a) How much work do you have to do on a 100-kg payload to move it from Earth's surface to a height of 1000 km?b) How much additional work must you do to put this payload into orbit at this altitude?

(a) WAB=8.48×108W_{AB}=8.48\$$times10^8$$WAB=8.48×108 J;(b) WBC=2.7×109W_{BC}=2.7\$$times10^9$$WBC=2.7×109 J

a) How much work do you have to do on a 100-kg payload to move it from Earth's surface to a height of 1000 km? b) How much additional work must you do to put this payload into orbit at this altitude?

Hey, guys. Hope you try this one on your own. So in this first part, we're being asked how much work do you have to do on a mass in order to lift it from the Earth's surface out to some height. So that's what we're gonna be working with here. So I wanna just draw a quick little diagram. We're working with, work, and that's what we're trying to find out. So we've got the Earth's surface right here, and that's gonna be big m. In this first part, we're gonna have a payload that's gonna be on the surface, and we're gonna give it some work so that it can reach some distance above the Earth's surface. We know that that height is equal to a 1000 kilometers. Then in the second part, what we need to do is we need to find out how much work we need to do to actually put this thing into a circular orbit. So I'm gonna just draw this little orbit right here. So I've actually got 2 different parts of this problem. So I wanna go ahead and label everything just so I don't get confused. So at the surface is gonna be a. When we actually get up to this height is gonna be b. And then finally, when we're in that circular orbit, it's gonna be part c. So now I've labeled everything. Now it's time to figure out what I'm trying to solve for in this part a. So I've asked how much work I need to do on this payload. So work is gonna be my target variable in order to lift it from the Earth's surface to this height. So, in other words, I have to do some work from a to b in order to get this thing actually up here, and that's what I'm trying to solve for. So, work AB. So I'm working with work, and I have distances and gravitation. So I'm gonna be using, energy conservation. So let's go ahead and set up my energy conservation. So for energy, we're gonna be doing k I. We're gonna be doing k initial plus, you know, potential initial. We're gonna be using this equation right here. Notice that even though I'm talking about circular orbits, I can't use this problem right here because in no part in the beginning or the end, so for instance on the surface or when I get out to this height, am I actually in orbit yet? So I'm just going from one spot to another spot. It's just like when we used to do in gravitational potential energy, just lift something from a to b. So I actually can't use that, equation just yet. So I'm gonna use the initial kinetic, which is k a, and the initial potential, plus any work that's done between a and b is equal to potential and kinetic at part b, right here. So let's go ahead and start simplifying. So let's see. When we're on the surface, we have no kinetic energy, so I'm gonna cancel that out. We do have still still have some gravitational potential energy that comes from the fact that we're still some distance. Right? We're still some distance right here, away from the center of mass. So we've got that r a is equal to just the radius of the Earth if we're standing at the surface. And so we still have some gravitational potential. We have no kinetic energy when we reach the top because, again, we're just going from one spot to another. We're not told there's any velocities or anything like that. So, basically, all you have to do is just set up the equations. I've got negative g m little m over r a, which we actually know is just equal to big R. The work that's done between A and B, and that's equal to negative GMM over, and that's gonna be RB. So I'm gonna move this stuff to the other side and make it positive. So, basically, I have that the work done between a and b is equal to Then I've got g m little m over r, because it comes positive, and then it becomes minus g m m over r b. So all I have to do is just figure out what this little r b is, because if you think about it, I actually have what all these constants are in this table right here. So, what this little r b distance is, I'm told that that is equal to Let's see. I'm told that we're at a height of a 1000 kilometers, so that's actually h. And if you wanna find r b at this point right here, all you have to do is just r plus h. So, we that's just equal to the oops. Let's not do that. 6.37 times 10 to the 6th plus a 1000 kilometers, which is actually 1 times 10 to the 6th in meters. Right? So, this is actually a 1000 kilometers in meters. If you go ahead and work that out, you're gonna get 7.37 times 10 to the 6th. That's our little r b. So, now I'm just gonna go ahead and plug in everything. One thing we could do is we can actually take these and pull these out as greatest common factors, just so we can make it easier and not have to plug a bunch of stuff into our calculator. So we got work between a and b is equal to now I've got 6.67 times 10 to the minus 11. Mass of the Earth is 5.97, 10 to the 24th. Payload is a 100. And now I can just do this big multiplication. I've got one over the radius of the earth minus one over this new distance right here. And if you go ahead and work this out in your calculator, you're gonna get the work that's done is equal to I've got 8.48 times 10 to the 8th in joules. So that is let's see. I've got 8. Now, notice how this work is actually positive. What that means is that we actually have to give it some work in order for this payload to go from the surface up to some height, and that makes sense. We have to give it some energy to do that. Now in the second part here, we're gonna be asked how much work we have to do once it gets up to the spot to actually push it into a circular orbit. So let's go ahead and work this out. So, again, we're working looking we're looking for an additional work, except this time it's gonna be from b to c, and so we're gonna set up our energy conservation equation. So we which which one can we actually use? We're gonna use this energy conservation equation for the first bit, but the thing is that when it's at its second point, when it's actually in orbit, we can use basically this equation, g m m over 2 r, instead of this k f plus u f. So we can actually sort of substitute in that in that. Because remember that this e circular is equal to k plus u. So, we can always put that in there if it's in circular orbit. Let's go ahead and work that out. So, we've got K initial, which is now at part B, plus U initial, plus the work that's done between B and C is equal to, and now we have e circular right here. So, I've know that when when I'm actually at this height at the initial point, I have zero kinetic energy. I'm gonna give it some work and then it's gonna go into a circular orbit. So this is my target variable, the work between b and c. So I've got negative GMM over now I've got RB which we actually know what that is, And then I've got the work done is equal to now negative whoops. Negative GM m over 2. Remember, it has that 2 in the denominator, and then the radius at part C. But let's take a look at the diagram. So now when we're at this height over here, we know that this entire distance is r b. So in the words, r b is this entire distance over here. And when it's at orbit, that orbit radius doesn't change. That orbital distance right here is still r c, and both of these things are equal to each other. So in other words, I know that RB in this case is equal to RC. And so I'm just gonna go ahead and call those little r's right here. So, now, I've gotta do basically the same exact thing, move it over, and then wbc becomes I'm just gonna skip a whole bunch of steps because we've already done this part right here. It's gonna be gmm over, over r. Right? Because these things become r's. And then I have 1 minus 1 half. So in other words, I've got one half of gmm over little r. So in other words, if you actually work this out, you plug in all these numbers, you should get one half of I've got 6.67 times 10 to the minus 11, 5.97 times 10 to the 24th. Then I got the mass, which is a 100, and then divide it by, let's see, the r distance, which is equal to 7.37 times 10 to the 6. If you go ahead and work this out, you should get the work done is equal to 2.7 times 10 to the 9th joules. Again, that number is positive because we actually have to give it some work in order for it to start moving in circular orbit. Let me know if you guys have any questions, and I'll see you guys in the next one.

Struggling with Physics with Algebra?

a) How much work do you have to do on a 100-kg payload to move it from Earth's surface to a height of 1000 km?b) How much additional work must you do to put this payload into orbit at this altitude?

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a) How much work do you have to do on a 100-kg payload to move it from Earth's surface to a height of 1000 km?
b) How much additional work must you do to put this payload into orbit at this altitude?