How Do You Calculate the Force Exerted on a Rocket at Takeoff?

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To calculate the force exerted on a rocket at takeoff, the mass flow rate of 1000 kg/s and the exhaust velocity of 60,000 m/s are crucial. The change in momentum can be expressed as Δp = Δm * v, where Δm is the mass expelled per second. Since the mass flow rate is constant, the force can be derived from the rate of change of momentum, leading to F = Δm/Δt * v. This means that for each second, the rocket gains momentum equal to the expelled mass multiplied by the exhaust velocity. Understanding these relationships allows for the calculation of the force at takeoff effectively.
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Homework Statement



Calculate the force exerted on a rocket, given that the propelling gases are expelled at a rate of 1000 kg/s with a speed of 60,000 m/s (at takeoff)

Homework Equations


p=mv
F\Delta{t} = \Delta{p}

The Attempt at a Solution



This is probably the first question I am posting where i don't reall know how to start. I am given a ratio of 1000 kg/s which represents no unit that i know of. Momentum is kg x m/s. I don't know what the initial velocity is given for. Presuming for the force at the exact time of takeoff..

I'd appreciate a couple hints..

Thanks,
Senjai
 
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In regards to your first equation, can you convince yourself that the change in momentum is equal to the change in mass if the velocity this mass is being expelled at is constant?That is, \Delta p = \Delta mv?

Now if you can convince yourself that is true, if you divide both sides by \Delta t, what result do you have?
 
I can't, i don't have a Delta t, yes the velocity the mass is being expelled from the rocket is constant, but wouldn't mass be inversely proportional to velocity in this case? so momentum wouldn't change? Uh.. i think i just said it myself.. are you treating 1000 kg as the mass?
 
*bump*
 
1000 kg/s is, as the unit suggests, the rate at which mass is being expelled (with time). hence, kilograms PER second. In other words, it is the mass flow rate.

Δm/Δt = 1000 kg/s = const.​

Which means that EACH second, a mass of Δm = 1000 kg is expelled at speed v = 60 000 m/s, thus carrying away with it an amount of momentum equal to Δm*v.

Conservation of momentum therefore dictates that EACH second, the rocket gains the same amount of momentum Δm*v in the opposite direction.

So now you know the change in momentum over a time interval of Δt = 1s. This means that you know the rate of change of momentum (= force).
 
F=dMV/dt=dM/dt*v .You have the information you need.

You beat me to it cepheid
 
Thanks bro... Much Appreciated.
 

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