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27

Start by considering what is seen by the people watching you from the Earth. Nothing can travel faster than the speed of light, $c$, so the quickest you could get to Kepler 186f would be if you were travelling at $c$ in which case it would take 490 years. In practice it would take longer than this because you have to accelerate from rest when you leave the ...


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This is a fun, high-quality qualifying exam question. The algebra is not hard; the physical insight takes some real thought; there are many ways to be partly right. Here's my take on it. Acceleration From Einstein's equation $E^2 = p^2 + m^2$ we have for each photon $E = p = hf_0$ (in the reference frame of the laser). We can use the power of the laser ...


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Might boron carbide perhaps work? Its pretty strong, neutron absorbing, 3,000 K melting temperature, Mohs hardness is around 9.5. Density is around 2.5 g/cm^2. It is also relatively easy to produce.


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The question needs to be turned around. Why is the combustion chamber bigger than the throat? This is to allow time for combustion before the gas exits the engine.


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As addition to @tpg2114 answer, I suggest also to read about de Laval nozzle and Rocket engine nozzle on wikipedia. Some typical values of the exhaust gas velocity for rocket engines burning various propellants are: 1.7 to 2.9 km/s (3800 to 6500 mi/h) for liquid monopropellants 2.9 to 4.5 km/s (6500 to 10100 mi/h) for liquid bipropellants 2.1 to 3.2 km/s ...


62

The whole point to the throat is to increase the exhaust velocity. But not just increase it a little bit -- a rocket nozzle is designed so that the nozzle chokes. This is another way of saying that the flow accelerates so much that it reaches sonic conditions at the throat. This choking is important. Because it means the flow is sonic at the throat, no ...


1

This concept has gotten consideration from NASA. In the NASA MagLifter concept, a 300-600 miles per hour speed on a superconducting magnetic levitation track appoximately 2.5 miles long and going up a mountain to about 10,000 feet is proposed. The option of using a helium filled tunnel to reduce drag was also proposed.


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4.Without air resistance, how much energy would be required to get a given mass to this altitude? Energy Needed ($E$) = Potential Energy at L1 ($V_{L1}$) - Potential Energy at Earth's Surface ($V_e$) $$V_e = -Gm(\frac{M_e}{r_e} +\frac{M_l}{LD - r_e}) $$ $$V_{L1} = -Gm(\frac{M_e}{d_{L1}} +\frac{M_l}{LD - d_{L1}}) $$ where $m$ is the transported mass, ...


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I assuming that the acceleration remains constant. So using Newton's equations- $$v=at$$ So after the fuel runs out, $$v=(10)(60)=600m/s$$ The distance it travelled in this time can be calculated by $$s=1/2at^2=(0.5)(10)(60)^2=18000m$$ Now, after the fuel runs out, the rocket will still go up a little bit, that can be calculated by- $$v^2-u^2=2gs$$ The $u$ ...



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