(III) Show that to slow down a spacecraft (perhaps to put it i...

Question

(III) Show that to slow down a spacecraft (perhaps to put it into orbit around a planet or moon) $v sub 0 sub x$ must be in the same direction as $\vec{u}$ and have magnitude greater than u. Show that this means the spacecraft will then pass in front of Jupiter. Draw a new diagram to replace Fig. 8–25b for this situation.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A space agency sends a 1200 kg space probe towards the planet Saturn at the mission control. Saturn appears to move along the x axis with a speed of 38 kilometers per second. Given that initially the probe approaches Saturn with an a an X axis speed of 40 kilometers per second relative to the mission control on earth. Calculate the change in kinetic energy after the probe passes Saturn, assume that along the direction Saturn is moving, the speeds are positive. Awesome. So it appears for this particular problem. The final answer that we're ultimately trying to solve for our end goal is we're trying to determine the change in kinetic energy after the probe passes Saturn. So now that we know that we're trying to figure out the change in kinetic energy after this probe passes Saturn, let's read off our multiple choice answers to see what our final answer might be noting that the kinetic energy is gonna be in units of jewels. So A is increases by 4.8 multiplied by 10 to the power of six B is decreases by 4.8 multiplied by 10 to the power of six C is increases by 1.8 multiplied by 10 to the power 11 and D is decreases by 1.8 multiplied by 10 to the power of 11. Awesome. So first off, in order for us to better visualize this problem, let's create a diagram of both the Saturn's frame of reference and the mission controls frame of reference or rather earth's frame of reference. So with that in mind, I've gone ahead and found a helpful diagram for us to refer to, to help us visualize this problem. So we have two figures here. The figure on the on the left, our diagram on the left represents the Saturn's frame of reference, which I'm gonna put a capital s up above to indicate that this is the Saturn's frame of reference. And our diagram on the right represents our mission control frame and earth's frame. So I mean, it's a MC. So mission control. So once again, our figure or diagram on the left is in Saturn's frame of reference. And our figure on the right is on the is in the mission controls frame of reference. So as you could see, we have a white circle with a black border representing our planets. Well, whereas the mission controls frame of reference. So basically we're we're dealing with. So it says given that initially the probe, a probe approaches Saturn with an x axis speed. So this is the space station is sending this probe towards planet Saturn. So at the mission control, Saturn appears to move around the x axis. So what as we could see the circle represents Saturn and as you could see the frame of reference changes depending on where you're at. So from Saturn's frame of reference, we could see that it has different variables going on compared to mission controls, frame of reference. So let's start with Saturn. So as we could see, we have Vsoxksovso or specifically vector VSO and Vsoy and then we have vs OX prime is equal to negative VSOX and we have vector V prime. So and then we have, we have V prime, so Vsoy prime is equal to Vsoy. Whereas mission control, we have the variables of Vixviy and vector V I and then we have vector V prime vector vs, then we have VY or VFY, so vfy and vector VF. And then most importantly, we have V FX equals VSOX prime plus VXVS. So these are our important variables and we'll discuss them more in depth in just a second. But this is just to visualize what's happening in these particular or for this particular problem for these two different frames of reference. Since this is a relativity based problem, it's very important to note what the what will change via change in perspective for the these two situations. So with that in mind, we can see that in Saturn's frame, the space probe comes in with the initial velocity of and that's what we're gonna be calling our initial velocity of vector VSO. So this is sa from Saturn's frame of reference, the space probe will come in with an initial velocity of vector VSO which will have an X component of VS OX and a Y component of VSOY. And we'll eventually exit with a final velocity of drum roll vector VSO prime as we see up above. So it's gonna be vector VSO prime and this vector VSO prime will have an X component of VSOX prime is equal to negative VSOX and we'll have a Y component of Vsoy Awesome. So as we should recall a note, the Saturn or the planet Saturn itself will move to the right along the X axis with a velocity of vs. On the other hand, when we consider the mission controls frame of reference, the space probe will come in with an initial velocity of vector V I and it will have an initial X component of vix and a Y component of Viy. And we'll eventually exit with a final velocity of vector VF uh And this is all we visualized or all plotted on our little figures here that we drew. And then we need to note that it will exit with the final velocity of vector VF and it will have an X component of V FX and I'll have a Y component of VFY. We also must recall and note that sense. Saturn is moving in the mission controls frame of reference with a velocity of vs we can go ahead and write safely. We could safely write that V FX is equal to V so X prime. So VSOX prime plus vs Awesome. So we can also go ahead and write that V FX is equal to negative Vsox plus vs Awesome. But we need to recall a note that since Saturn is moving with a velocity of V like vs that's important, Saturn is moving with a speed of, VS, we can go ahead and write that V FX is equal to negative vix minus vs plus vs which we can also write that VFX is equal to negative vix plus two vs Awesome. So now at this point, we need to con consider the change in kinetic energy. So now we need to recall and use the kinetic energy formula to help us figure out the change in kinetic energy. So let us note that the change in kinetic energy which is gonna be denoted as delta K. So the change in kinetic energy using the kinetic energy equation will be equal to one half multiplied by the mass M multiplied by the final velocity minus the initial velocity. So when we plug in our known variables. For this particular problem considering the variables we have at work. For this problem, we can rewrite our change in kinetic energy as one half multiplied by M multiplied by V fx squared plus vfy squared minus vix squared minus viy squared. And don't forget your bracket. But we need to recall a note that the Y component of velocity does not change. Therefore, we can state that VFY squared is also equal to viy squared. So therefore, we can go ahead and safely write that delta K or change in kinetic energy is equal to one half multiplied by M. And don't forget your parentheses multiplied by V FX prime, sorry, not prime. So V FX squared, not prime, it's V FX squared. So one half multiplied by M multiplied by V FX squared minus vix squared. Awesome. So now what we're gonna be doing is we need to note that we can go ahead and write that delta K is equal to one half multiplied by M multiplied by negative vix. So negative vix plus two vs and this is when we're plugging in our known variables. So it's equal to parenthesis, negative vix plus two multiplied by vs all to the power to minus vix squared. So now we need to take this expression and we need to simplify it down to its most simplest or the simplest form. And when we do that, I'm gonna skip some steps. But when you've done your algebra correctly, the most simple form of this expression for delta K or change of kinetic energy will be equal to two multiplied by M multiplied by, vs all multiplied by vs minus vix fantastic. So now all we need to do is plug in our known variables to so the numerical value of delta k. So when we do that, we will find that delta K is equal to two. And we need to take two, we need to multiply it by the mass which is 1200 kg. And then we need to multiply it by 38,000 m per second. Because as we should recall, the speed in the pro itself was 38 kilometers per hour. But we need to convert kilometers to meters per second. So that's how we get from roll 38,000 m per second. So 38,000 meters per second. Once again, we had to convert kilometers per second to k to meters per second. And then we need to multiply this by 38,000 meters per second minus. And once again, we have to convert all the kilometers to meters per second. So kilometers per second to meters per second need to emphasize that. So minus 40,000 m per second. Fantastic. So let me plug that into our calculator and we round to one decimal place, writing our final answer and scientific notation, we will find that delta K, the change of kinetic energy is approximately equal to negative 1.8 multiplied by 10 to the power of 11 jules. And that's it. That is our final answer. Hooray. We did it. And it's important to recall and note that the negative sign in this case indicates that it is decreasing. So therefore, without a doubt, are correct. And final answer has to be multiple choice answer letter D which states that it decreases by 1.8 multiplied by 10 to the power of 11 jewels. Thank you so much for watching. Hopefully that helped and I can't wait to see in the next video. Bye.

Key Concepts

Relative Velocity

Orbital Mechanics

Trajectory and Path of Motion

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