A two-stage rocket is fired vertically up and is above the...

A two-stage rocket is fired vertically up and is above the atmosphere when the first stage burns out and the second stage separates and ignites. The second stage carries $1200 \mathrm{kg}$ of fuel and has an empty mass of $200 \mathrm{kg}$. Upon ignition the second stage burns fuel at the rate of $5.2 \mathrm{kg} / \mathrm{s}$ and has a constant exhaust velocity of $3000 \mathrm{m} / \mathrm{s}$ relative to its nozzle. Determine the acceleration of the second stage 60 seconds after ignition and find the maximum acceleration and the time $t$ after ignition at which it occurs. Neglect the variation of $g$ and take it to be $8.70 \mathrm{m} / \mathrm{s}^{2}$ for the range of altitude averaging about $400 \mathrm{km}$.

A two-stage rocket is fired vertically up and is above the atmosphere when the first stage burns out and the second stage separates and ignites. The second stage carries $1200 \mathrm{kg}$ of fuel and has an empty mass of $200 \mathrm{kg}$. Upon ignition the second stage burns fuel at the rate of $5.2 \mathrm{kg} / \mathrm{s}$ and has a constant exhaust velocity of $3000 \mathrm{m} / \mathrm{s}$ relative to its nozzle. Determine the acceleration of the
second stage 60 seconds after ignition and find the maximum acceleration and the time $t$ after ignition at which it occurs. Neglect the variation of $g$ and take it to be $8.70 \mathrm{m} / \mathrm{s}^{2}$ for the range of altitude averaging about $400 \mathrm{km}$.

Key Concepts

Net Acceleration in Rocketry

Net acceleration is determined by the balance between the upward acceleration produced by thrust and the downward acceleration due to gravity. In rocketry, it is calculated by dividing the thrust by the instantaneous mass of the rocket and then subtracting the gravitational acceleration. This concept is crucial for evaluating different flight phases, such as determining specific moments of peak acceleration during ascent.

Variable Mass Systems

In the analysis of rockets, the system's mass changes continuously as fuel is burned and expelled, making it a variable mass system. This requires a special treatment within Newtonian mechanics, where the changing mass must be accounted for in the momentum conservation equations and the application of Newton’s second law.

Tsiolkovsky Rocket Equation

This equation is a fundamental principle in rocket propulsion that relates the change in velocity of a rocket to its exhaust velocity and the ratio of its initial to final mass. It provides insights into how much velocity increment can be achieved by expelling a given mass of fuel at a certain exhaust speed, making it central to understanding and designing multi-stage rockets.

Thrust Calculation

Thrust is the force generated by the rocket engine and is computed as the product of the mass flow rate (the rate at which fuel is expelled) and the exhaust velocity of the gases relative to the rocket. This concept is key in determining how much force is available to accelerate the rocket, especially in variable mass systems where the mass continuously decreases as fuel is burned.

Transcript

00:01 So here on this problem we are given this two -stage rocket and the second stage carries fuel of mass 1200 kg and the mass of the ampity rocket is 200 kg now the fuel is burned at the rate we are given to be as 5 .2 kg per second and the exhaust velocity you we are given to be as 3 ,000 meter per second now we have to calculate two things first the acceleration of the rocket after 60 seconds from the ignition so we have to calculate the acceleration after 60 second and the maximum acceleration so these two things we have to calculate and for that we'll be first let me first show the forces on the rocket so this would be the mg force that would be acting in the downward direction and we will be taking this direction as our y direction positive y axis and acceleration is like this for the rocket now to solve this problem i will be using the variable mass approach and by using this approach we can write that net force sigma if in the y direction will be equals to m .u plus mv dot here m.

01:38 M dot is the mass flow rate based this is the rate at which the fuel is burned and this m is the mass total mass of the rocket at the instant we have to calculate the acceleration so mass would be equals to 200 plus 1200 minus the mass is being reduced since the fuel is burned at the rate this much so for any time t though the total mass of the rocket would be this much this expression would tell the total mass of the rocket at any time t.

02:15 Suppose you want to find the time at t equal to 60 seconds so we can plug t as 60 here and we can find the mass of the rocket at that instant and so on now m -dash sorry m dot will be equals to negative m -dash because fuel is being burned and it is equals to minus 5 .2 kg per second now how we can put the values in this formula now sigma f5 we have only one force in the wide direction that is m g and that too acting in the downward direction and we have taken positive as in the vertically upward direction so we have to take this force to be negative so minus mg and just putting it here is equals to now m dot so m dot we have negative m dash which is minus 5 .2 and times u so you we have 3 ,000 plus now this mass can be written as this much this is the mass so which is 1400 minus 5 .2 times this v dot means the acceleration so rate of change of velocity is acceleration so we can write v dot as a we can put the values m g we can put it as this much so 1400 minus 5 .2 t times g is equals to this would become minus 5 .2 times 3 ,000 plus this one as it is 1400 minus 5 .2 times t multiplied by a now can put the values so 1400 as it is minus 5 .2 times t now this time t is the time at which we want to find the acceleration which is 60 seconds so we have to put t as 60 second here times g is the acceleration to gravity which we are given the equation to be 8 .70 is equal to this one as it is 5 .2 times 3000 plus this one would be 1 ,400 minus 5 .2 times 60 seconds and multiplied with acceleration so from here we can easily calculate the acceleration a and it comes out to be as nearly 5 .64 meter per second square so this is acceleration at t equals to or you could say after 60 second from the ignition so this is the acceleration of the rocket.

04:58 Now next i will be calculating the maximum acceleration.

05:02 Now to calculate the maximum acceleration, we have to first calculate the time at which it occurs.

05:09 So initially we have 12 kg of fuel...

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