Hey, guys. Let's see if we can work up this problem together. So we have a moon in orbit around a planet. Right? So let's go ahead and draw a diagram. We have a planet like this and a moon that is orbiting around in a circular orbit. Just pretend that's perfectly circular. Right? And this moon has a speed around its planet, and we're told that the velocity is equal to 75,100. And we're also told that it takes 28 hours to go all the way around. So, all the way one rotation, we know that \(t = 28\) and that is hours. So, using that information, how can we find out what the mass of this planet is? We're finding the mass of the planet. That's the thing in the center. So, we're really looking for capital \( M \). So, how do we go about doing that? Well, let's see. We've got a whole bunch of equations involving forces and gravitational acceleration, but we don't have any information about forces or gravitational acceleration. But we do have the motion of a satellite, so we're going to use all these equations over here. So let's see. I've got orbital velocity and I also have orbital period. So really I can use any one of these equations to start off. I'm going to use the \( v_{\text{sat}} \) equation. So we've got \( v_{\text{sat}} = \sqrt{\frac{g \times M}{r}} \). And let's see. I know what the velocity of the satellite is, and I'm looking for big \( M \), which is my target variable. Gravitational constants is the number. So, if I can figure out what this little \( r \) distance is, I could find that. But I actually don't have any information about the orbital distance or height or anything like that. So let's see. Maybe this isn't the approach. Let's see if I can start off with my \( t_{\text{sat}}^2 \) equation, with the \( t^2 \) one. So let's start out with the \( t_{\text{sat}}^2 \) equation. Let's see if we have better luck there. So you've got \(\frac{4\pi^2 r^3}{g \times M}\). Again, that big \( M \) is my target variable here, so I know what the orbital period is. But again, there's that \( r \) that I don't have, that orbital distance. So in both of these approaches here, both of these equations that I've seen, I ended up with the same unknown variable. I have too many unknowns. So there's got to be something else I can do to solve for this little \( r \) distance. Let's see. What's the one equation I haven't used yet? So, I'm going to go over here and solve for a little \( r \). The one I haven't used yet is this one, the \( v_{\text{sat}} = \frac{2 \pi r}{t} \). So I've got \( v_{\text{sat}} = \frac{2 \pi r}{t} \). Now in this situation for this equation, I have two knowns and only one unknown. So now I can use that to solve for my little \( r \). So let's see. I've got, moving everything over to the left side. I'm going to get \( v_{\text{sat}} \times t \div 2 \pi \) is going to give me little \( r \). So that means that that little \( r \) here, if you just go ahead and plug everything in, is going to be, let's see, 75,100 times the period. Now, that \( t \) is expressed in hours. So first I need to multiply by 3,600 to get into seconds. If you do that, you should get \( 1.008 \times 10^5 \), that's seconds. So now I'm just going to go ahead and plug that in. \( 1.008 \times 10^5 \). Now, I'm just going to divide by \( 2 \pi \). So, if I divide by \( 2 \pi \), I'm going to get the orbital distance, that little \( r \). That's \( 1.2 \times 10^8 \). So now what I can do is now, I can take this little \( r \) that I found and I can plug it back into either this \( t^2 \) equation because now I only have this unknown, or I can plug it back into the \( v_{\text{sat}} \) equation so that I still only have one known. So, really the choice is up to you. I'm going to just keep going with the \( v_{\text{sat}} \) approach. So I've got now my little \( r \) distance. So I've got \( v_{\text{sat}} = \sqrt{\frac{g \times M}{r}} \). So now I'm just going to go ahead and solve for that big \( M \) because I already have everything else. Right? So I've got \( v_{\text{sat}} \) and I'm going to square both sides because I want to get rid of the square root. So I've got \( v_{\text{sat}}^2 = \frac{g \times M}{r} \), and then I just move the \( r \) over everything over to the left side. So I get \( v_{\text{sat}}^2 \times r \div g = M \). And if you plug all that stuff in, what you're going to get is 75,100 squared times the radius, 1.2 times \( 10^8 \), and then we've got to do divided by the gravitational constant, so times \( 10^{11} \). And if you do that, you should even get the mass of the planet which is \( 1.01 \times 10^{26} \) and that's in kilograms. That's it for this one. Let me know if you guys have any questions with that.
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Satellite Motion: Speed & Period
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