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Quote from: alright1234 on 07/05/2019 04:55:52While your at it tell me what is causing the halve circle path of the CSML and the CSML is propagating with a constant velocity since why would you use more fuel? Also, Space X thought that a rocket engine has a 50% efficiency.If you are asking why why it follows a curved trajectory, then the answer is Earth's and Moon's gravity. Early on the Earth's gravity dominates, but as it moves further away this effect becomes weaker as the Moon's becomes stronger. The craft would have to burn fuel in order to maintain a constant speed. As it is, it is on a ballistic trajectory and has to give up velocity in exchange for climbing higher in the Earth's gravity field. Upon the beginning of the trans-lunar trajectory it is moving at something over 10.9 km/sec ( not that much shy of escape velocity) . By the time it gets to near Moon space, it will have given up almost all of that speed during the long climb against Earth's gravity.
While your at it tell me what is causing the halve circle path of the CSML and the CSML is propagating with a constant velocity since why would you use more fuel? Also, Space X thought that a rocket engine has a 50% efficiency.
Quote from: Janus on 07/05/2019 06:16:00Quote from: alright1234 on 07/05/2019 04:55:52While your at it tell me what is causing the halve circle path of the CSML and the CSML is propagating with a constant velocity since why would you use more fuel? Also, Space X thought that a rocket engine has a 50% efficiency.If you are asking why why it follows a curved trajectory, then the answer is Earth's and Moon's gravity. Early on the Earth's gravity dominates, but as it moves further away this effect becomes weaker as the Moon's becomes stronger. The craft would have to burn fuel in order to maintain a constant speed. As it is, it is on a ballistic trajectory and has to give up velocity in exchange for climbing higher in the Earth's gravity field. Upon the beginning of the trans-lunar trajectory it is moving at something over 10.9 km/sec ( not that much shy of escape velocity) . By the time it gets to near Moon space, it will have given up almost all of that speed during the long climb against Earth's gravity.Checkmate king two.
So, now that you know you are wrong, are you going to apologise?
Quote from: Janus on 07/05/2019 15:58:42Quote from: Petrochemicals on 07/05/2019 07:04:10Quote from: Janus on 07/05/2019 00:35:44The math is quite simple: An object in a circular orbit has an orbital energy of E= -um/2r , where u is the gravitational parameter for the body it is orbiting, m is its mass, and r it orbital radius. This is the same as the Sum of its kinetic energy mv^2/2 and its gravitational potential energy -um/r or E= mv^2/2-u/r . It reduces to the first equation when you substitute sqrt(u/r), the circular orbital for v in the last equation.For an object sitting on the surface of the Earth, its energy is just E= -um/re, where re is the radius of the the Earth. Thus the energy difference between an object sitting on the surface of the Earth and one in a circular orbit is:E = -um/2r-(-um/re) = um/re-um/2r = um(1/re-1/2r)u = 3.987e14 for the Earth, re= 6378,000m and r = 6578000m if m=1000kg, then energy difference isE= 3.987e14(1000)(1/63378000-1/(2(6578000)) = 3.22e10J This is a bit more than the value I gave earlier as that value was just for the transfer from surface to LEO, and neglected the boost needed to circularize the orbit. It's both rocket science and orbital mechanics. Well if you do mgh to geosync orbit, at 35000km G=0.75 @~r=20000km you get something like 3x10 per tonne. This still seems a bit high for me, suppose it has to do with the fact that once out of the atmosphere the object already has significant velocity due to the fact it is orbiting at the speed of planet rotation and the planets solar orbit etc etcAgain, you are neglecting the energy needed for the craft to attain orbital velocity. If you were to assume no obstacles or atmosphere, you could put an object in orbit around the Earth just above its surface, but you would have to accelerate it up to a bit over 7.9 km/sec n order to do so. This, in of itself, would require 3.13e10 joules per 1000 kg of mass. mgh doesn't even come into play because you never raised the object any height in Earth's gravity field. To attain GEO requires 5.78e10 joules per 1000 kg, 2.56e10 joules per 1000 kg more than the 200 km LEO orbit. But this, like the mgh discrepancy creates a problem. The energy content of diesil fuel being 4.9x10^10per tonne, halved ish for the oxidiser giving 2.5x10^10 per fuel tonne burned. You would never ever make orbit if it required any x10^10 number due to inefficiency and container weights. Same goes for the mgh to the moon !
Quote from: Petrochemicals on 07/05/2019 07:04:10Quote from: Janus on 07/05/2019 00:35:44The math is quite simple: An object in a circular orbit has an orbital energy of E= -um/2r , where u is the gravitational parameter for the body it is orbiting, m is its mass, and r it orbital radius. This is the same as the Sum of its kinetic energy mv^2/2 and its gravitational potential energy -um/r or E= mv^2/2-u/r . It reduces to the first equation when you substitute sqrt(u/r), the circular orbital for v in the last equation.For an object sitting on the surface of the Earth, its energy is just E= -um/re, where re is the radius of the the Earth. Thus the energy difference between an object sitting on the surface of the Earth and one in a circular orbit is:E = -um/2r-(-um/re) = um/re-um/2r = um(1/re-1/2r)u = 3.987e14 for the Earth, re= 6378,000m and r = 6578000m if m=1000kg, then energy difference isE= 3.987e14(1000)(1/63378000-1/(2(6578000)) = 3.22e10J This is a bit more than the value I gave earlier as that value was just for the transfer from surface to LEO, and neglected the boost needed to circularize the orbit. It's both rocket science and orbital mechanics. Well if you do mgh to geosync orbit, at 35000km G=0.75 @~r=20000km you get something like 3x10 per tonne. This still seems a bit high for me, suppose it has to do with the fact that once out of the atmosphere the object already has significant velocity due to the fact it is orbiting at the speed of planet rotation and the planets solar orbit etc etcAgain, you are neglecting the energy needed for the craft to attain orbital velocity. If you were to assume no obstacles or atmosphere, you could put an object in orbit around the Earth just above its surface, but you would have to accelerate it up to a bit over 7.9 km/sec n order to do so. This, in of itself, would require 3.13e10 joules per 1000 kg of mass. mgh doesn't even come into play because you never raised the object any height in Earth's gravity field. To attain GEO requires 5.78e10 joules per 1000 kg, 2.56e10 joules per 1000 kg more than the 200 km LEO orbit.
Quote from: Janus on 07/05/2019 00:35:44The math is quite simple: An object in a circular orbit has an orbital energy of E= -um/2r , where u is the gravitational parameter for the body it is orbiting, m is its mass, and r it orbital radius. This is the same as the Sum of its kinetic energy mv^2/2 and its gravitational potential energy -um/r or E= mv^2/2-u/r . It reduces to the first equation when you substitute sqrt(u/r), the circular orbital for v in the last equation.For an object sitting on the surface of the Earth, its energy is just E= -um/re, where re is the radius of the the Earth. Thus the energy difference between an object sitting on the surface of the Earth and one in a circular orbit is:E = -um/2r-(-um/re) = um/re-um/2r = um(1/re-1/2r)u = 3.987e14 for the Earth, re= 6378,000m and r = 6578000m if m=1000kg, then energy difference isE= 3.987e14(1000)(1/63378000-1/(2(6578000)) = 3.22e10J This is a bit more than the value I gave earlier as that value was just for the transfer from surface to LEO, and neglected the boost needed to circularize the orbit. It's both rocket science and orbital mechanics. Well if you do mgh to geosync orbit, at 35000km G=0.75 @~r=20000km you get something like 3x10 per tonne. This still seems a bit high for me, suppose it has to do with the fact that once out of the atmosphere the object already has significant velocity due to the fact it is orbiting at the speed of planet rotation and the planets solar orbit etc etc
The math is quite simple: An object in a circular orbit has an orbital energy of E= -um/2r , where u is the gravitational parameter for the body it is orbiting, m is its mass, and r it orbital radius. This is the same as the Sum of its kinetic energy mv^2/2 and its gravitational potential energy -um/r or E= mv^2/2-u/r . It reduces to the first equation when you substitute sqrt(u/r), the circular orbital for v in the last equation.For an object sitting on the surface of the Earth, its energy is just E= -um/re, where re is the radius of the the Earth. Thus the energy difference between an object sitting on the surface of the Earth and one in a circular orbit is:E = -um/2r-(-um/re) = um/re-um/2r = um(1/re-1/2r)u = 3.987e14 for the Earth, re= 6378,000m and r = 6578000m if m=1000kg, then energy difference isE= 3.987e14(1000)(1/63378000-1/(2(6578000)) = 3.22e10J This is a bit more than the value I gave earlier as that value was just for the transfer from surface to LEO, and neglected the boost needed to circularize the orbit. It's both rocket science and orbital mechanics.
I calculated the efficiency using the Saturn rocket and got .004%.
While your at it tell me what is causing the halve circle path of the CSML
Quote from: alright1234 on 07/05/2019 20:22:09I calculated the efficiency using the Saturn rocket and got .004%.Show the math.Quote from: alright1234 on 07/05/2019 21:34:15While your at it tell me what is causing the halve circle path of the CSMLThe same thing that causes all orbits.
The earth's gravity has a negligible effect on the 50,000 kg CSML.
Then why are astronaut in the ISS weightless?
Why does not a penny in the ISS stick to the walls?
OK, that makes sense except for the expressed-as-power part. It should be expressed as total work. Power is the rate that the work is performed, which is often quite low for the most efficient engines.
You are making the same mistake as some eminent scientist(I can't remember who)did when trying to claim that a rocket could never escape the Earth. He worked out how much energy would be needed to accelerate 1 kg to escape velocity, and compared to to amount of energy that could be released by 1 kg of Nitro-glycerine, the most powerful explosive of the time. He showed that this was less than the energy needed to get the 1 kg of nitro-glycerine up to escape velocity. He argued that if nitro-glycerine couldn't even get its own mass up to escape velocity, how could you get a rocket up to it? What he failed to realize is that with a rocket, you don't need to get the entire mass of your fuel up to escape velocity. The vast majority of the fuel mass gets left behind near the surface of the Earth. For a rocket it is all about having enough thrust to lift your rocket and the exhaust velocity. Exhaust velocity determines what kind of mass-ratio you need in terms of fuel to payload. It comes down to the rocket equation dV = Ve ln(MR) dV is the velocity change of your rocket, Ve is its exhaust velocity, MR is the mass ratio ( Fueled rocket mass/ dry mass of rocket). In addition, the RP-1 rocket fuel (kerosene equivalent) was only used in the first stage of the Saturn V, the remaining stages used liquid Hydrogen, which allow for an ISP of 421 sec vs, the lower ISP of 263 sec for the first stage ( to get exhaust velocity from ISP, multiply it by g.)Regardless of what you think, this is the actual science behind this.
Quote from: alright1234 on 07/05/2019 22:21:53Then why are astronaut in the ISS weightless?The same reason that a ball dropped off of the Empire State Building is weightless: because they are in free fall.