How Do You Calculate Rocket Exhaust Velocity and Total Acceleration at Liftoff?

[AbigailM]

Still preparing for a classical prelim. Not sure if my solution is correct. Any help is more than appreciated.

Homework Statement

According to specifications, the five Saturn V booster engines collectively supplied a liftoff thrust of $34 \times10^{6}N$. The specific impulse was 2580 N-s/kg. The initial mass of the rocket was $3.0\times 10^{6}kg$. Given these specifications, what was the exhaust velocity, and what was the total acceleration including the acceleration of gravity, felt by the astronauts on liftoff (in g's)?
*The specific impulse is the ratio of the thrust to the rate of consumption of propellant. It is nearly constant.

Homework Equations

$v_{ex}=I_{sp}g \hspace{5 mm}\dot{m}=\frac{F_{th}}{I_{sp}}$

$a(t)=\frac{\dot{m}v_{ex}}{m_{0}-\dot{m}t}-g=\frac{F_{th}g}{m_{0}-\dot{m}t}-g$

where $m(t)=m_{0}-\dot{m}t \hspace{5 mm}m(t)a(t)=\dot{m}v_{ex}$

The Attempt at a Solution

$\dot{m}=\frac{3400\times 10^{4}N}{2580\frac{N.s}{kg}}=1.3\times 10^{4}\frac{kg}{s}$

$a(t)=\frac{3400\times 10^{4}N}{3.0\times 10^{6}-1.3\times10^{4}\frac{kg}{s}t}g -g$

Note: Something that confused me was that that the $I_{sp}$ given is in m/s.

Thanks for the help.

 

[Filip Larsen]

When specific impulse is given in unit m/s (which is equal to N-s/kg) it really represent exhaust velocity, as related by your first equation. That is, the exhaust velocity of 2580 m/s corresponds to a specific impulse of 263 s.

Notice, that your second equation is not correct (it misses a "g", or, you should replace I_sp with V_ex). However, since you actually used V_ex in your calculation you ended up with the correct result for the mass rate.

The expression for the rockets acceleration looks correct, but if you want a(t) to represent how much acceleration the astronauts feel (including gravity) you want consider what a(t) should be equal if the rocket has zero acceleration. Also, as I read the last question, it seems you only need to calculate the value of a(0), that is, the felt acceleration at lift-off.

 

[AbigailM]

As always Filip thanks for the help.

Yep it does say a(0) lol, hope I don't miss something like that on an exam.

There ended up being an error in my a(t) equation, because I should have used $v_{ex}=I_{sp}\frac{g}{g}$. When I use that and crunch out the liftoff acceleration ignoring gravity, I get roughly 1.15g's, which is close to the 1.13g's experimental.

And thanks for the gravity hint, I should have put + instead of -. The acceleration vector on the crew is in the same direction as gravity, even though the rocket's acceleration vector is upwards. Newton's third law :(.