b. A rocket with a total mass of 330,000 kg when fully loaded bur...

Question

b. A rocket with a total mass of 330,000 kg when fully loaded burns all 280,000 kg of fuel in 250 s. The engines generate 4.1 MN of thrust. What is this rocket's speed at the instant all the fuel has been burned if it is launched in deep space? If it is launched vertically from the earth?

Accepted Answer

Everyone in this problem. We have a rocket scientist developing a rocket to be launched vertically from the surface of the earth. It has a total mass of 275,000 kg, which includes 175,000 kg of fuel. The rocket engine is going to generate 3.5 multiplied by 10 to the exponent six newtons of thrust and will consume all of its fuel in 340 seconds at a constant rate. And we're gonna consider that all of the fuel has been consumed and we wanna find the speed of the rocket during this vertical takeoff, ok? And secondly, we want to find the speed again if gravity is negligible. Ok? So if we're in an area of space where that effect of gravitation is negligible. So we have four answer choices here options. A through D each of them containing a different speed in meters per second or part one and part two of this problem. OK. We're gonna come back to those answer choices once we work through this. So let's think about what we're doing here. We're trying to find this speed during the vertical takeoff. We have this rocket ship problem recall that the speed of V can be given by the following V is gonna be equal to V not the initial speed minus vex. The exhaust speed multiplied by the one M divided by M not M not is our initial mass case when we have full fuel and M is gonna be the mass at whatever instant we're interested in. And then we're gonna subtract that effect of gravitation, which is gonna be the acceleration due to gravity. G multiply by the time elapsed T. Yeah. So we not, this is being launched from rest. So we know the initial velocity or the initial speed. What about this exhaust velocity? We aren't really given information about the exhaust velo exhaust velocity. So let's think about how we could rewrite that. Hey, we're given a force of thrust. Well, recall that the force thrust F we equal to minus the exhaust velocity not supplied by the change in mass over time DM divided by DTF is equal to negative vex multiplied by DM divided by DT. What this allows us to do is right, the exhaust speed vex as negative F divided by the rate of change of mass DM divided by DT. All right. So if we call this equation two and equation one is the first equation we wrote for the speed V and we can substitute equation too into equation one. And that's gonna allow us to rate the speed V in terms of the thrust force that we were given instead of having that exhaust speed that we don't know. Ok. So now we have that the speed V we're looking for is going to be V not minus oops negative F divided by DM, divided by DT multiplied by one of M divided by M not minus G multiplied by D. All right. So we have minus and negative. So that's gonna turn into a positive there where we're adding that second term. And now let's think about what the values are for all of these variables that we need to use. OK. Let's start with the mass Femina. OK. Now, the initial mass we're told is 275,000 kg, that total mass 275,000 kg. That is a heavy spaceship. OK. What about the mass at? Ok. This is the mass at the time we're interested in. We recall that we're told that we're gonna consider that all of the fuel has been consumed. OK. So the mass we have is gonna be the initial mass minus the mass of the F. All right. Well, we're told them that the initial mass is 275,000 kilograms. And we're also told that the mass of fuel is 175 that was in telegrams, which means that when all of the fuel is exhausted, we have 100,000 kg left over. Ok? So the spaceship itself without any fuel is 100,000 kg. All right. What about the force F? And we're told that this thrust force is 3.5 multiplied by 10 to the exponent six Newton's. We know that the time elapsed is 340 seconds. Ok. That's how long it takes to exhaust all of that fuel. So we're looking at the instant that the fuel is completely gone. The last thing we need to look at is this rate of change of math DM divided by TDT. Now we're told this is a constant rate and so we can look at the change. Ok? Now the change in mass is 175,000 kg. We have 100 and 75,000 kg of fuel mass that we exhaust and we do that in 340 seconds. So we divide by 340 seconds. Ok? So DM divided by TDT 175,000 kg divided by 340 seconds. That's that rate that we're reducing the fuel. And we've done this equation again because they told us that it's a constant ri so now that we have all of this, we can get to our answers. So for part one, we're gonna assume that we have gravity on earth just like normal. So part one, our speed V is going to be equal to that initial speed. Ok? We're launching. So we're starting from rest. So it's gonna be a zero minus this thrust force. So we have 3.5 times 10 to the exponent six Newton divided by our rate of change 175,000 kilograms for 340 seconds. Now, I have a negative here. The equation has minus the negative. OK. So what happened? Well, what happened is that our rate of change for our mass should actually be negative. Hey, we're losing mass as time goes on, we're reducing that mass. So this rate of change is negative. So when we go to our equation, we have minus a negative that gives us a positive. Now we're dividing by a negative, we get back to a negative, right? So we do have a negative here but it was the result of three different negatives multiplied together. All right. So that's the first term. Let's put some brackets here. So we don't get mixed up. Now, we're multiplying by law of 100,000 kg divided by 275,000 kg. And we're gonna subtract the acceleration due to gravity 9.81 m per second squared multiplied by the time it takes 340 seconds. And while we work all of this out, we have that our speed is gonna be about 3543.48 6 2 m per second. OK. So we have part one that let's move to part two and part two is going to be very, very similar. Remember, the only difference is now we're in an area of space where the effect of gravitation is negligible. And if the effect of gravitation is negligible, let me go back to our equation, then what we're gonna do is ignore this last term in our equation. OK. So the negative GT is just going to be equal to zero gravity is negligible. So for part two, our equation is going to become B is equal to zero minus 3.5 multiplied by 10 to the exponent six newtons divided by 175,000 kg, divided by 340 seconds multiplied by lawn of 100,000 kg, divided by 275,000 kg. OK. So the first part of our calculation is exactly the same, we just don't have the extra term the gravitation at the end. And this is gonna give us a speed V of 6800 78 8 8 6 2 m per second. And this makes sense. We found that in part two, if the effect of gravitation is negligible that the speed is larger, that means that this is moving faster. And that makes sense. OK. We don't have gravity pulling it back down towards the earth and so it is going to be going faster. All right. So we figured out the speed of a rocket ship, let's take a look at our answer choices. Now, these are rounded to the nearest meter per second and we can see that the correct answer is going to be option B for part one, we have 3543 m per second. And for part two, we have 6879 m per second. Thanks everyone for watching. I hope this video helped you in the next one.

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b. A rocket with a total mass of 330,000 kg when fully loaded burns all 280,000 kg of fuel in 250 s. The engines generate 4.1 MN of thrust. What is this rocket's speed at the instant all the fuel has been burned if it is launched in deep space? If it is launched vertically from the earth?

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