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Team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing experiments, one of their tasks is to d...

Question

Team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing experiments, one of their tasks is to demonstrate the concept of the escape speed by throwing rocks straight up at various initial speeds. With what minimum initial speed Vesc will the rocks need to be thrown in order for them never to "fall" back to the asteroid? Assume that the asteroid is approximately spherical, with an average density p = 2.41 X 106 glm'

team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing experiments, one of their tasks is to demonstrate the concept of the escape speed by throwing rocks straight up at various initial speeds. With what minimum initial speed Vesc will the rocks need to be thrown in order for them never to "fall" back to the asteroid? Assume that the asteroid is approximately spherical, with an average density p = 2.41 X 106 glm' and volume V = 1.48 X 1012 m? . Recall that the universal gravitational constant is G = 6.67 X 10-H1 N-m? /kg? Vesc mls



Answers

A spherical asteroid has a radius of $365.1 \mathrm{~km}$. The escape speed from its surface is $319.2 \mathrm{~m} / \mathrm{s}$. What is the mass of the asteroid?

As we all know that he is equal to understood to gM. By uh simplifying it further, I can write the value of R is equal to two Gm by V E square. So just putting the value here, I can write to multiplication 6.67 multiplication at 10 to the power minus 11, multiplication, 1.769 Multiplication 10. to depart 20 by 2 7 3.7. Holy square on solving it further, finally, I get the value of R is equal to 315017 m. So changing it in kilometer 315 .017 km is the radius of the destroyed. So I hope you understand how I calculated the ideas of the destroyed. I just use the simple formula to calculate the value.

In this question, we are told that asteroids have average densities of about 2500 kg per meter. Cute and radi I from 407 kilometers down to less than a kilometer and were asked to estimate the radius of the largest asteroid from which you could escape simply by jumping off, assuming that the asteroid has a spherical symmetric mass distribution and then in part B were asked to calculate something a little bit different. So let's go ahead and start off with Part Aid and we were given a hint. The hint is that we can estimate are jump speed by relating it to the maximum height that you can jump on Earth. So in order to estimate ah, the largest asteroid that we could escape from, we need to know the biggest escape speed that we can generate with our legs. Right. Um, so this is what this hint is about, right? If so, we're looking at, you know, a person and they jump off the ground with some speed. That speed is going to be the escape speed. And on earth, that speed is directly related. Thio, you know how high you can jump, right, So maybe you jump with a speed V and you reach a height h. And so if we can estimate the height that we can typically jump, then we can use conservation of energy to estimate what that speed was that generated that height. So think about what a typical height would be that somebody could maybe jump. Um, I think you know, if you investigate this on your own, you know how high you can jump. Most people are going to be, you know, a foot off the ground, maybe 2 ft. Um, so I'm gonna estimate that is about 30 centimeters or 0.3 m. And what speed do you need on earth to generate? Uh, that height, right. How fast you need to be leaving the ground. In order to get to that height, we can calculate that using conservation of energy, right, because all of the cannon energy at the beginning of the motion should be, ah, a transmitted into gravitational potential at the height of the motion. So the kinetic energy is one half MV squared, and the gravitational potential is MGH. So we can just cancel off our EMS multiplied by two and take the square root. Okay, so here we have our estimate for the speed the maximum speed that someone can generate with their leg muscles. So we're just gonna plug everything in here. And from that calculation, I get about 2.43 m per second. So answers air going to vary here, depending on you know what you estimate for that height. So now we're going to use this as the escape speed. We're going to assume that this is the largest escape speed. Um are the largest speed that someone congenital Wait and see how that corresponds to the mass of the asteroid that can be escaped from. So let's look at the formula for escape speed. So that's gonna be equal Thio to GM over r. And you can take a look, at example 12.5 in the book. If you want to know more about this formula, Great. The problem with this formula is that we were given an average density for a particular um or sorry. We were given an average density for asteroids. Ah, but we weren't given a radius or a mass. In fact, the masses what we're trying to r the radius. The story is what we're trying to find. So what we need to do is rewrite this formula to get rid of mass because that's something that we're not calculating and it's not given to us. What we can dio is we can rewrite Ah, the mass in terms of the density, right? So we know that mass is equal to density times volume and we can rewrite volume as four pi over three r cubed So the mass becomes four pi over three ro r cubed And so we can just substitute into the formula to get a new formula for escape speed. So the four pride becomes a pie when it gets multiplied by two. We still got RG there. We've got a factor of row and then the r cubed can't, um, gets divided by r in the denominator. And so we're left with r squared. So, in fact, we can pull that are squared out of the square root and it just becomes our everything else is still inside the square root. We're just pulling out the r squared, taking this square root so that becomes our So now we're off to the races here I have a nest emit for my escape speed. That's 2.43 and I'm going to use that typical density that was given 2500 kg. So once you plug this factor into the calculator with G and row, we get 1.18 times 10 to the minus. Three times are and then we confined our just by taking 2.43 and dividing by 1.18 So this is an estimate. And once we do this division, we get a value. That's about 2000 kilometers. Okay, so the biggest asteroid, Uh, sorry. Not 2000 kilometers, 2000 m or two kilometers. So the biggest asteroid that we could escape just merely by jumping off and using the power of our legs is about two kilometers. Answers are going to vary here, depending on what you estimated. For the height you can jump on Earth and the escape speed here. So this is the final answer, Thio part A. But again, answers can vary in part B. We're told that Europa is one of Jupiter's four largest moons and it has a radius of 15 70 kilometers and an acceleration due to gravity of 1.33 years per second square and were asked to calculate the average density. Eso Let's go ahead and do that. So we're looking for density. We've got a radius of 15 70 kilometers or a million, 1.5 million m and an acceleration due to gravity 1.33 So for density, typically we need a volume so we can calculate that using the radius and a mass the mass is what we don't have here. And that's what we can use G to find so little g is equal to G um over r squared and so we can find mass. Um, by rearranging this formula, G r squared over big G. So plugging all that in we get a mass for this asteroid or not asteroid but moon of 4.92 times 10 to the 22 kg. Which makes sense because that's about you know, 1 1/100 of the mass of the earth and then using this mass as well as the radius, we can calculate the density so dense is going to be mass over volume again, we're gonna assume pretty much spherical, so it's going to be four pi over three R cubed for the volume. And when we plug that in to the calculator, we get a value about 3300 33,000 and 32 kg per meter cubed. So this is the final answer for Part B. It might be kind of hard to see how A and B are related at first, but I think the point is trying to make Is that the density of this moon Europa? I'm presumably other moons. Um, it's quite similar to the average density for an asteroid. So maybe the point here is that, you know, we could be looking at a source of asteroids in moods, right? Or perhaps that they have a similar source material to one another.

For this problem, on the topic of momentum and collisions, we are told that a satellite approaches a large planet with speed of 13.5 kilometers per second and a mass of 274 kg if the planet is moving at a speed of 10.5 kilometers per second in the opposite direction. And we treat the interaction as an approximate elastic collision. In one they mentioned, we want to know the speed of the satellite after the so called collision. Now we can use the conservation of energy and the conservation of momentum in this problem, from the conservation of momentum, we get the initial momentum of the system must equal the final momentum, which means that the mass of the satellite, M one times its initial velocity, the I one plus the mass of the planet, M two times its initial velocity. The I too, must equal the final momentum for the satellite. M one, the F one plus the final momentum for the planet M two, the F two, which means that M one into the I one minus the F one is equal to M two into the F to minus V F B I to. We'll call this equation number one. Now, since we can treat this as an elastic collision, we can we can conserve the energy as well. So the initial kinetic energy of the system must equal the final kinetic energy system. So that's a half M. One. The I one squared plus a half. M. To the I two squared is equal to a half and one. The F one squared plus a half M. To the if two squared. Now with some mathematics we get this the equation to simplify to be I one less. The F one is equal to M two into the F to minus the I two, divided by M one into V. I one minus V. F one times BF two plus the I too. Now if we use this and we use expression one. So we go back into the conservation of the momentum equation above. We get by simplifying some more VF one, which is what we want to calculate to be. M one minus M two, divided by M one. Bless him too. Into the I one less two. M two divided by in one. That's M two, V I two. So the F one is the final speed of the satellite, which you want to calculate. But we can also make the assumption that the mass of the planet M two is much greater then in one. And so this means that the first term which has M one minus M two of uh M one plus M two. And this can be approximated two minus one. And similarly, the second term we have to M two over M one Bless em too. And this can also be approximated to be too, which means that the final speed of the satellite, V. F one can be approximated to b minus the I one last two, V I two. And we I one is the initial speed of the satellite 13.5 kilometers per second. V two. Here is the speed of the planet minus 10.5 kilometers per second. And so, if we substitute our values, we get this to b minus 13.5 kilometers the second last two into minus 10.5 kilometers the second, which gives us the final velocity of the satellite to b minus 34.5 kilometers the second. And the negative sign makes sense, since it should be moving in the opposite direction from its original direction.

So we have Hey and I destroyed dislocation Traveling at ve even as 660 meters per second towards the earth. The radius of the earth is 6.38 times Turn to the bar six So it is traveling toward the earth here and they're asking what is the velocity of the asteroid belt and reaches the surface of the arc? Here it works speed Will it hit dirt? So we assumed that the mechanical energy is conserved and because there is no friction. So ah, friction is ignored in this case, the total energy at this location e one equal sort of energy the surface of the earth. So hee won and this location is equals to have and even squared. And we have Bless the the pretension and damages equals to G m e. M m is a massive asteroid divided by our one now are one is actually equals. Two five dames turn to the par six kilometers which is basically equals 25 times. Turn to the part six meters. That's five times for the bar. Nine meters. Bless our e. So the distance of this asteroid is actually taken from the center of the earth. So therefore the distance are one is equals. 25 times 10 to the power nine meters. Bless our eat. Okay, the energy at location into where did this location is equals to have am Vito squared? Where Vito is the velocity off the asteroid act The surface of the and the G p e minus m e m over our two and our do when the astral reached the surface of the earth is radius. So we're just equal to Ari. Basically. Now even should be equals to eat too. So we ride the equation half M V one squared, minus g m pm by our one That should be equals to have m v two squared minus g m e m by our two. To simplify that, I'll multiply both the sites by to buy em. Mister, get rid of this M So therefore we'll get V even squared. Minus g m e by are one is equals. Two video squared minus two g. M e by our two. So I hold this equation is clear. And if he quit this, we need to find the value of Vito. So Vito squared is equals to be one squared less do g mass off the ark one by our do minus one by our one. So that is the equation which we're gonna use from the next speech. So I'll just plug in the value here. So we two squared is equals 26 60 squared plus two G 6.67 times turn to the par negative 11 times mass off the order, which is 5.98 times turned to the part of 24 dimes. Just continue here, run over. Our two now are two is 6.38 times 10 to the power off six plus five times turned to the par of nine This is our two minus are 11 divided by a 6.38 times turning to the park off six. So that is our valley. If you plug it in the calculator and you find out the value of Vito and the square root of that, you will get 1.12 times turned to the part off four meters per second, which is the velocity of the asteroid as it reaches the surface


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