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From Wikipedia's article on Vostok 1:
Path of Gagarin's complete orbit; the landing point is west of the
takeoff point because of the eastward rotation of the Earth.
The flight was at least one orbit of the Earth (note the latitude of the landing is higher than the launch, see @DavidHammen's answer), and beyond that the mission involved firing ...
31
New Horizons went into Earth parking orbit first, so it doesn't count.
For a suborbital direct ascent trajectory, some early lunar probes (USSR's Luna-1 for example) would hold this record. Otherwise, early vertical research probes included the Blue Scout Junior, one of which reached 44400 km on 1961 Dec 4
(mission O-2) - another may have reached 225000 km ...
29
Using the times of injection and retrofire in this diagram:
and the orbit information from NASA, I get that Yuri did about 272° to 273° of a 360° orbit. So about 3/4ths of an orbit. I am not including the ten minutes it took to get from the launch pad to orbit as being in orbit, nor am I including the 30 minutes it took from the deorbit burn to landing as ...
29
Hill sphere is the radius at which gravity of the body (a planet) becomes too weak to pull an object back in, and the object enters a solar orbit. In case of Earth, that is 1.5 million km. In reality this will be a bit less, because at the very edge of the Hill sphere the body moves quite slowly, is susceptible to other influences (like Moon gravity) and it ...
22
I'm going to make some educated guesses and grossly oversimplify the problem:
First, we'll ignore the change in gravitational acceleration over the 200km fall (it's only a 5% difference, so who cares). Next we'll assume that our spacecraft encounters negligible atmospheric resistance until it reaches the Karman line at 100km. Finally we'll assume that it ...
21
So he landed 17 degrees west of the launch site. Sure sounds like a (long) sub-orbital flight.
You are looking at the wrong parameter. The Earth rotated underneath the orbiting spacecraft during the 108 minute flight. A better parameter to look at is latitude. He launched to the northeast and landed 5.35 degrees north of the launch site. He passed the ...
16
Not sure if this counts, but New Horizons was launched directly into an escape trajectory and did not enter orbit. It made it to Pluto and beyond. From Wikipedia:
New Horizons was launched from Cape Canaveral Air Force Station
directly into an Earth-and-solar escape trajectory with a speed of
about 16.26 kilometers per second
That might be considered ...
16
There is absolutely no reason to ever even conceive of flying it again. It was retired because it successfully did the sole task it was designed to do.
SpaceShipOne was unsafe1, barely met its requirements2 (which was certainly good enough) and was designed specifically and solely to win the prize.
The tiny crew compartment meant that there was zero ...
15
One interesting answer to your question is Copenhagen Suborbitals. They're almost a back-yard project.
Our mission is very simple. We are working towards launching a human
being into space.
This is a non-profit suborbital space endeavour founded and led by
Kristian von Bengtson and Peter Madsen, based entirely on sponsors,
private donators and ...
14
Could you? Well, maybe. Let's take a look at one instance where people were encouraged to do that very feat, for $10 million. I'm of course talking about the Ansari X-Prize of 10 years ago. This did a few things that were difficult for the time, giving a cash incentive, and helping with some of the permits and other required items to make it happen. Most of ...
14
There were only four manned Russian programs:
Vostok
No suborbital flights were made. See
Voskhod
No suborbital flights were made. See
Soyuz
There was a two suborbital mission: a failed Soyuz launch in 1975, Soyuz 7K-T No.39
Its apogee was 192km high, for a flight time of 21min.
Made it to space but clearly short of orbit.
The failure was due to an ...
12
It would be extremely unstable. There are 2 things that weight a lot on an empty rocket, the capsule and the engine. With the two of them on opposite sides, the rocket would become extremely unstable. Separating the two allows for the entire system to be more stable. This is particularly important when it is on the ground, where it could easily tip over if ...
10
The altitudes are not what I would call ridiculous. Though you seem prone to using unjustified superlatives in your questions.
The optimum, i.e. minimum injection velocity sub-orbit was provided by HopDavid in this answer. With some manipulation, you get that the maximum altitude of the optimal trajectory is:
$${r\over 2}\left(\sin{\alpha\over 2}+\cos{\...
10
Yes, the minimum $\Delta V$ ballistic trajectory between two points on the Earth's surface will have a negative altitude periapsis, which of course is never reached. For London to Sydney, that trajectory has a periapsis about $800\,\mathrm{km}$ below the surface.
That "orbit" is elliptic with an eccentricity of 0.12. The $\Delta V$ at the surface is $7.85\...
10
I agree with @OuNelson Mangela on the 4,000km figure.
I got 4,052km on flightclub.io, but it was a bit of a struggle. I had to glide quite a bit between entry and landing burns to bleed off velocity that I didn't have the fuel to remove propulsively, and that extended my range quite a bit.
That said, there may be flight profiles that are more efficient ...
10
I can't find anything on this online. It would seem to be a useful technique as it would lower the delta-v for the vehicle that lifts the payload from the ground to the rendezvous point relative to fully making orbit. (NOTE: Yes, you'd have to get extra fuel to the tug to perform this manoeuver but that's potentially a smaller problem.)
This is the opposite ...
9
If the sounding rockets go past the Kármán line at 100 kilometres (62 mi) above the mean sea-level that's now officially the edge of space since the Outer Space Treaty sort of ratified that, there is no difference, and they're all sub-orbital spaceflight. Meaning, they don't reach required orbital velocity (around 8 km/s) to stay in orbit and their flight is ...
9
No
From Ars Technica, the failure came around the T+2:00 mark, which would have been close to when staging happens
What we know is that at about two minutes, there was some sort of failure with the first stage of the rocket and/or its strap-on boosters. (Rumors are circulating that perhaps one or two of the boosters didn't properly separate from the first ...
9
SpaceShipOne was retired so quickly because it was a prototype.
One of the key lessons-learned over the last seventy-plus years was that the waterfall model does not work when applied to creating something that is substantially new.
Another of the key lessons-learned in the same timeframe was that prototyping is a very good way of what I call "debugging ...
8
The goal of Project Mercury as stated in 1958 was to put a man into orbit safely.
In order to do that, NASA wanted to send animals, first. The Russians had done exactly that, and NASA preferred using a chimpanzee instead of a dog. The X-15 had to be piloted, you couldn't put an animal into it. Thus, the Mercury capsule was needed to put a chimpanzee into ...
8
Impossible to say without heavy modelling.
If the Earth was a solitary, stationary object, then the highest possible altitude of a sub-orbital flight would be nigh-infinite, since it would always be possible to construct an elliptical orbit which has its perihelion underneath the Earth's surface; the higher the aphelion of the orbit is, the more eccentric ...
7
First calculate the semi-major axis $a$ where $\mu$ is standard gravitational parameter of the planet that in orbit around:
$$E=\frac{v^2}{2}-\frac{\mu}{r}$$
$$a = -\frac{\mu}{2E}$$
Then with the eccentricity vector:
$$ e = {\left |v \right |^2 r \over {\mu}} - {(r \cdot v ) v \over{\mu}} - {r\over{\left|r\right|}}$$
Can now calculate:
$$r_p=a(1-\left|e\...
7
There can be many reasons:
Test of technology
Doing in-situ measurements in the mesosphere, which is too high for balloons, too low for satellites
Performing microgravity experiments where parabolic flights have too short segments of microgravity but where launching a spacecraft is too expensive
This list is almost certainly incomplete.
7
A minimum energy ellipse between departure and destination corners of a Lambert space triangle is described on page 65 of the 1993 edition of Prussing and Conway's Orbital Mechanics textbook.
In this particular Lambert Space triangle, both r1 and r2 would be the radius of the earth, 6378 km. The 3 points of the triangle would be earth's center, Sidney and ...
7
Some sounding rockets are sub-orbital. Sub-orbital merely means going into space with less than orbital velocity. Going into space: That's a bit arbitrary, but a person who goes 50 miles or higher (~80km) gets astronaut wings; another definition is the nice round figure of 100 kilometers (~62 miles) in altitude; that's the Kármán line. Yet another is the ...
6
Perigee (and apogee) can be defined in two ways - by orbital radius around the barycenter, or by distance above surface.
from the Free Online Dictionary
per·i·gee (pĕr′ə-jē) n.
The point nearest the earth's center in the orbit of the moon or a satellite.
The point in any orbit nearest to the body being orbited.
If the perigee is defined as orbital radius,...
6
The "rocket shape" dart has a significantly lower drag coefficient than the cylinder or the shown 45° cone. It's still above the teardrop, but those drag coefficients are for subsonic regimes.
Once one gets into transonic and supersonic flight regimes, the major issue isn't atmospheric flow, but shockwave turbulence. HARP's projectiles were well into the ...
6
Yes, it's been done:
In the Project HARP a U.S. Navy 16 in (410 mm) 100 caliber gun was used to fire a 180 kg (400 lb) slug at 3600 m/s or 12,960 km/h (8,050 mph), reaching an apogee of 180 km (110 mi), hence performing a suborbital spaceflight.
Or, if you follow this link from the Kickstarter campaign you mention (Graf's excellent history of the HARP ...
6
There's 2 major reasons: safety and practicality
Safety: If you look at the history of booster landings it is not a proven technology, the reliability is not good enough for safety. Don't get me wrong, it's amazing that they can land a booster for re-use, and it works a good portion of the time, however when it goes wrong it goes very, very wrong very fast. ...
6
We can get an best-case estimate by just solving the equations of motion:
$$s = \frac{1}{2}(u + v)t $$
Where
$$t = \frac{v-u}{a}$$
Since we want $v = 0ms^{-1}$ at $s = 100,000m$, we get:
$$s = \frac{u^{2}}{2a} \Rightarrow u = \pm \sqrt{2as}$$
which means that a single instantaneous impulse giving a velocity of ~1400$ms^{-1}$ would get us to the Kármán ...
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