringing them in from infinity where there is no force, one by one. First we bring in number one, which requires no work, since no other objects are yet present to exert force on it. Next we bring in number two, which does take some work, namely . Now, and this is an important point, suppose we bring in the next object to position three. At any moment the force on number can be written as the sum of two forces—the force exerted by number and that exerted by number . Therefore the work done is the sum of the works done by each, because if can be resolved into the sum of two forces,



then the work is



That is, the work done is the sum of the work done against the first force and the second force, as if each acted independently. Proceeding in this way, we see that the total work required to assemble the given configuration of objects is precisely the value given in Eq. (13.14) as the potential energy. It is because gravity obeys the principle of superposition of forces that we can write the potential energy as a sum over each pair of particles.

—  Feynman • Leighton • Sands , The Feynman Lectures on Physics, Lecture 13

Username160611000000 (talk) 14:35, 13 October 2017 (UTC)[reply]

Thanks for the reminder. I know it's been more than the week I promised, but I've kept that problem in mind though I've not taken the time to put pencil to paper yet. I will try to do so soon. -- ToE 15:09, 13 October 2017 (UTC)[reply]

I think I am close to understanding. Such a formula as appears in MEPhI Solutions is a formula for stationary 2 objects: the sun and the earth (it is what Feynman is talking about at the end of Ch. 13-3) , or simply the probe have initial speed zero. So The Solutions' authors have found for the case when the probe was not moving . And then they have jumped to moving reference frame. Username160611000000 (talk) 12:11, 25 October 2017 (UTC)[reply]

UP. Username160611000000 (talk) 16:09, 2 November 2017 (UTC)[reply]

I'll look at this tonight and either get something to you by tomorrow or suggest a refined question to ask on WP:RD/S. -- ToE 16:54, 2 November 2017 (UTC)[reply]

OK, let's start with a simpler problem and calculate the total system escape velocity v_te, as described in Escape velocity#Multiple bodies, from the surface of the Earth, escaping the Solar System. This a simplification of your Exercise 14-21 where we are seeking a residual speed of 0 instead of 10 mi/sec. This calculation will take into account the orbital speed of the Earth about the Sun, but as with the standard calculation of a single body escape velocity, it will not take into account the rotational speed of the Earth.

As before, I will use:

v_{e_Earth} = 11.2 km/sec = 6.96 mi/sec

v_{e_Sun@1AU} = 42.1 km/sec = 26.2 mi/sec

v_Orbital@1AU = 29.8 km/sec = 18.5 mi/sec

and v_∞² = v_initial² - v_e², the formula from Escape velocity relating initial speed to the hyperbolic excess speed.

Note that v_{e_Sun@1AU} = sqrt(2) v_Orbital@1AU. This is used in the derivation of the formula for v_te in §Multiple bodies, but we will just use the numbers.

To achieve the 26.2 mi/sec Solar escape velocity v_{e_Sun@1AU}, we need residual speed upon escaping Earth of v_{e_Sun@1AU} - v_Orbital@1AU = 26.2 mi/sec - 18.5 mi/sec = 7.7 mi/sec. So our initial speed from the surface of the Earth needed to be sqrt( (7.7 mi/sec)² + (6.96 mi/sec)² ) = 10.4 mi/sec. That converts to 16.7 km/sec, close enough to the 16.6 km/sec listed in §List of escape velocities given the number of significant digits we are working with.

This is the correct answer, and is equivalent to the method used to come up with the 11.8 mi/sec answer to Exercise 14-21 given in Exercises.

Now let's try the naive energy balance approach as used in Solutions for Exercise 14-21.

They would argue that the Sun's specific gravitation well is -0.5 v_{e_Sun@1AU}² and the Earth's specific gravitation well is -0.5 v_{e_Earth}². If the probe's initial velocity, measured in the Sun's reference frame, is v_i, then its specific kinetic energy is 0.5 v_i². To escape the Sun with 0 excess velocity, our energies should sum to zero. This can be arranged as

v_{e_Sun@1AU}² = v_i² - v_{e_Earth}²

which resembles the formula above for relating initial velocity to hyperbolic excess velocity. (And this would be a correct formulation for total system escape velocity were the Earth not moving with respect to the Sun.) Continuing,

v_i = sqrt(v_{e_Earth}² + v_{e_Sun@1AU}²) = 27.1 mi/sec

But that is in the Sun's reference frame, so to calculate the probe's launch speed with respect to the Earth, we subtract v_Orbital@1AU

27.1 mi/sec - 18.5 mi/sec = 8.6 mi/sec

But we know that it wrong, as if it 1.8 mi/sec shy of the correct value of 10.4 mi/sec, just as the answer given in Solutions to Exercise 14-21 was 1.2 mi/sec shy of the correct value given in Exercises. But why?

Correct: v_te = sqrt( v_{e_Earth}² + (v_{e_Sun@1AU} - v_Orbital@1AU)² )

Wrong: v_te = sqrt(v_{e_Earth}² + v_{e_Sun@1AU}²) - v_Orbital@1AU

... cont ...

So it is correct that if the Sun, the Earth and the probe all have zero speed , then the probe must be supplied with energy to have zero speed at infinity. Also it is correct that if the Sun and the probe (in the Sun reference frame) both have zero speed, but the Earth have a speed 30 km/sec, then also the probe must be supplied with energy to have zero speed at infinity. If it is correct, then I have 2 hypotheses:
1. The mistake is that the initial velocity of the probe is not taken into account. And as we know the kinetic energy needed to accelerate the probe from 0 m/sec to v₂ m/sec does not equal the energy for acceleration from v₁ to v₂ m/sec.
2. The mistake is that after overcoming the gravity field of the Earth, the probe should have a speed of 30 km/sec. But if we subtract we don't achieve .
Username160611000000 (talk) 15:56, 4 November 2017 (UTC)[reply]

I have found that if I use the conservation of momentum law , I can rearrange the kinetic energy formula like a sum of energies, but no longer a sum of velocities PNG. This also correlates with ex. 10-7, 10-5.
Username160611000000 (talk) 05:19, 6 November 2017 (UTC)[reply]

For single bodies, direction of travel is irrelevant to the calculation of escape velocity. From Escape velocity#Overview:

Escape velocity is actually a speed (not a velocity) because it does not specify a direction: no matter what the direction of travel is, the object can escape the gravitational field (provided its path does not intersect the planet).

If the Earth were not in motion about the Sun (say, held in place by an ideal rigid massless rod), then the two methods of calculation yield the same total escape velocity.

v_{te_Unmoving_Earth} = sqrt(v_{e_Earth}² + v_{e_Sun@1AU}²) = 43.6 km/sec

But for calculations where the Earth is moving in its orbit, direction of travel is important, not only for determining the speed in the Earth's reference frame, but also the speed in the Sun's reference frame. Consider two cases, a prograde launch (as from before) where the probe takes off so that it departs Earth tangentially to Earth's orbit, in the forward direction of the orbit, and a retrograde launch, where the probe takes off so that it also departs Earth tangentially to Earth's orbit, but in the opposite direction of the orbit.

In both cases, the naive energy balance method (as suggested in Solutions) gives the same 43.6 km/sec total escape speed in the Sun's reference frame, and simply adds or subtracts the Earth's orbital velocity, giving:

Wrong: v_{te_Prograde} = v_te = sqrt(v_{e_Earth}² + v_{e_Sun@1AU}²) - v_Orbital@1AU = 13.8 km/sec

Wrong: v_{te_Retrograde} = v_te = sqrt(v_{e_Earth}² + v_{e_Sun@1AU}²) + v_Orbital@1AU = 73.4 km/sec

The correct, two-step method (as suggested in Exercises) gives:

v_{te_Prograde} = sqrt( v_{e_Earth}² + (v_{e_Sun@1AU} - v_Orbital@1AU)² ) = 16.6 km/sec

v_{te_Retrograde} = sqrt( v_{e_Earth}² + (v_{e_Sun@1AU} + v_Orbital@1AU)² ) = 72.8 km/sec

Converted to the Sun's reference frame:

v_{te_Prograde_Sun's_RF} = 46.4 km/sec

v_{te_Retrograde_Sun's_RF} = 43.0 km/sec

So the naive energy balance method underestimates the required velocity for the prograde departure but overestimates it for the retrograde departure. Why?

Consider two probes taking off simultaneously, one prograde at 13.8 km/sec from the Earth, and the other retrograde at 73.4 km/sec from the Earth, so that both start traveling at the same 43.6 km/sec in the Sun's reference frame. The first probe is being chased by the Earth, taking longer to escape the Earth's gravitational drag, while the second is speeding away from Earth and its gravitational influence much more quickly. This is similar to gravity assist maneuvers, where total energy is conserved only when you take into account changes in the planet's orbital energy. For the first probe, that negative gravity assist maneuver means that it will not have quite enough energy to escape the Sun, and it will enter a highly elliptical orbit, while the second probe will escape the Earth/Sun system entirely, with a 7.2 km/sec hyperbolic excess velocity. -- ToE 12:23, 8 November 2017 (UTC)[reply]

Hi, I got 10.4 mi/s too. I used the sum of the probe's initial kinetic energy, initial potential energy from the Earth's gravity, and initial potential energy from the Sun's gravity, then equated the sum to the final kinetic energy of the probe. --Bob K31416 (talk) 00:11, 30 November 2017 (UTC)[reply]

Hello Toe, thanks for that perceptive observation. No, I hadn't made the connection---at least, not explicitly. My thinking was along the lines of "soak up the odd bits in the rest of the sequence by flipping the first one", but this is pretty ugly and parity is the Right Way (tm) to think about it. Best wishes, Robinh (talk) 08:45, 19 December 2017 (UTC)[reply]

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Thanks for standing up for my honor ( ;) ) here. I find it fascinating that someone signing their name "Tamas" would affect offense at being referred to as "this guy", but what can you do? I like "God has ten fingers." --JBL (talk) 21:46, 6 March 2018 (UTC)[reply]

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Awesome

Ten years!

--Gerda Arendt (talk) 10:37, 3 September 2019 (UTC)[reply]

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Hello Thinking of England,

Hope all has been well. You are probably already aware, but if you are not, I have put a template (Template:Calendar widget) on your "Links to all archives" subpages of Wikipedia:Reference desk (almost half a year later.) The use of the template reduces the size of the pages by almost half a megabyte, and may make it easier to replicate the next year when necessary. Hope this works pretty well. Steel1943 (talk) 23:15, 27 November 2019 (UTC)[reply]

On 15 April 2020, In the news was updated with an item that involved the article Jacques Blamont, which you updated. If you know of another recently created or updated article suitable for inclusion in ITN, please suggest it on the candidates page. Stephen 01:54, 15 April 2020 (UTC)[reply]

Thanks for cleaning up that massive wall of text related to the discussion mergers. It was much needed as that "wall" was very distasteful in appearance. — Mr Xaero ^☎️ 14:29, 4 May 2020 (UTC)[reply]

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Sagittarian Milky Way (talk) 02:25, 25 June 2020 (UTC)[reply]

At the RD you wrote "0 ≤ a < b < n". Did you mean "0 < a − b ≤ n"?  --Lambiam 15:11, 12 August 2020 (UTC)[reply]

Thanks Lambiam, I corrected the order of my inequality.

If the questioner is who I think it is, then I suspect they aren't intending ask about congruence classes despite the use of the vinculum. In either case, your answer has it fully covered. -- ToE 15:41, 12 August 2020 (UTC)[reply]

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For the information regarding the heat of the sun:

https://en.wikipedia.org/w/index.php?title=Wikipedia:Reference_desk/Science&oldid=1010649846

--IBBishops (talk) 20:39, 16 May 2021 (UTC)[reply]

I was messing around with Quarry, and saw that your sandbox was one of the largest user pages on the site. Looking at it, I decided to run a SQL query using Quarry, and determined that there are (currently) 35,797 reference desk archives. Wowie! jp×g 11:52, 11 November 2021 (UTC)[reply]

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