Classical Mechanics: Rocket Propulsion Calculation

[I_Try_Math]

Homework Statement

78. A rocket takes off from Earth and reaches a
speed of 100 m/s in 10.0 s. If the exhaust speed
is 1500 m/s and the mass of fuel burned is 100
kg, what was the initial mass of the rocket?
79. Repeat the preceding problem but for a rocket
that takes off from a space station, where there
is no gravity other than the negligible gravity
due to the space station.

Relevant Equations

Conservation of momentum

My textbook says the correct answer for #79 is 1551 kg but I get 1600 kg.

I just attempted to solve it using conservation of momentum. Can't see where the math is incorrect.

 

[haruspex]

The given fuel exhaust velocity is relative to the rocket. Its velocity relative to the ground reduces as the rocket speeds up.
Use the rocketry equation.

 

[I_Try_Math]

The given fuel exhaust velocity is relative to the rocket. Its velocity relative to the ground reduces as the rocket speeds up.
Use the rocketry equation.

Okay now I see, which if I understand correctly means my equation is wrong because the exhaust velocity in it is incorrect. I'll try to derive the rocket equation myself and use it.

 

[Orodruin]

Arguably though, the given answer has way too many significant digits given the input and should be rounded to 1600 kg … a little bit depending on how many significant digits you consider the input to have ..

 

[I_Try_Math]

The given fuel exhaust velocity is relative to the rocket. Its velocity relative to the ground reduces as the rocket speeds up.
Use the rocketry equation.

I'm still having a bit of trouble wrapping my head around why it can't be found this way. Is it incorrect to say that at t=10 seconds, relative to the space station the velocity of the exhaust gas would be -1400 m/s?

 

[I_Try_Math]

I'm still struggling with this problem. I haven't been able to figure out why it isn't possible to solve it with a simple conservation of momentum calculation in the way I tried to above.

 

[jbriggs444]

I'm still having a bit of trouble wrapping my head around why it can't be found this way. Is it incorrect to say that at t=10 seconds, relative to the space station the velocity of the exhaust gas would be -1400 m/s?

The velocity of the most recently expelled exhaust gas would be -1400 m/s in that frame, yes. However, the exhaust stream as a whole does not share that single velocity. The exhaust stream is not a rigid rod. It is an expanding cloud of gasses.

 

[I_Try_Math]

The velocity of the most recently expelled exhaust gas would be -1400 m/s in that frame, yes. However, the exhaust stream as a whole does not share that single velocity. The exhaust stream is not a rigid rod. It is an expanding cloud of gasses.

Thank you for the reply. I don't doubt your statement is true. It still seems counterintuitive. I guess my question would be if the most recently expelled gas is going -1400 m/s what force acts on it change this? It must be from the gas expelled after?

 

[jbriggs444]

Thank you for the reply. I don't doubt your statement is true. It still seems counterintuitive. I guess my question would be if the most recently expelled gas is going -1400 m/s what force acts on it change this? It must be from the gas expelled after?

We can safely assume that the exhaust velocity of -1500 m/s is always measured relative to the then-current velocity of the craft.

At the beginning of the burn, the craft is moving at 0 m/s relative to the station and the exhaust is moving at -1500 m/s relative to the station.

At the end of the burn, the craft is moving at 100 m/s relative to the station and the exhaust is moving at -1400 m/s relative to the station.

As you can plainly see from the numbers, the far end of the exhaust stream (emitted first) is moving rearward faster than the near end of the exhaust stream (emitted last).

To be clear, there is no more gas emitted after the last gas is emitted. That final bit of exhaust gas has the velocity it has because that is the velocity with which it was emitted. No subsequent forces need act to change that velocity. It is just that the rocket nozzle from which it was expelled was moving when that gas was expelled.

 

[I_Try_Math]

We can safely assume that the exhaust velocity of -1500 m/s is always measured relative to the then-current velocity of the craft.

At the beginning of the burn, the craft is moving at 0 m/s relative to the station and the exhaust is moving at -1500 m/s relative to the station.

At the end of the burn, the craft is moving at 100 m/s relative to the station and the exhaust is moving at -1400 m/s relative to the station.

As you can plainly see from the numbers, the far end of the exhaust stream (emitted first) is moving rearward faster than the near end of the exhaust stream (emitted last).

To be clear, there is no more gas emitted after the last gas is emitted. That final bit of exhaust gas has the velocity it has because that is the velocity with which it was emitted. No subsequent forces need act to change that velocity. It is just that the rocket nozzle from which it was expelled was moving when that gas was expelled.

Very true. I was having trouble visualizing it but that way of breaking it down makes it a lot easier for me to grasp.

 

[Orodruin]

I mean, as has already been mentioned in #2, what you really need here is the Tsiolkovsky rocket equation.
$$
\Delta u = - v_{\rm ex} \ln(m_f/m_0)
$$
Its derivation requires a minimal amount of calculus but nothing major. It is available on the Wiki page.