A rocket accelerates by burning its onboard fuel, so its...

A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at liftoff (including its fuel) is $m,$ the fuel is consumed at rate $r,$ and the exhaust gases are ejected with constant velocity $v_{e}$ (relative to the rocket). A model for the velocity of the rocket at time $t$ is given by the equation $$v(t)=-g t-v_{e} \ln \frac{m-r t}{m}$$ where $g$ is the acceleration due to gravity and $t$ is not too large. If $g=9.8 \mathrm{m} / \mathrm{s}^{2}, m=30,000 \mathrm{kg}, r=160 \mathrm{kg} / \mathrm{s},$ and $v_{e}=3000 \mathrm{m} / \mathrm{s}$ , find the height of the rocket one minute after liftoff.

A rocket accelerates by burning its onboard fuel, so its mass
decreases with time. Suppose the initial mass of the rocket at
liftoff (including its fuel) is $m,$ the fuel is consumed at rate $r,$
and the exhaust gases are ejected with constant velocity $v_{e}$
(relative to the rocket). A model for the velocity of the rocket
at time $t$ is given by the equation
$$v(t)=-g t-v_{e} \ln \frac{m-r t}{m}$$
where $g$ is the acceleration due to gravity and $t$ is not too
large. If $g=9.8 \mathrm{m} / \mathrm{s}^{2}, m=30,000 \mathrm{kg}, r=160 \mathrm{kg} / \mathrm{s},$ and
$v_{e}=3000 \mathrm{m} / \mathrm{s}$ , find the height of the rocket one minute
after liftoff.

Key Concepts

Integration of Kinematic Equations

Integration plays a central role in kinematics, particularly when computing displacement from a given velocity function. In systems where velocity is provided as a function of time, integrating that function over a specified time interval yields the net change in position, making this method crucial for tracking the trajectory of moving bodies like rockets.

Gravitational Effects on Motion

Gravitational acceleration is the constant acceleration due to gravity acting on any mass, which significantly influences the motion of a rocket during its ascent. Accounting for this acceleration is critical in accurately determining the net effect of the rocket's propulsion and the resulting trajectory, especially when additional forces such as thrust are present.

Variable Mass Systems

Variable mass systems involve objects whose mass changes over time, such as a rocket burning fuel during its ascent. This concept is essential because the reduction in mass directly affects the momentum and acceleration, and requires modifications to the traditional Newtonian dynamics which assume constant mass.

Tsiolkovsky Rocket Equation

The Tsiolkovsky rocket equation is a fundamental result in rocket science that relates the change in velocity of a rocket to the effective exhaust velocity and the ratio of its initial and final mass. It is derived from the conservation of momentum in a system with continuously decreasing mass, forming the backbone of analyzing rocket propulsion.

Transcript

00:01 To find the height of our rockets, and since we know that our velocity is a rate of change of height, we can use the fundamental theorem of calculus, so using the fundamental theorem of calculus.

00:31 So you use that of the part two.

00:37 So we have the integral from 0 to a of our velocity, which is equal to the integral from 0 to a, the derivative by h, which is our height, dt, and this is equal to h a h minus h zero.

01:03 Zero so since the initial height here we know it's equal to zero then this implies that our h a would be equal to the integral from zero to a or three t the t the t so this gives us the height of the rockets at eight seconds after launching so then the integral v t d t is equal to the integral of you have minus g t minus v e lene you have one minus r divided by m t the t so this then it's equal to the integral minus g t the t minus integral v e we have length of one minus r divided by mc t so applying the sum rule here so we substitute we make this to be equal to so let's see we are saying this is equal to you so then it's implies that the so if this is equal to you then it's implies that you then it's implies that you have minus r divided by m d t to be equal to the u so then this implies that your d t is going to be equal to minus m divided by r d u okay so then from here you have so this integral so this is equal to this will give us minus g t squared divided by two you have plus v em divided by r your integral in u d u this as well you can solve using integration by parts here so we solve this using integration by parts so we have this to be equal to negative g t squared divided by two plus vem divided by r.

04:50 So this will give us u -lain u minus u plus c.

05:01 So we have this to be equal to negative g -t squared divided by 2 plus vem divided by r you have u plain u minus one plus c which is our constant so in the last equation we wrote just c so we wrote just see instead of this times c because it's still a constant when you multiply that's before a constant you are still a constant you are still getting a constant you are still getting a that is why we just root c here.

05:54 So then we substitutes u back.

05:59 So if you substitutes your u back, this is going to give you.

06:06 So this is going to be minus g, t squared divided by two, plus b, e, m divided by r...

Please give Ace some feedback