Hot answers tagged

29

Wikipedia gives $0.51 {km \over s}$ or $510 {m \over s}$ escape velocity, so, no, no leaving Ceres by jumping. Following my earlier calculations, an asteroid of the radius of Ceres would have orbital speed at near-surface orbit of about $336 {m\over s}$, which is way beyond jump strength of anyone as well. Gravitational acceleration on the Moon is $1.6249 {...


27

Its final orbit will only be 375 km above Ceres, but you have to give it time. Dawn is powered by xenon ion engines, which are extremely efficient, but very weak. The usual comparison is that they are pushing the craft forwards about as much as a sheet of paper pushes down on your hand. Their advantage is that they can do this for a very long time. This is ...


26

Planning travel within the solar system doesn't work quite like you assume. A spacecraft typically uses its rockets for a few minutes at the start and end of the journey and coasts the rest of the way, but while it is coasting the gravity of the planets it leaving or approaching, and much more importantly the Sun act to change its direction and velocity. To ...


25

Dawn has several mission objectives, including to continue testing the Ion Thruster. But why Ceres? Ceres and Vesta were chosen, because they have contrasting content, one icy and one rocky. Also, they are among the protoplanets that remain intact since formation, which (hopefully) leads to a better understanding of the formation of our solar system, ...


22

Does it have any additional thrusters? Not to thrust towards its targets. For that, it's 100% ion thruster propelled. It does also have a set of 12 MR-103G variable thrust (0.9 N maximum) RCS (Reaction Control System) hydrazine monopropellant thrusters that launched with only 46 kg of propellants (read: total thrust of its RCS doesn't provide the spacecraft ...


18

Let's go back our old friend the Pork chop plotter. Earth to Jupiter using minimum fuel takes around 2 years and you get one opportunity per year, more or less, to get there. You can shorten the journey to perhaps 20 months with minimal extra fuel. The delta-V required at Earth (over and above escape velocity) is about 9.3 km/s (you can in theory aerobrake ...


17

This hinges as bit on what "as easy as" means. We clearly can't go to Mars today because we don't have the technology. We do have all the bits and pieces in theory, but we haven't build anything that can actually do it. SpaceX is famously trying to do exactly that with Starship/BFR. The discussed paper is Low-thrust trajectories for human missions to Ceres. ...


16

This answer is outdated. The Dawn flyby in February 2015 added a lot of information which was not known at the writing of this answer. We don't actually know much about Ceres. All we know about it is from earth-based or earth-orbit-based observation. Until now it wasn't visited by a probe, but the Dawn spacecraft will do so in February 2015. This will ...


13

FYI, the relevant equation is to set the kinetic energy equal to the gravitational potential energy. This is written on a per-mass basis, because both kinetic energy and gravitational potential energy is proportional to mass. $$ \frac{ G M m }{ r } = \frac{ 1 }{ 2} m v_{\infty}^2 $$ For Ceres: $$ v_{\infty} = \sqrt{ \frac{ 2 G M }{r} } = \sqrt{ \frac{ 2 ...


12

You'd have to go a lot smaller to achieve escape velocity with musclepower alone. According to xkcd, you can escape from Deimos using a bike and a ramp...


12

NASA has produced a topographical features map of Ceres, with names for some craters. The map was produced in 2015. This one has few more details,


11

Why was Dawn placed into an orbit that would only be stable for “decades” In order to simultaneously obey the rules and maximize science. Maximizing science means staying at Ceres and collecting data as absolutely long as possible. Using the last remaining hydrazine and Xenon for attitude control and propulsion to leave Ceres would have reduced the time at ...


11

I whomped up a spreadsheet to compare scenarios like this: Hohmann.xls. Typing Earth into departure planet cell and Mars into destination planet I get Launch windows open each 2.14 years (synodic period) Trip time .71 years Delta V Low Earth Orbit to Low Mars Orbit: 5.7 km/s Typing Mars into departure planet and Jupiter into destination: Launch window: ...


10

Landing on Ceres would probably be much like landing on the Moon or Mars--it's mostly flat probably with a few craters, so it wouldn't be that novel of an event. And we've sent unmanned missions far further than Ceres, so propulsion obviously isn't an issue. The single biggest engineering obstacle that would have to be worked out would simply be figuring ...


10

Well, I'd say that "We can go to Ceres as easily as Mars" isn't quite true, but it's really a matter of perspective. On the one hand, measuring by fuel expenditure, specifically rocket delta-v, a journey to Ceres would be rather similar to a journey to Mars. The difference in delta-v requirement to go to Ceres is only around 400 m/s more than going to Mars ...


9

The oblateness of Ceres seems to indicate water. Moreover water geysers were recently detected by Herschel Space telescope. Hopefully we'll learn more when the Dawn spacecraft arrives February of 2015. Planetary chauvinists like to point out mass of the Main Belt is small compared to Earth or Mars. But most of a planet's mass is inaccessible. As we burrow ...


8

Fear not. If all goes well, Dawn will get down to 375 km. You can read more details in the blog entry, but Dawn's lifetime will be limited by its use of hydrazine, which is in fact driven by the failure of two of the four reaction wheels. Update 3+ years later: Dawn is currently on it's way to a 35 km altitude orbit! (Not a typo.)


8

The USGS Astrogeology Science Center's Astropedia is an excellent source for derived mapping data products (though only 7 Ceres products). Here is a "Ceres Nomenclature" data product:


7

We're a long way away from being able to do this. Is it theoretically possible? Let's see if we have enough delta-V: Mass is 1021 kg, with 1/3 available as propellant. Using the Rocket Equation: $$ \Delta v = I_{sp} \ g \ ln\left(\frac{m_i}{m_f}\right) $$ Isp = 104 ? = 105 * ln (1021/7*1020) ? = 35,000 Delta-V to get from Ceres to Mars is in the ...


6

There is one wheel still operable (at least it was operable the last time they operated it), but Dawn no longer uses it. The third wheel failure was in April of this year (2017 for those reading this answer thousands of years from now). Yes, gravity tugging on the very long lever arms called solar panels is a significant driver on hydrazine usage, made much ...


6

There is a lack of solid, science-based study into realistic colonisation options because until Dawn arrived very little was known for certain. Much more is known now but even so, we know probable orders of magnitude more about the Moon/Mars and we still would need more before colonisation could be tackled seriously. Anything serious that may have been ...


6

In addition to Steve Linton's excellent answer there's a simple pattern: To get somewhere for the minimum fuel generally takes one half the orbital period of the slower of the launch and target orbits--and when you get into the realm where this breaks down you're also in the realm where you're going not going to be using a simple minimum-fuel trajectory ...


6

Indeed, you are correct, it could reach escape velocity. The M110 can reach speeds of over 700 m/s, which is well above the escape velocity. Most guns actually don't need oxygen to work either, as the gun powder has the oxygen needed. So yeah, be extra careful where you fire a gun on an asteroid.


5

Wikipedia has a decent summary of the current state of thinking on the bright spots in Occator crater: on 29 June 2016, scientists reported the bright spot to be mostly sodium carbonate (Na2CO3), implying that hydrothermal activity was probably involved in creating the bright spots. Hydrothermal activity may be ongoing, some photos of Occator show a ...


5

Ceres has a surface gravity on of 0.029G and an escape velocity of 500ms. So there would be a down, standing up would be possible and it would be impossible to accidentally escape into orbit. Falling over (or dropping something held) would take around 10 seconds, anything kicked up or thrown would be aloft for much longer. So jumping your mining truck wouldn'...


4

If you could power a "massless" propulsion method (like a bicycle that climbed out of the gravity field), you could easily escape Ceres, but this highlights the need for huge power thrust at the early stage, because propulsion needs mass that it can "throw backwards", and if you accelerate slowly, that means you have to carry all that mass with you for ...


4

This is not a complete answer but I can offer an example of an orbit being in resonance with the rotation rate of a central body. Astrophysicists indeed sometimes use the word "resonant" to describe orbit synchronization with a rotating primary. That synchronization doesn't have to be 1:1: it can be 2:1, 5:4, any ratio of integers. The literature involved ...


3

A hydrazine monopropellant rocket would have a specific impulse of around 230-240 seconds and so an exhaust velocity around 2000 m/s. The rocket equation then tells us that for a delta-V of 262 m/s that means a mass ratio of about 1.13, so maybe 130 kg of hydrazine for a 1 ton payload. The advantage of a monopropellant is simplicity and reliability. Once ...


3

Formulation Brandon A. Jones wrote an excellent summary of the different formulations and methods to compute spherical harmonics in his 2010 PhD thesis available here. Chapter 2 is especially of interest. You may also want to read Fantino & Casotto 2008 "Methods of harmonic synthesis for global geopotential models and their first-, second- and third-...


3

What you're missing is the descriptions of the contents of that file. That's in another file, right next to the one you found. The data for the Ceres gravity model are in JGDWN_CER18C_SHA.TAB and the descriptions are in JGDWN_CER18C_SHA.LBL. Each harmonic has four floats. Two could be the sine and cosine terms, but would the third and four float be their ...


Only top voted, non community-wiki answers of a minimum length are eligible