How much energy is required to move a 1000-kg object from Earth's surface to a height twice Earth's radius?

4.17×10104.17\$$times10^{10}4.17×1010 J$$

How much energy is required to move a 1000-kg object from Earth's surface to a height twice Earth's radius?

Hey, guys. So in this problem, we're asked to find out how much energy is required to lift something from here to here on the earth. There's a lot of potentially confusing words here, so I'm gonna go ahead and draw a diagram. Now we're working with energy, so we're working with energy here. So we need an initial and a final, position. So I've got the initial position right here, which is on the surface of the earth. That's my initial. And I'm told that that initial position is basically on the surface. So in other words, the initial distance is going to equal the radius of the earth. And then I'm gonna lift this object with by doing some work or by giving it some energy, and I want it to be at a height that is equal to twice the Earth's radius. So I know that the height, the distance above the surface, is going to be 2 times the radius of the earth, and this is going to be my final position. So because I'm working with energy, we're supposed to set up our energy conservation equations. So let's go ahead and do that. So I've got the kineticinitial plus potentialinitial, and then any work that's done is equal to the final kinetic end potential. So we're told that we're basically just lifting this thing from rest, or we're supposed to imply that it's from rest. And it's finally going to end up at some height above the surface. There's no velocities or anything like that. So we can go ahead and just assume that the kinetic energies don't exist. We're just going to lift this thing from point a to point b. So that means that actual variable that we're looking for is the work that's done by non conservative forces. First, it has some potential energy because it's sitting on the surface, and then it has some potential energy because it's at some distance. And the work that's done is gonna do that. So that's what we're looking for. So let's go ahead and write out these expressions. So our initial gravitational potential energy is, we know, gmm over rinitial. Then we got the work that's done by non conservative forces, and then the final gravitational potential energy is going to be gmm over rfinal. So let's go ahead and just isolate the, variable that we're looking for. So with the work that's done by non conservative forces, it's just going to be I'm going to move this thing over to the other side. So I've got g m m over r initial, because this being becomes positive, minus gmm over r final. So let's just go ahead and see if we can simplify this equation and also we just need to take inventory of all of our variables. I've got these are just constants here. I have the mass of the earth and the mass of the object here. We're told that it's 1,000 kilograms. I have the initial position and, I got the same variables over here. The only thing I need to know is what the final position is. I'm told what the height of the final distance is, but that's h, not r. So I just have to remember that this r is actually just, big R, the radius of the earth, plus the final height. And, actually, we do have both of those things here, so we do have r f. So now I'm just gonna go ahead and simplify this expression. I've got this gmm is actually just a constant, or sorry, it's a greatest common factor because it appears in both terms. So I've got work done is equal to what I can do is, oftentimes, you can do this in these kinds of problems. You can, like, pull out the gms and stuff like that as greatest common factors. So I'm going to have one over rinitial minus 1 over and then rfinal is just going to be, let's see, we've got the radius of the earth plus the height, which is 2 times the radius of the earth. So that means in total, this is going to be 3 times the radius of the earth. And this initial distance right here is just going to be the radius of the earth. And, actually, we have what those numbers are. We know that these radii right here are just this one in this table. So we can go ahead and just basically plug everything in. So I've got the work done is equal to, and I've got 6.67 times 10 to the minus 11. Then I've got the mass of the earth, 5.97 times 10 to the 24th, and then I've got 1,000. And now I have to just do this whole parenthesis right here. One over over 6.37 times 10 to the 6th minus 1 over, and now we've got 3 times 6 0.37 times 10 to the 6th. Make sure you plug that in parentheses, and then this whole thing is in parentheses right here. And then you should get that the work done is equal to 4.17 times 10 to the 10 joules. And that's a positive number. And that makes sense, because we actually have to do some work in order to lift this thing up. So, anyway, that's the answer. And let me know if you

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How much energy is required to move a 1000-kg object from Earth's surface to a height twice Earth's radius?

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