# Tag Info

1

In addition to the good answers above, I'd like to add a point about the nuclear reactor losing mass. It's true, it does, albeit at a really tiny rate. The reason that this doesn't contribute to propulsion is that the mass is loss in all directions so the net effect is zero.

2

No, This device would just oscillate (vibrate) When the first two electromagnets switched on the metal ball (which is presumably made out of a ferromagnet) would accelerate towards them but the craft would also accelerate towards the ball (At a lower acceleration as it has more mass). It would then switch to the next electromagnet pair and do the same ...

0

The thrust a rocket can generate is proportional to the mass flow (thus $\frac{dm}{dt}$) and the velocity at which this mass leaves the rocket, often called the effective exhaust velocity, $v_e$. So the sum of all forces acting on the rocket, when also neglecting the fictitious Coriolis and centrifugal forces and assuming a pure vertical ascent, will be ...

3

The main phenomena which limit land vehicle speed are aerodynamic drag, lack of stability, and lack of power. Drag increases (approximately) with the frontal area of the vehicle. Stability increases with the weight, length and width of the vehicle. Finally, power increases (approximately) with the volume of the vehicle. Let's say you keep the shape of the ...

0

At any given moment the thrust is equal to the force experienced by the rocket at that moment. That's the definition of thrust. To get the equation of motion right, you need to consider the instantaneous force (variable) and mass (also variable). The acceleration is then given by $$a(t)=\frac{F(t)}{m(t)}$$

3

Think of shooting a shell from a battleship at an instrument, and designing a device that allows it to hold on and continue working while it is transported by the shell. That's probably easier than doing the same with an asteroid. If you take a look at this table of near-earth asteroid approaches, you'll see that the relative velocity column has units of ...

4

The sense in which a comet could be a "free taxi" is that it is a big source of volatile compounds. A probe that lands on a comet could, in principle, use these to refuel. A comet may be easier to use for this purpose than a large icy moon because of its low gravity.

2

As earlier answers have stated this method doesn't work to save fuel directly, however it might still be useful to hitch a ride on a comet using it for a gravitational assist that the probe would not survive without more shielding than is practical. Placing the body of the comet between the Sun and the probe would presumably shield it from much of the solar ...

1

Fuel, or better energy indeed is the limiting factor. You can steal energy using a gravity assist, but comets simply don't have enough gravity. Jupiter is much more massive, and a gravity assist around Jupiter would in fact make sense for an interstellar probe. Still, it's only a single boost as you fly by Jupiter, and you have to spend the fuel to reach ...

12

As @ChrisWhite has already stated, catching up with Halley's comet offers no appreciable benefit since you'd already need to be traveling at the same speed in a frictionless environment. Where you could get a big boost is if your probe was positioned in the comet's path, either coming across it at an oblique angle or simply stationary, relative to the ...

69

There seems to be a fundamental misunderstanding as to how movement in space works. In space there is no air friction, that is, once you are moving toward your destination, you don't need a continuous source of power to keep going. Landing on a comet doesn't buy you anything, since in order to land you must first match the comet's orbit, at which point the ...

3

Conservation of energy means that it would take you just as much energy to catch up with the comet and make a soft landing (in the ideal case where you ignore the extra fuel costs for landing) as you would need to launch the spacecraft in a similar orbit as the comet. What you could do is let the comet give a bounce to the spacecraft, but I don't think this ...

3

Well, since unburned fuel is not distinguishable from dry mass, we could replace the above equation with $$\Delta v = v_{ex}\ln(\frac{m_i}{m_i - m_b}) - gt$$ where $m_b$ is the mass of the burned fuel. Assuming constant burn rate $r_b$, it would just be $$\Delta v = v_{ex}\ln(\frac{m_i}{m_i - r_bt}) - gt$$

1

prerequisite Thrust produced by a nozzle can be given by $$F_T = \dot{m} V_e + (p_e - p_0)A_e$$ Thrust component in a nozzle can be split into two component that is pressure thrust ($(p_e - p_0)A_e$) and momentum thrust ($\dot{m} V_e$). In most of the nozzles we try to achieve exit pressure equal to ambient pressure, this phenomena is called fully expanded ...

0

The most important equation to remember to use is: $F\Delta t = m\Delta v$ Where "F" is the magnitude of the thrust and "t" is for how long (in this case) constant force is being applied. If the thrust changes based on time, you have to use: $\int F dt = m \int dv$ EDIT: Kyle pointed out a flaw in my answer, so I will try to remedy it. Mass depends on ...

4

It does follow from the conservation of momentum. Consider the diagram (from Wikipedia) of a rocket expelling gas of mass $\Delta m$: At $t=0$, the initial momentum is $$p(t=0)=\left(m+\Delta m\right)V\tag{1}$$ but at $t=\Delta t$, we've lost some mass and gained some velocity,  p(t=\Delta t)=m\left(V+\Delta V\right)+\Delta m \left(V-v_e\right) ...

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