Astronomy C

[JCicc]

Well - maybe just a few boos. More like a collective groan. The 2015 exams should be up soon.

[syo_astro]

I saw they were uploaded, they just need to be transferred to the test exchange. Regardless of what the groans say, I know people on this forum (or just people that browse for useful resources) find it useful! In case you have not seen me say it before , your tests being put up online alone have really done a deal to influence my test writing style and studying, which I thank you very much for. I hope to supervise until I drop dead too .

[EastStroudsburg13]

I always found it simultaneously disappointing and incredibly satisfying to hear that sort of sigh (we would hear it at regionals as well, even though that test was definitely at an easier level than the states ones. If I recall, for three years it was Athens and me, and then a huge dropoff for everyone else ). Disappointing because I knew it was a fantastic test, and if the other teams had legitimately prepared (and seen some of the really cool stuff happening in the event), the could have done significantly better, but also incredibly satisfying because I instantly knew I had placed higher than the majority of teams.

[Techsam]

I was looking over the test from the PA states tournament and I cant seem to get the same answer as the key for #59 on section three of the Astronomy test.

Basically you are given the orbit of a planet around a star and you are asked to calculate the total mechanical energy of the system. I got the first two questions which asked me to find the Semi-major axis using the periastron and the apastron (you average them). I also found the eccentricity which is found using:

Now based on what I knew I said:

And then I simplified that by saying

Now based on balanced forces we know that which simplifies to

You can put that all together to get that where R = semi-major axis

Now I plug in all the masses and the semi-major axis and I get the answer to be -3.697 * 10^36 J

[JCicc]

Not sure why your answer does not match. I just did it again and got the reported answer. Maybe your masses are different? I used 1.99E30 kg for the sun, 1.898E27 kg for MJ. My semimajor axis is 7.39E10 m.

[Techsam]

Ahhh, the semi-major axis we are using are different. According to the key, the semi-major axis is .716 AU or roughly 1.07E11 meters

[Crazy Puny Man]

I was looking over the test from the PA states tournament and I cant seem to get the same answer as the key for #59 on section three of the Astronomy test.

Basically you are given the orbit of a planet around a star and you are asked to calculate the total mechanical energy of the system. I got the first two questions which asked me to find the Semi-major axis using the periastron and the apastron (you average them). I also found the eccentricity which is found using:

Now based on what I knew I said:

And then I simplified that by saying

Now based on balanced forces we know that which simplifies to

You can put that all together to get that where R = semi-major axis

Now I plug in all the masses and the semi-major axis and I get the answer to be -3.697 * 10^36 J

Hang on - doesn't centripetal force (m1v^2/r) only apply to perfectly circular orbits? Or can you apply them with elliptical orbits as well, if you use the semi-major axis?

[Techsam]

I was looking over the test from the PA states tournament and I cant seem to get the same answer as the key for #59 on section three of the Astronomy test.

Basically you are given the orbit of a planet around a star and you are asked to calculate the total mechanical energy of the system. I got the first two questions which asked me to find the Semi-major axis using the periastron and the apastron (you average them). I also found the eccentricity which is found using:

Now based on what I knew I said:

And then I simplified that by saying

Now based on balanced forces we know that which simplifies to

You can put that all together to get that where R = semi-major axis

Now I plug in all the masses and the semi-major axis and I get the answer to be -3.697 * 10^36 J

Hang on - doesn't centripetal force (m1v^2/r) only apply to perfectly circular orbits? Or can you apply them with elliptical orbits as well, if you use the semi-major axis?

Exactly, since you are using the semi-major axis you can use this equation. The total energy stays constant in an elliptical orbit, but the amount potential and kinetic energy chances along the elliptical orbit.

[JCicc]

That's my issue. It seems that I changed something and forgot to change the original data. I will update as soon as I can.

[Techsam]

That's my issue. It seems that I changed something and forgot to change the original data. I will update as soon as I can.

Don't worry! I actually just looked on wikipedia and the actual semi-major axis is as you said about .495 AU

But just for clearing my mind, is the semi-major axis just the average of the periastron and the apastron?