The 12,000-kg Lunar Command Module is in a circular orbit above the Moon's surface. If it spends ¼ of its fuel energy (−1.74×109-1.74\times10^9 J) bringing it to a circular orbit just above the surface, how high was its original orbit?

1.9×1051.9\$$times10^51.9×105 m$$

The 12,000-kg Lunar Command Module is in a circular orbit above the Moon's surface. If it spends ¼ of its fuel energy ( − 1.74 × 10 9 -1.74\times10^9 J) bringing it to a circular orbit just above the surface, how high was its original orbit?

Hey, guys. Hopefully, you're gonna try this one on your own. So we got the lunar command module that's above the Moon's surface, and we're gonna go from a circular orbit. It's gonna spend some fuel and some energy, bring it down to another circular orbit that's just above the surface, and we're supposed to figure out how high is the original orbit. So we know that we're solving for h. That's the height. And every time we're solving for h, we're always gonna solve for we're always gonna solve for little r and then relate that using big r+h. So if you can figure out what this little r is, we can figure out what this h is. Now we're talking about lunar orbits changing from one circular orbit to another, so we know we're gonna use energy. And so we're gonna set up our energy conservation equation. We've got the initial orbit plus any work that's done by non conservative forces is equal to the final energy orbit. And the reason we use this equation is because we're changing from one circular orbit to another. So I just wanna draw like a quick little diagram of what's going on first. We've got the moon's surface that's gonna be right over here. Whoops. That's m. And we've got a circular orbit that is, like, far above the lunar surface somewhere, like, around there. And then we're gonna spend some fuel, and we're gonna basically bring it to lunar an orbit that's just above the surface right here. So we're starting off with some from some place, some initial distance away. That's gonna be r I, and I have some initial energy of that orbit. Right? And that's given by this equation right here, g m m over 2 r. So then what happens is I'm expending some energy or by by doing work. The thrusters are actually doing work, so that's that work done. We finally have a different amount of energy when we're at the final orbit. So that's basically what's going on in the problem. Let's go ahead and see if we can figure that out. So I'm gonna go ahead and write out those these equations. So I've got negative g, the big m, little m over 2, our initial. Then I've got any work that's done is equal to negative gm, little m over 2 r final. Right? Because now I'm at a different circular orbit, so I'm gonna use that equation and I have 2 different r's to solve for. Now what I'm basically asked for is how high is the original orbit. So me ask for h. I'm gonna solve for little r and I'm gonna solve for the initial r distance right there. So that's basically what I'm gonna go ahead and solve for. So all I need to do is just bring this over to the other side. Let's see. I know what the All these constants are gm and little m. I know the work that's done by the non conservative forces. I'm told that that is a negative number, and then I know what these distances are. And just above the surface means that this little r distance right here at the very end when it's at its final orbit, what this means over here is that little r final is actually equal to the radius of the Moon itself. So we actually know what that number is from this table right here. So let's go ahead and solve. We're gonna bring the work done to the other side. So we have negative g m little m over 2 r initial equals. Then we got the negative work done by non conservative forces. We got minus g m little m over 2 r final. So now what I'm gonna do is basically just plug everything in. So one thing I wanna point out is that we wanna We don't wanna solve for this r initial right here. So you might be tempted to start rearranging all of these things, but the problem is that this isn't the denominator. We're gonna have to flip this whole entire equation, move stuff over. It's gonna get really ugly. So instead, what we're gonna do is we're actually gonna solve for what the numbers of these things are, and then we're gonna go ahead and manipulate the equation. It's gonna be a lot easier that way. So I'll show you what I mean. So basically just gonna evaluate what this thing is by plugging in all the variables. So I've got negative g m, little m over 2 r initial equals I've got negative of the work that's done by non conservative forces. That's the negative 1.74. So be careful because there's actually 2 negative signs right here. 1.74 times 10 to 9th minus, then I've got g which is just this constant over here. We've got the mass of the moon which is 7.35 times 10 to the 22. The mass of the spacecraft is 12,000, and then we've got divided by 2 r final. So in other words, 2 times the radius of the moon, which is 1.74 times 10 to the 6th. So now what happens is we're just gonna get a number for this on the right side. So we're gonna get negative g m, little m over 2 rinitial is equal to, and what I get is negative 1.52 times 10 to the 10th and that's in jewels. Now notice that on this equation that we've gotten, I've got a minus sign and a minus sign over here, so you just cancel those out. And then now what we're gonna do is we're solving for this r initial right here. So I'm gonna bring it over to the other side, and then this is gonna trade places with this guy. They're just gonna swap. So in other words, I've got negative g. Oops, not 6. So I've got negative Oh, wait. Actually, the negative goes away. So I've just got g which is at 6.67 times 10 to minus 11. The mass of the moon, 7.35 times 10 to the 22. The mass of the spacecraft, 12,000 divided by, we've got 2 times, and now I've got 1.52 times 10 to the 10th. And if you go ahead and plug that in, you're gonna get the r initial. And what that is is 1.93 times 10 to the 5th, and that's in meters. Yeah. Actually, sorry. That's 6th. So 1 times 10 to the 6th meters. But we're not quite done yet because we actually have to figure out what the height is. That's our target variable. So instead, now we have to do is rinitial equals big R+Hfinal, whereas this is really what our target for variable is. High. So we're gonna go ahead and plug that stuff in. So, basically, I've got our initial minus the radius of the Moon is equal to h I. So in other words, let's see. Finish things off. We got 1.3 93 times 10 to the 6th minus the radius of the Moon, one point 74 times 10 to the 6th. What you're gonna get is 1.9 times 10 to the 5th. That's in meters. So that's actually around 190 kilometers, and that's our final answer. That's how high above the surface that initial orbit was. Let me know if you guys have any questions with this.

The 12,000-kg Lunar Command Module is in a circular orbit above the Moon's surface. If it spends ¼ of its fuel energy (−1.74×109-1.74\$$times10^9 J) $$bringing it to a circular orbit just above the surface, how high was its original orbit?

1.9×1051.9\$$times10^51.9×105 m$$

1.6×1061.6\$$times10^61.6×106 m$$

1.8×1061.8\$$times10^61.8×106 m$$

3.7×1063.7\$$times10^63.7×106 m$$

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Multiple Choice

The 12,000-kg Lunar Command Module is in a circular orbit above the Moon's surface. If it spends ¼ of its fuel energy ( $- 1.74 \times 10^{9}$ J) bringing it to a circular orbit just above the surface, how high was its original orbit?

$1.9$\times$10^5$ m

$1.6$\times$10^6$ m

$1.8$\times$10^6$ m

$3.7$\times$10^6$ m

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The 12,000-kg Lunar Command Module is in a circular orbit above the Moon's surface. If it spends ¼ of its fuel energy (−1.74×109-1.74\(\times\)10^9 J) bringing it to a circular orbit just above the surface, how high was its original orbit?

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The 12,000-kg Lunar Command Module is in a circular orbit above the Moon's surface. If it spends ¼ of its fuel energy ( $- 1.74 \times 10^{9}$ J) bringing it to a circular orbit just above the surface, how high was its original orbit?