cameron marical Posted February 2, 2009 Share Posted February 2, 2009 if i were to launch a rocket out my backyard when would it have to be if it was to get to the moon. i know it would blast me and my neghborhood to bits:doh: but i just want an equation and an answer. lets say i live in new york. the rocket goes 17,000 mph, straight, no stoping till the moon. it has just enough fuel to reach the moon. it weighed 1.5 tons. (if that matters) please, if you can, i want an equation with the answer so i can play with this for a while. and if im missing anything about the imagenary rocket or setup, just make something up and list it. thanks, if your cool enough to do it. Link to comment Share on other sites More sharing options...
CaptainPanic Posted February 2, 2009 Share Posted February 2, 2009 The escape velocity of the earth is 11000 m/s (11km/s). The velocity you wrotre, 17000 mph (or 7.6 km/s) won't be enough. You're asking about the time to launch to get the right trajectory, I guess? Here's a picture of a trajectory to the moon (keywords in google: trajectory apollo, then go for pictures). You can see that there are in fact 2 launch moments: 1. The lift off, to get into the right orbit. 2. The second launch to get out of orbit, and go to the moon. And sorry for not providing equations for this. I don't know them. The mass of the rocket definitely matters a lot. You wrote 1.5 tonnes (1500 kg, please use SI/metric units, rockets have crashed because units were mixed up). Is that the mass including fuel, or excluding? More than 90% of the mass of the rocket is fuel (google for saturn V to get an idea: a >100 m tall rocket, for such a small lunar lander). So the calculations for the thrust of the rocket must take into account the changing mass. Keywords are "specific thrust" and "specific impulse". I think you have to make a force balance, including the thrust, and gravity. Then solve that along the trajectory including the changing mass (I've never done this myself, but I think you'll end up with a differential equation that is dF/dx, with F = force, x = trajectory, and also ballistics equations - I ask a physics expert to comment on this if possible). Probably, you should include 3 length units, either in the xyz, or polar coordinates. Then you get a set of equations which need to be solved simultaneously... Anyway... it's too much for me to solve here (I'm supposed to be working on different topics than this ). Good luck, if you get some result, please post it! And if anyone thinks that they have a better answer, I thank you in advance. Here are some links for specific thrust and impulse: http://www.grc.nasa.gov/WWW/K-12/airplane/specth.html http://en.wikipedia.org/wiki/Specific_impulse http://exploration.grc.nasa.gov/education/rocket/ienzl.html http://en.wikipedia.org/wiki/Specific_thrust Link to comment Share on other sites More sharing options...
cameron marical Posted February 3, 2009 Author Share Posted February 3, 2009 thanks man ill get to work. 1 Link to comment Share on other sites More sharing options...
CaptainPanic Posted February 3, 2009 Share Posted February 3, 2009 thanks man ill get to work. That is just the coolest reply on this forum yet. I explain that you need a 100m tall rocket, going 11000 km/s, two launch moments, nightmare calculations, and you just reply: thanks man ill get to work. Awesome attitude. Link to comment Share on other sites More sharing options...
John Cuthber Posted February 3, 2009 Share Posted February 3, 2009 thanks man ill get to work. "We choose to go to the moon in this decade and do the other things, not because they are easy, but because they are hard". Link to comment Share on other sites More sharing options...
DrP Posted February 4, 2009 Share Posted February 4, 2009 You're asking about the time to launch to get the right trajectory, I guess? Here's a picture of a trajectory to the moon (keywords in google: trajectory apollo, then go for pictures). You can see that there are in fact 2 launch moments: 1. The lift off, to get into the right orbit. 2. The second launch to get out of orbit, and go to the moon. And sorry for not providing equations for this. I don't know them. You're over complicating matters - see diagram: Link to comment Share on other sites More sharing options...
CaptainPanic Posted February 4, 2009 Share Posted February 4, 2009 LOL Note to self: Don't ever ask DrP to fix a bike. Link to comment Share on other sites More sharing options...
Janus Posted February 5, 2009 Share Posted February 5, 2009 You're over complicating matters - see diagram: Actually, a straight from Earth surface to Moon shot is more complicated then the two step method mentioned. The launch window would have to be really precise. Not to mention having to account for effects such as wind aloft throwing your launch trajectory off. Link to comment Share on other sites More sharing options...
Airbrush Posted February 5, 2009 Share Posted February 5, 2009 This place is amazing. Someone asks what I felt was a silly question, and knowledgeable people bend over backwards trying to help. Generous folks. Link to comment Share on other sites More sharing options...
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