# Tag Info

0

Why would you need a rail? If you're launching off of a mountain, it would reduce the amount of fuel to reach orbit. That in itself may be a more cost-effective investment.

-1

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.

-1

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, ...

1

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 $u$ ...

0

To calculate the energy required, use the law of preservation of energy, which dictates that: Potential energy at earth's surface + kinetic energy at the surface + invested energy = potential energy in L1 + kinetic energy in L1 Potential energy = $-\frac{GM_{earth}m}{r_{earth}} -\frac{GM_{moon}m}{r_{moon}}$ Kinetic energy at earth's surface = $\frac 1 2 m ... 0 The recoil? From massless photons? How do you figure? a) Instantaneous Acceleration:$0\ m/s^2$b) Power:$P_0$c) It will never reach$0.9c$, and if it did it would have the same rest mass,$m_{rocket}$Let's use the Tsiolkovski Rocket Equation:$\Delta v = v_\text{e} \ln \frac {m_0} {m_1}\$ https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation ...

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