For Questions That Don't Merit Their Own Thread

[Bill Phil]

15 minutes ago, 0111narwhalz said:

Can the normal vector be anywhere but the horizon?

Orbital physics, of course.

I think so. But only in orbits which aren't circular.

[kunok]

14 hours ago, 0111narwhalz said:

Can the normal vector be anywhere but the horizon?

Orbital physics, of course.

In 2 body interactions (or patched conics that KSP uses)? In stable orbits no, parabolic and hyperbolic can have other orientations.

But remember that the patched conics is a huge simplification

[Green Baron]

It's not a question, but it doesn't merit it's own thread and i found it worth to share here:

https://www.scientificamerican.com/article/eurocreationism/

This is concerning, especially the case where a ministry of education forbids to teach evolution in school ... fortunately this time he was chased out of his chair.

In order to not step on forbidden ground i hold myself back with further comments, draw your own conclusions.

 

Edited November 23, 2016 by Green Baron

[NSEP]

I would say: Believe what you want to believe, but dont force others to believe what you believe. 

Im not going to say more, i dont want it to get so for that the thread is going to get locked.

[0111narwhalz]

How might one go about determining the focal length of an obscenely large lens without burning things?

[peadar1987]

2 hours ago, 0111narwhalz said:

How might one go about determining the focal length of an obscenely large lens without burning things?

Laser pointers? Three or four of them ought to do, point them at the lens, see where the beams intersect.

[p1t1o]

12 hours ago, 0111narwhalz said:

How might one go about determining the focal length of an obscenely large lens without burning things?

You could work it out mathematically if you can accurately describe the geometry of the lens and know the refractive index of the material.

[Green Baron]

15 hours ago, 0111narwhalz said:

How might one go about determining the focal length of an obscenely large lens without burning things?

Autocollimation is an experimental method, but could be impractical with large lenses.

Large lenses (>30cm, for refracting telescopes) where made in the 19th century. They are difficult to produce and from a certain size on they start to deform under their own weight. Only few refractors with modern material and apertures >20cm exist, namely made by american Astro-Physics (oil-spaced triplet APO) and russian LZOS (air-spaced triplet APO). Image quality is superb compared to a reflector, and so is the price ...

 

Edited November 25, 2016 by Green Baron

[todofwar]

So I always have trouble with this, and I can't ever find a good answer on the Google but maybe I just don't know where to look. Anyway, how do you calculate potential energy of something so high above earth that mgh doesn't cut it anymore? Like, the total energy required to get to geostationary by traveling straight up.

[p1t1o]

7 hours ago, todofwar said:

So I always have trouble with this, and I can't ever find a good answer on the Google but maybe I just don't know where to look. Anyway, how do you calculate potential energy of something so high above earth that mgh doesn't cut it anymore? Like, the total energy required to get to geostationary by traveling straight up.

Looks like it is this:

V(x) = - ([GM]/x)

Where V(x) is the gravitational potential at x metres from point mass M, G being the gravitational constant, measured in Joules-per-kilogram.

So for a 1000kg satellite at GSO:

x = 35786km + 6371km (Earth radius) = 42157km

M = 5.979e24kg

V(x) = - ([6.674e-11*5.979e24] / 42.157e6)

V(x) = -9465532.6

V(x) = - 9.465MJ/kg

So: it would take 9.465*1000= 9465 MJ to lift a 1000kg satellite to GSO, which seems about right.

***edit***

Oh actually that is the energy required to lift it from the centre of the Earth, so you'd have to work out the potential at 6371km and subtract that for the result from sea-level.

**edit2**

Ummm now Im not so sure as the result fro V(x) at sea level is around 62.6MJ/kg - perhaps this is the energy required to lift a mass from this altitude to infinity?

So, the true answer to lift a satellite to GSO would be 62.6MJ/kg - 9.46MJ/kg = 53.17MJ/kg

That sound right to anyone else?

***edit3***

https://en.wikipedia.org/wiki/Gravitational_potential

***

Yup, looks like it is defined as "The amount of work needed to move the mass from infinity to the altitude specified", it gets a negative value meaning that you get energy back as you move the object closer in from infinity. The converse of which means the positive-sign vale is the energy needed to move the object from the specified height *to* infinity.

So 53.17MJ/kg would appear to be correct for sea-level to GSO.

 

Edited November 29, 2016 by p1t1o
woops

[peadar1987]

6 hours ago, todofwar said:

So I always have trouble with this, and I can't ever find a good answer on the Google but maybe I just don't know where to look. Anyway, how do you calculate potential energy of something so high above earth that mgh doesn't cut it anymore? Like, the total energy required to get to geostationary by traveling straight up.

You have to take an integral.

g=(G*M)/r^2

Ep=m*g*r

So Ep=m*((G*M)/r)

Integrate this from r=6000 to r=? and you get the change in potential energy

[todofwar]

@p1t1o @peadar1987 How does centrifugal force play into that (yeah yeah, fictitious force, but we are in a rotating reference frame after all)?

[p1t1o]

Just now, todofwar said:

@p1t1o @peadar1987 How does centrifugal force play into that (yeah yeah, fictitious force, but we are in a rotating reference frame after all)?

In the example given, it doesn't, we aren't dealing with anything in orbit here, just how hard it is to lift something out of a gravity well (the satellite and GSO were just example for mass and altitude, our satellite in this case would fall back to Earth when released.)

If you want to bring orbital dynamics into the equation, a good place to start is here: https://en.wikipedia.org/wiki/Specific_orbital_energy

[todofwar]

23 hours ago, p1t1o said:

In the example given, it doesn't, we aren't dealing with anything in orbit here, just how hard it is to lift something out of a gravity well (the satellite and GSO were just example for mass and altitude, our satellite in this case would fall back to Earth when released.)

If you want to bring orbital dynamics into the equation, a good place to start is here: https://en.wikipedia.org/wiki/Specific_orbital_energy

I'll take a look, thanks! But before I dive too far down a rabbit hole, the main question I'm trying to answer is, would it take more or less fuel to reach GSO by traveling straight up from the equator? So, a space elevator without the elevator. Can't find an answer to this question, but this whole week has been a long stretch of google fails on my part so I might just be asking the question in the wrong way

[Kryten]

32 minutes ago, todofwar said:

I'll take a look, thanks! But before I dive too far down a rabbit hole, the main question I'm trying to answer is, would it take more or less fuel to reach GSO by traveling straight up from the equator? So, a space elevator without the elevator. Can't find an answer to this question, but this whole week has been a long stretch of google fails on my part so I might just be asking the question in the wrong way

Travelling straight up with a rocket is always a bad idea, it maximises your gravity losses

[p1t1o]

1 hour ago, todofwar said:

I'll take a look, thanks! But before I dive too far down a rabbit hole, the main question I'm trying to answer is, would it take more or less fuel to reach GSO by traveling straight up from the equator? So, a space elevator without the elevator. Can't find an answer to this question, but this whole week has been a long stretch of google fails on my part so I might just be asking the question in the wrong way

I see what you're getting at.

No, you dont get away with things that easily!

The twist [no pun intended] is angular momentum - in order to "go straight up" to GSO, you'd actually have to carve a huge helix spiral (relative to a fixed point in space), which would in essence, boil down to the sort of orbital insertion we are all familiar with.

The classic example - the ballerina spins faster as she pulls in her arms, and slower as she extends them. As you travel straight up, you are "extending", thus you rotate around the Earth more slowly, thus your trajectory curves backwards (relative to a fixed point on the Earth's surface). This is essentially the coriolis force - angular momentum asserting itself.

You could travel straight up (relative to a fixed point on the Earth's surface) but you would also have to thrust sideways somewhat as you were ascending in order to gain the angular momentum you need (and to keep station with the fixed point) - which is what we do already. Sortof. (Remember how most of a launch to orbit is sideways thrusting? The vertical component is the easy part)

(If it helps clear things up - the example above with the calculation of gravitational potential? That represents rising from a planet that is not rotating.)

**edit**

Fun Fact - if we did build a space elevator, cargo would experience, and the cable would have to be built to withstand, significant sideways forces for precisely this reason. It is also the reason that if a space elevator cable broke, it doesnt just fall to Earth and actually it gets quite complicated.

Edited November 30, 2016 by p1t1o
helix, spiral whatever

[Guest]

How would molecular bonds work in Thomson's model (Plum pudding)?

Edited November 30, 2016 by Guest

[K^2]

On 11/29/2016 at 9:16 AM, todofwar said:

How does centrifugal force play into that (yeah yeah, fictitious force, but we are in a rotating reference frame after all)?

If you want to know how much energy it'd take to lift something from surface to GEO straight up, while always staying above the same point on the surface, just use conservation of energy. Total energy at each point is mv²/2 - MGm/r. At surface, v is sidereal velocity and r is Earth's radius. At GEO, v is orbital velocity and r is semi-major axis. M and G are Earth's mass and Gravitational constant respectively. You can look up all of these numbers in Wikipedia, take the difference of energy at GEO and at surface, and this will be the energy you need to supply. As p1t1o points out, in real world, things get a little more complicated because of Coriolis force, but you can ignore that for an estimate.

The neat thing about energy conservation approach is that it completely accounts for all forces, including centrifugal, without you having to even think about it.

Edited December 5, 2016 by K^2

[todofwar]

51 minutes ago, K^2 said:

If you want to know how much energy it'd take to lift something from surface to GEO straight up, while always staying above the same point on the surface, just use conservation of energy. Total energy at each point is mv²/2 - MGm/r. At surface, v is sidereal velocity and r is Earth's radius. At GEO, v is orbital velocity and r is semi-major axis. M and G are Earth's mass and Gravitational constant respectively. You can look up all of these numbers in Wikipedia, take the difference of energy at GEO and at surface, and this will be the energy you need to supply. As p1t1o points out, in real world, things get a little more complicated because of Coriolis force, but you can ignore that for an estimate.

The neat thing about energy conservation approach is that it completely accounts for all forces, including centrifugal, without you having to even think about it.

I know, I just didn't know how to take into account the change in gravitational force and the centrifugal force. Rotating reference frames were always the death of me in physics for some reason. But apparently I get to even ignore that by using the energetics of the orbits themselves. I'm working on an amateur launch system involving a large floating rail gun type assembly and a very week rocket. If I ever get something remotely feasible on paper I'll start a thread where plenty of people can helpfully shoot it down.

[K^2]

17 hours ago, todofwar said:

I know, I just didn't know how to take into account the change in gravitational force and the centrifugal force. Rotating reference frames were always the death of me in physics for some reason. But apparently I get to even ignore that by using the energetics of the orbits themselves. I'm working on an amateur launch system involving a large floating rail gun type assembly and a very week rocket. If I ever get something remotely feasible on paper I'll start a thread where plenty of people can helpfully shoot it down.

You don't need to take into account forces if you are dealing with energy, so long as all forces involved are conservative. (Gravity and centrifugal force are.)

[TheDestroyer111]

When the Rosetta spacecraft descended onto the asteroid it orbited, it was stated that its descent speed was 2 m/s. But then they lost contact with it. But why? 2m/s onto an asteroid is not much, I don't think it could destroy the spacecraft... I can't find any source as to why did they lose contact with the spacecraft.

[andrewas]

There would have been no more science to do after the landing since Rosetta lacked instruments to sample the surface, so it was programmed to shut down after landing. 

[Aghanim]

In Star Wars 7, in opening, we saw Rey mixing some sort of food tablet thingy into water and it magically turns into bread:

At first I thought that it is just some sort of CGI effect, then when I randomly browse the internet, I found this: http://www.mtv.com/news/2728173/star-wars-rey-bread/  that claims that it is a practical effect, not CGI.

So, assuming that we can use any chemical, because the website states that the bread is probably not edible, how can we make something like this?

[p1t1o]

7 hours ago, Aghanim said:

In Star Wars 7, in opening, we saw Rey mixing some sort of food tablet thingy into water and it magically turns into bread:

 

So, assuming that we can use any chemical, because the website states that the bread is probably not edible, how can we make something like this?

My first thought was that they filmed a loaf of bread decomposing and simply reversed the footage - bread grows out of goo!

Or that could literally be a loaf of bread being baked, but the footage is speeded up.

Obviously both methods come with a healthy dose of CGI to make it look just right (the backgrounds, colours etc I dunno) and possibly some other practical touches.

[shynung]

What are the biggest hurdles in designing a working fusion power reactor today, in ELI5 terms?

Edited December 23, 2016 by shynung