Rocket Motion: Solving the Rocket Equation for Lift-Off

AI Thread Summary
To determine when a rocket will lift off, the initial mass is 70,000 kg, with a fuel burn rate of 250 kg/s and an exhaust velocity of 2,500 m/s. The correct rocket equation to use is v = 2500 * ln(70000/(70000 - 250t)) - 9.8t. The rocket will lift off when the velocity becomes positive, which occurs at approximately 47.9 seconds. An alternative approach involves calculating the thrust force and equating it to the weight of the rocket, leading to a lift-off time of about 25 seconds. Both methods highlight the importance of correctly applying the rocket equation and understanding the forces involved.
SlickJ
Messages
6
Reaction score
0
This question was posed in class the other day for extra credit:

A rocket with initial mass 70000kg burns fuel at a rate of 250kg/s; it has an exhaust velocity of 2500m/s. If the rocket is at rest, how long after the engines fire will the rocket lift off?

I've been trying to solve it using some variation of the rocket equation:
v=v(exhaust) * ln(M(0)/M(t)) - gt
but to no success.
Any hints or help would be greatly appreciated
 
Physics news on Phys.org
There is a problem with the equation you are using
it is r (rate of burning the fuel) for g you used, also it is dimensionally incorrect we can have log of only natural numbers.
If it works that's Ok, otherwise try by equating force by the ejected gass to mass of rocekt at the moment.
 
SlickJ said:
This question was posed in class the other day for extra credit:

A rocket with initial mass 70000kg burns fuel at a rate of 250kg/s; it has an exhaust velocity of 2500m/s. If the rocket is at rest, how long after the engines fire will the rocket lift off?

I've been trying to solve it using some variation of the rocket equation:
v=v(exhaust) * ln(M(0)/M(t)) - gt
but to no success.
Any hints or help would be greatly appreciated

Be careful. The equation of your rocket is:

v=2500\cdot ln\Big(\frac{70000}{70000-250t}\Big)-9.8t

There is an interval of velocities 0<t<47.9 s in which v<0. The rocket will start to lifting off when v>0 or t>47.9 s. (solve numerically the equation v=0, you will obtain t=0 and t=47.9 s).
 
Thanks both of you for your help, very much appreciated.
 
I think you're all making this much too complicated.
Just calculate the force the rocket produces using the equation force = change in momentum / time, with knowledge of the fact that momentum is mass times velocity. The weight of the rocket is given by acceleration due to gravity x (original mass of rocket - (rate at which the rocket loses mass x time). Set the weight of the rocket to equal the its force of propulsion (which you already calculated) and solve for time. At that time, the force of thrust will balance the rocket's weight and immediately afterwards it will start lifting off.
 
Last edited:
I get about 25 seconds.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top