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Rocket Launch Speed 
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#1
Jul2010, 03:24 PM

P: 3

Hey Guys,
I have a problem, Say I have a rocket that has a mass about 350kg and produces 1250kg of thrust and I launch it horizontally, I know that 1250/350 gives me the acceleration in m/s2 (roughly 3.6 m/s2) but how do I then work out the speed? Also if I where to launch the rocket vertically would I need a different equation to incorporate the force of gravity and if so how would I do that? Thanks for your help! Andrew 


#2
Jul2010, 05:26 PM

P: 42

the expression of the rocket propulsion is that
V_{f}  V_{i} = V_{e} ln([tex]\frac{Mi}{Mf}[/tex] ) I hope it helps 


#3
Jul2010, 05:26 PM

P: 42

yea Ve is the escape velocity



#4
Jul2110, 12:42 AM

P: 3

Rocket Launch Speed
Thanks again for your help! Andrew 


#5
Jul2110, 04:44 AM

Sci Advisor
P: 2,470

Why should it be 3.6m/s² and not, say ft/s²? Or miles/hr²? Or any other unit of acceleration? If you take 1250kg/350kg = 3.6. There are no units. You get a unitless quantity for something that should have a unit. That should be your first warning sign. You cannot have 1250kg of thrust. Thrust is a force, and kg are units of mass. Two entirely different quantities. What you really have is the amount of thrust equivalent to weight of 1250kg under normal Earth gravity. That's 1250kg * 9.8m/s² = 12,250N of thrust. Now the acceleration is F/m = 12,250N/350kg = 35m/s². That's your acceleration. Note that this is roughly 3.6G, which is really what you get when you divided the two numbers above. If the thrust and the mass of the rocket remain constant, then the final velocity is just acceleration * time. So if your rocket was running for 5s, the final velocity would be 35m/s² * 5s = 175m/s. Notice that the units still work out. However, a real rocket produces thrust by burning fuel, so its mass will decrease. That's why you generally need to use the equation given by the_storm. In that equation, M_{i} = Initial mass of the rocket. M_{f} = Final mass of the rocket. v_{e} = Exhaust velocity. Also known as Specific Impulse. v_{i}v_{f} = Change in velocity you get between M_{i} and M_{f} states of the rocket. ln(x) = Natural logarithm. The problem, of course, is finding v_{e} for your rocket. It's not always given. However, most rocket engines give total impulse. For a rocket with constant thrust, total impulse can be found by multiplying thrust by amount of time the engine runs. Then, if you take that total impulse and divide it by (M_{i}M_{f}) you will get exactly v_{e}. Note that if total impulse is given, this will work even if total thrust is not constant, because thrust usually varies due to the rate at which the fuel is burnt, and not due to the changes in v_{e}. If you are just curious, I hope that clears a few things. If you have a more specific problem in mind, you'd probably have to share some more details before we can help you better. 


#6
Jul2110, 06:16 AM

P: 42

all of what you said is correct :).
physicsadict listen to K^2.  Best regards 


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