# Tag Info

8

Does an ion thrust engine consume more energy as it speeds up? The answer to this question is no. So when it hits the top speed what is the bottle neck? The bottleneck is that the vehicle runs out of propellant. The problem is described by the rocket equation, $$\frac {\Delta v}{v_e} = \ln\frac{m_{\text{initial}}}{m_{\text{final}}}$$ Where ...

5

$$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v$$ So the differential equation with x as the independent variable becomes $$v(v'+1)=1$$

5

That page is not well written. The 90km/s speed is the exhaust velocity of the engine. It is not the maximum speed of the spacecraft. There is no maximum speed of the spacecraft, short of the speed of light. They make the mistake again when they say: "While a chemical rocket's top speed is limited by the thermal capability of the rocket nozzle" ADDED: ...

4

It isn't clear from your question exactly what you are integrating and how, but this is the way to tackle problems like this. You know that: $$\frac{dv}{dt} = -kv$$ The way to solve equations like this one is to rearrange it by dividing both sides by $v$ and multiplying both sides by $dt$ to get: $$\frac{1}{v}dv = -k\,dt$$ Now we can integrate both ...

3

It's pretty natural to think that a star can have velocity - there's no reason a star shouldn't be able to move. The first thing you need to know is "velocity relative to what?" Stars in our galaxy are all in some kind of orbit around the galaxy, so you can talk about velocity in galactic coordinates. Binary stars orbit each other, so you can talk about ...

2

Place the two objects on a Cartesian $(x,y)$ coordinate system, as shown below: $v_1$ and $v_2$ are the scalar values (the magnitude, if you prefer) of the velocity vectors, $\alpha$ and $\beta$ the angles between the respective vectors and the $x$ axis. We can now calculate the $x$ and $y$ projections (components) of the vectors: $x$ projections: ...

2

You throw the ball upwards with velocity $v$ and it returns to your hand with velocity $-v$. Let's draw a graph showing the velocity as a function of time: Acceleration is defined as: $$a = \frac{dv}{dt}$$ so it is the gradient of the line in this graph. The velocity-time line is straight so the gradient is constant which means the acceleration is ...

2

Physically, it tells you the direction in spacetime that the object is going. Because it is a unit vector and it points the direction in spacetime the object is going. Recall how in 3d you can represent the direction of a velocity vector with a unit vector? You could choose any orthonormal basis you like and then the components were ...

2

When calculating any change, the proper order is new - old: $$\Delta S = S_\text{new}-S_\text{old} \text{ or } S_\text{final}-S_\text{initial}.$$ $S$ represents any scaler or vector quantity.

1

Firstly, it's hard to imagine the first particle alone getting all the kinetic energy: the most obvious scenario is with the atom at rest, so the total momentum of all the fragments in the center of mass frame is nought. However, supposing it were possible for the first particle to get all the kinetic energy, then the total energy of the first particle is ...

1

The limit in either case is when you run out of fuel. By the Tsiolkovsky rocket equation, if all else is the same, the top speed of a rocket is proportional to the exhaust velocity. So, the faster a rocket ejects its (often burning) exhaust, the higher the final speed when the rocket runs out of fuel. As a metaphor, imagine being in a boat filled with ...

1

Lorentz boosts and inner products First, it helps to know a little bit about relativity: relativity is a slight tweak of our normal understanding of physics, where if I put an infinite grid of clocks around you and carefully desynchronize them so that, from your perspective, you see them all "tick" at the same time and show the same amount no matter how far ...

1

You can probably assume that you can neglect air friction and that the acceleration due to gravity a constant -9.8 m/s^2. Also, from the wording of the question, it sounds like you can assume that the maximum height reached was 2 miles. Using your intuition, do you see how all this information results in a unique trajectory? Write down the general equation ...

1

Equivalence requires acceleration to curve space the same way gravitating mass does. Hermann Weyl, Zur Gravitationstheorie, Annalen der Physik, 54, 117, (1917) argued that kinetic energy should curve space just as gravity and electromagnetic fields do by entering into the stress energy tensor. The concept of inertial mass increasing at high speed has been ...

1

An accelerating object has a changing velocity. Obviously so since the object starts with zero velocity and the velocity increases with time according to the SUVAT equation: $$v = u + at$$ So your equation 1.1 is no use here. It calculates the average velocity. This could actually be used to calculate the acceleration, but the working is a bit involved ...

1

Your equation 1.1 can be used with constant velocity. Here you have to use the $2^{\text{nd}}$ equation. ie $a = 2d/(t^2)$. So, the answer is $118.4 \, \text{cm}/s^2$.

1

A photon is a quantum mechanical elementary particle and follows quantum mechanical formulae, not classical ones. In quantum mechanics the only way an elementary particle can change direction is through an interaction with another elementary particle or field. The interaction is shown with feynman diagrams which give the integrals that have to be calculated ...

1

When you shoot the ball upwardly, gravity acts on it with a force $mg$ where $m$ is the mass of the ball and $g=9.81 ms^{-2}$ the Earth's gravitational acceleration. If the initial upward velocity was $v_0$ then the instantaneous velocity $v$ is given by: $v=v_0-gt$, so after some time $t=\frac{v_0}{g}$ the balls's velocity becomes $v=0$. However, we know ...

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