Alright, guys. Let's work this one out together. So we have a comet that's traveling around the sun, travels once every 2000 years, and we're told some information about its closest approach. And in part a, we're supposed to figure out how far what is its farthest distance from the sun. So you should recognize that as the apoapsis. Let's just go ahead and draw a quick little diagram. So we've got the sun right there, and I'm going to have this highly elliptical orbit like this. Now if we're trying to figure out the farthest distance that corresponds to that length right there, that's apoapsis, and that's what I'm really looking for. Now there are only 2 other pieces of information I know about this problem. The first is that the entire orbital period for the whole thing is equal to 2000 years, so I've got that. And I've got that the apoapsis distance, the closest that it gets to the sun, which is r_p, is just given by some number, 3 times 10 to the 8th kilometer. So I've got that. So let's look through my equations. Now you might be tempted to sort of use the equation which we compare 2 parts of an elliptical orbit. That's v₁r₁ equals v₂r₂. The problem with this equation is that we if you look through, we don't have any of the velocities, and we're going to be solving for one of the distances. So that gives us with 3 unknowns for this problem. There's no way we're going to be able to use that, so it can't be this. We're going to have to look for some other equations to solve for r_a. So let's go ahead and look through my equations involving r_a. So let's see. I've got this equation over here, and I've got let's see. The tsat doesn't work because there's no r_a in there. These two equations relate the semi major axis and the eccentricity. Both of those values are unknown, so I can't use either of those. So it's going to have to be this equation over here, which is that the semi major axis is equal to r_a + r_p divided by 2. Cool. So I've got what r_p is, and if I'm trying to figure out the apoapsis is, there's only one equation or one unknown I need to solve for, and that is the semi major axis. Okay? So, unfortunately, I'm going to need another equation for this. So let me go ahead and just move this over here. So how do I figure out what the semi major axis is? Okay. Let's go back to our equations. Now I can't go for this equation because this is the equation I'm just coming from. So obviously, it can't be that one. Again, I don't have any information about the eccentricities. So that means it only leaves me with one option. I'm going to have to use Kepler's 3rd law for elliptical orbits. Now this is really useful because I'm actually told some information about the orbital period, and that's the variable t. So I'm definitely going to use this equation. So I've got t t squared is equal to, 4π squared a 3 divided by g m. So all we have to do is just rearrange everything to the other side. The g m is going to go up, the 4 π squared is going to go down, and I'm going to get g m t 2 4 π 2 is equal to a cubed. So now if I rearrange, I'm just going to get that a is equal to the cube roots of that whole entire mess over there. So I've got 6.67 times 10 to the minus 11. Now I've got 2 times 10 to the 30th for the mass of the sun. Now I need t squared. Alright? The problem is that the t I'm given is in years, it's not in seconds. So I'm going to have to first convert. So we've got t is equal to 2,000. Now I've got, that's going to be in years, so I have to multiply by 365 days per year to cancel that out. Now, I've got to multiply by 86,400 because that's how many seconds there are in one day. So let me let me, sort of write that out again. So we've got seconds per day. If you didn't know that off the top of your head, that's totally fine. But the years are going to cancel with the years, the days are going to cancel with the days, and then t is just going to be 6.31 times 10 to the 10 seconds. That's how many years there are in 2000 or so that's how many seconds there are in 2000 years. Great. So we're just going to plug that into this equation over here, 6.31 times 10 to¹⁰, and now we have to square that. Don't forget that. And now we have to divide this whole entire expression by 4 π squared. Don't forget to put that in parenthesis. Anyway, so if you plug all of this huge mess into your calculator, you're going to get a semi major axis of 2.38 times 10 to¹³, and that's going to be in meters. Right? So we can actually plug this all the way back into our initial equation and then figure out what r_a is. So let's just go ahead and do that really quick. If we were trying to figure out what r_a is by isolating it so let's see. We're going to get that, 2 a. Right? We're going to multiply this up, and then we're going to have to subtract the r_p over. So 2_a minus r_p is going to be our _a. Right? That's going to be our expression, and that just comes directly from this equation right here. Great. So we've got r_a is just going to be Let's see. We've got oura is going to be 2 times we've got 2.38 times 10 to the¹³ minus the periapsis distance, which is in kilometers. So you have to be very careful because this unit right here is given in kilometers. So this is actually 3 times 10 to¹¹ in meters. Right? So this has to be in meters, and this has to be meters. So that means that our apoapsis distance over here, what I get is I get 4.73 times 10 to the¹³ in meters. And that's the answer to part a. So it goes very, very far out and then it comes in close. Usually, that's what comets do.

Alright. So for part b, now we're figuring out what is the ratio of the comet's speed at its closest point to its farthest point. What does that mean? We're looking for a ratio, so that means it's going to be something divided by something else. Now remember, the velocity at its closest point is going to be v_p, and its velocity at its farthest point is going to be v_a. So we're looking for this ratio over here. Now because we're comparing the speeds of 2 different points in the orbit, now we can act