# Tag Info

51

The problem is what Konstantin Tsiolkovsky discovered 100 years ago: as speed increases, the mass required (in fuel) increases exponentially. This relation, specifically, is $$\Delta v=v_e\ln\left(\frac{m_i}{m_f}\right)$$ where $v_e$ is the exhaust velocity, $m_i$ the initial mass and $m_f$ the final mass. The above can be rearranged to get  ...

19

TL;DR: This answer arrives at roughly the same conclusion as Kyle Kanos', i.e. in addition to payload considerations, the difficulty lies in stuffing a small rocket with a mass of fuel exceeding to the mass of the rocket itself. This answer, however, is more rigorous in how the $\Delta v$ budget is treated. Developing a relationship between rocket and ...

6

Courtesy of the book Carl found we have an answer! Consider the element of the liquid helium at a height $h$ above the fluid surface and distance $y$ from the wall. To raise that element above the fluid surface costs an energy $mgh$, but because there is a Van der Waals attraction between the helium atoms and the wall you get back an energy $E_{VdW}$. ...

3

Because most payloads are quite heavy. I am not sure what kind of payloads you had in mind, I am no expert on this, but I think that most launches contain satellites, which might be heavier then you think, for instance the satellite in this BBC Documentary weighs 6000 kg. And according to Wikipedia, miniaturized satellites weigh less than 500 kg (so heavier ...

2

This is an experimentalist's answer and yes, accelerated charged particles either in stable circular orbits or in linear acceleration do radiate. Classically, any charged particle which moves in a curved path or is accelerated in a straight-line path will emit electromagnetic radiation. Various names are given to this radiation in different contexts. For ...

2

Possibly helpful: http://arxiv.org/ftp/arxiv/papers/1103/1103.0517.pdf www.paper.edu.cn/download/downPaper/200812-856‎ The bizarre behaviour of superfluids! Climbing up walls and geting out of glass beakers EDIT: A googlebooks excerpt seems more useful:

2

There are some very interesting subtleties here. Let's analyze the situation very carefully. Let's choose our system to consist of the block, spring, and Earth. By choosing the Earth and block to be in our system, we will have a change in gravitational potential energy. In the beginning, the (massless) spring hangs vertically with a block of mass $m$ ...

1

The difference in potential energy is due to different definitions of what $x=0$ means. Since from the perspective of the spring this would be when the spring is not compressed nor stretched (rest length). However in the case of a mass-spring system in a gravity field (assumed to be a constant acceleration, $g$) this position is often chosen to be the ...

1

If you call $\chi$ the exergy (availability) then $\chi = U + p_o V - T_o S$ where $p_o, T_o$ are the pressure and temperature of the environment (and are assumed to be constant). To find the maximum amount of useful work that can be extracted form the system it is sufficient to analyze reversible processes only so that $dU=TdS-pdV$ and then the exergy ...

1

They're not the same thing. They have very different implications. You can imagine Force and thus Momentum as the "push" that will happen to the target, while Kinetic energy is the damage it causes. E(k) is equal the Work the object will perform, let it be penetration, fracture, etc. As soon as the object hits the target, the E(k) applies (i.e. the ...

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