A rocket needs to hover stationary over the ground

[DanielA]

Homework Statement

A rocket (initial mass ##m_o##, constant exhaust velocity ##v_{ex}## needs to use its engines to hover stationary, just above the ground.
a) If it can afford to burn no more than a mass ##\lambda m_o## of its fuel (##\lambda \lt##), for how long can it hover?
b) If ##v_{ex} = 3000 m/s ~\text{and} \lambda \approx 0.10## for how long could the rocket hover just above the Earth's surface?

Homework Equations

The question didn't give any explicit equations, but the question should use $$\dot p = m dv + dm v_{ex} = F^{ext} dt $$ where ##dm v_{ext} = thrust##
which I believe is the rocket equation for a rocket moving vertically. I got it from my book, though they continued the example with no gravity or air ressitance so ##F^{ext} = 0##

The Attempt at a Solution

To begin, I believe I have a finished solution, but since I'm used to having a professor around to check my work, I'm not confident I'm doing it right as it seems too simple.
a)
Since there is no motion (and we haven't learned air resistance for a rocket) the only force ##F^{ext}## is gravity and ##F^{ext} = mg## where m is the mass of the rocket.

So, we want to solve for t. Let's use my equation above and integrate it .
$$
mgdt = mdv + dm v_{ex}
\\ \int_{v_0}^v dv = \int_0^t g\, dt - v_{ex} \int_{m_0}^m {\frac {dm} m}
\\ v-v_0 = gt - v_{ex} \ln {\frac m {m_0}}
\\ v = 0, v_0 = 0, m = \lambda m_0
\\ \text{since the rocket is motionless throughout all of this and the maximum mass loss is given in the question}
\\ t = \frac {v_{ex}} g \ln (\lambda)
$$
b)
This part is just plug and chug using the above equation. My answer is 704.15 seconds (g = -9.81 m/s), which seems excessively long to me, but I don't have any frame of reference on what is realistic for this.

 

[TSny]

Welcome to PF!

$$ v-v_0 = gt - v_{ex} \ln {\frac m {m_0}}$$

There is a sign problem here. The left side is zero. But the right side is positive overall since ##gt## is positive and ##-v_{ex} \ln {\frac m {m_0}}## is also positive. Did you take into account the direction of the external force?

Also, the final mass is not ##\lambda m_0##. Review how ##\lambda## is defined in the problem statement.

 

[DanielA]

Welcome to PF!There is a sign problem here. The left side is zero. But the right side is positive overall since ##gt## is positive and ##-v_{ex} \ln {\frac m {m_0}}## is also positive. Did you take into account the direction of the external force?

Also, the final mass is not ##\lambda m_0##. Review how ##\lambda## is defined in the problem statement.

I now made ##F^{ext}## negative at the beginning (##F^{ext}## = -mg) to account for the negative direction and changed the final mass to ##m_0 - \lambda m_0## since that is the final mass of the rocket after losing ##\lambda m_0## fuel. My answer changed to $$t = \frac {-v_{ex}} g \ln{(1-\lambda)}$$
This gives a much more reasonable 32.22 seconds of hovering when fuel equal to 10% of the original rocket mass is used

 

[CWatters]

Just a thought... The numbers might be available from the Apollo 11 moon landing. They had to "hover" a bit longer than expected and almost ran out of fuel.

 

[TSny]

I now made ##F^{ext}## negative at the beginning (##F^{ext}## = -mg) to account for the negative direction and changed the final mass to ##m_0 - \lambda m_0## since that is the final mass of the rocket after losing ##\lambda m_0## fuel. My answer changed to $$t = \frac {-v_{ex}} g \ln{(1-\lambda)}$$
This gives a much more reasonable 32.22 seconds of hovering when fuel equal to 10% of the original rocket mass is used

All of this looks correct to me.

 

[DanielA]

Just a thought... The numbers might be available from the Apollo 11 moon landing. They had to "hover" a bit longer than expected and almost ran out of fuel.

The question actually had a bit about how a similar situation occurred to Apollo 11. However, he specifically said it was similar in the question. The landing in Apollo 11 took 90 seconds which is much longer than my answer here

 

[DanielA]

All of this looks correct to me.

Thank you. I hope over time I'll gain more confidence in my ability. I'll end up having many more questions in the future though so look out for them

 

[TSny]

Thank you. I hope over time I'll gain more confidence in my ability. I'll end up having many more questions in the future though so look out for them

Sounds good!