A simple challenge for those who claim they like topics related to space exploration

[rodsky]

you'd be surprised to find out that 95% of my class had these lasting negative impressions. im a bit thankful that the tides of change has arrived.. but i'm on the *merciful* side (kuno), and find that particular teacher a bit pathetic and pitiful because most of the students who know of his *reputation* will not enrol anymore again in his class. but i wouldn't go so far as call that teacher *evil*, as I also see the good points in him.. but i believe it can be delivered without the *intimidation* aspect.

I've had my share of "bad" or "terror" teachers, and could admit I suffered and failed under some of these teachers as well. But as the years went by, I realized that, if I simply didn't mind how they behaved socially or in-class, and instead, focused on the subject matter they were trying to teach, then perhaps I could've learned more than I actually did.

Going back to your answer.

1. There is no such thing as a "weak point" in an elliptical orbit around a gravitating body.

2. Yes, the answer has something to do with fuel and thrust, but you didn't explain properly how and why an elliptical orbit would involve less/more thrust and less/more fuel.

If I posed this question under a geometry/trionometry/topology thread, then perhaps your answer would have more merit, despite its lapses, because yes, like you said, it all boils down to mathematics. However, this is NOT a thread about the number-crunching to arrive at a solution--this is thread about space exploration, ergo, you have to describe in detail (kung pwede pa gani, dapat mag post ug diagrams, the same way I do them...you said you are a teacher, and I can only praise of the invaluable use of VISUAL AIDS to explain a topic/point) the mechanics behind why this or that should be done.

-RODION

[bluedes]

I've had my share of "bad" or "terror" teachers, and could admit I suffered and failed under some of these teachers as well. But as the years went by, I realized that, if I simply didn't mind how they behaved socially or in-class, and instead, focused on the subject matter they were trying to teach, then perhaps I could've learned more than I actually did.

oh, i've learned that in college, never to rely solely on the teacher. and that terror teacher i mentioned could not fail me because he simply cannot. i aced his exams and occasionally sleep in class. but i feel sympathy for my fellow classmates who really felt intimidated.. that is why i dont intimidate my students at all. on the contrary, i challenge them. but its hard changing the culture, so its one step at a time.

Going back to your answer.

1. There is no such thing as a "weak point" in an elliptical orbit around a gravitating body.

2. Yes, the answer has something to do with fuel and thrust, but you didn't explain properly how and why an elliptical orbit would involve less/more thrust and less/more fuel.

fine, fine.. let's get technical then.. but i will still steer clear of the semantics. I do not major in astronomy. It is only my personal interests that I have studied astronomy.

anyway, its not a *weak point* per se, but in mathematics, there are points of entry in an elliptical path where it requires the least energy burst to achieve orbit. this is because an elliptical orbit has two centers of gravity defined by their locii, calculate those coordinates wherein it has the nearest distance to the chosen center of gravity (therefore the least angular velocity required to achieve orbit), and you can start launching the satellite from that point of entry. compare that with a circular orbit wherein you only have one center of gravity, and everywhere is just the same.

you can supply the terms required to make this answer more astronomically sounding. i come from a mathematical background, you see and in mathematics, words are just symbols used to convey an abstract idea, and we (mathematicians) are not *overly-dramatic* with such words.. as long as the concepts are understood. professionally though, a convention of terms are important. but this is a public forum, not the least professional. the approach i believe then is more of an informational tone rather than a debating attitude on the terms being used.

If I posed this question under a geometry/trionometry/topology thread, then perhaps your answer would have more merit, despite its lapses, because yes, like you said, it all boils down to mathematics. However, this is NOT a thread about the number-crunching to arrive at a solution--this is thread about space exploration, ergo, you have to describe in detail (kung pwede pa gani, dapat mag post ug diagrams, the same way I do them...you said you are a teacher, and I can only praise of the invaluable use of VISUAL AIDS to explain a topic/point) the mechanics behind why this or that should be done.

-RODION

i will not go into details of the mathematics behind it, but the basis of the answer is actually the mathematics. you can't prove science without the math.

anyway, i'm just here to post my ideas in textual format.. but i'll take your challenge and help post some images here too.. let me google up some images muna..

[brownprose]

Rodsky, indulge my answer being only just an amateur in Physics.

Based on the illustration given, the shuttle doesn't need to obtain more thrust since outer space is almost frictionless unlike in aerospace. Once the shuttle reaches low-earth orbit as in the ISS, the shuttle does not need to approach directly or follow the ISS as it is already orbiting the earth in speeds of around 28,000/hr per your estimate. Given the speed of orbit, the ISS can come back in the same position in approximately 90 minutes (following the angular velocity of the earth). So the shuttle only needs to compute a point of rendezvous (instead of catching the ISS) where the shuttle can safely dock and follow the same orbital path of the ISS.

[gregggy_ph]

wazzap 5th graders??hehe

thank you master rodsky for the beautifully explaining the solution to the problem..

will go back to mababang paaralan of timbuktu and ask officials der to affiliate with dost..hehe

[rodsky]

Rodsky, indulge my answer being only just an amateur in Physics.

Based on the illustration given, the shuttle doesn't need to obtain more thrust since outer space is almost frictionless unlike in aerospace. Once the shuttle reaches low-earth orbit as in the ISS, the shuttle does not need to approach directly or follow the ISS as it is already orbiting the earth in speeds of around 28,000/hr per your estimate. Given the speed of orbit, the ISS can come back in the same position in approximately 90 minutes (following the angular velocity of the earth). So the shuttle only needs to compute a point of rendezvous (instead of catching the ISS) where the shuttle can safely dock and follow the same orbital path of the ISS.

Sorry brownprose, that is incorrect. The shuttle still has to do some catching up, considering the fact that the ISS is over 300 kilometers away, and the only way to do this is:

a) by first turning retrograde and then doing a burn (as per my previous explanation, see illustrations) until the shuttle's orbital radius is smaller than the ISS'...

b) then waiting until it gets closer to the ISS, and THEN turning prograde and then doing another burn, to enlarge its orbital radius (until it gets close to the ISS' orbital radius size ) to slow down or be equal in speed to the ISS.

c) then doing small RCS thrusts (if the ISS is 100 meters away or less), until it drifts toward a position for station-keeping and eventually, docking.

-RODION

[rodsky]

anyway, its not a *weak point* per se, but in mathematics, there are points of entry in an elliptical path where it requires the least energy burst to achieve orbit. this is because an elliptical orbit has two centers of gravity defined by their locii, calculate those coordinates wherein it has the nearest distance to the chosen center of gravity (therefore the least angular velocity required to achieve orbit), and you can start launching the satellite from that point of entry. compare that with a circular orbit wherein you only have one center of gravity, and everywhere is just the same.

A good answer, but insufficient explanation to clearly illustrate the situation. I agree that mathematics can answer the question, no doubt about that, but if you ask the person next to you here in this thread, if he understood what you meant, do you think he/she would?

There lies the difference between you and me. You want to show the answer, but are reluctant to show to others (here in iStorya) why the answer should be that answer (and instead offer the "nice" excuse of "it's all in the math"), and on the other hand, I also want to show the answer, by via the use of various methods of analogy and illustrations/diagrams designed for those who are not technically inclined, nor well-versed with the math aspect of the problem.

I'm still waiting for randzg's answer. By 6pm today, I shall post my answer + diagrams/illustrations to demonstrate it.

-RODION

[brownprose]

Sorry brownprose, that is incorrect. The shuttle still has to do some catching up, considering the fact that the ISS is over 300 kilometers away, and the only way to do this is:

a) by first turning retrograde and then doing a burn (as per my previous explanation, see illustrations) until the shuttle's orbital radius is smaller than the ISS'...

b) then waiting until it gets closer to the ISS, and THEN turning prograde and then doing another burn, to enlarge its orbital radius (until it gets close to the ISS' orbital radius size ) to slow down or be equal in speed to the ISS.

c) then doing small RCS thrusts (if the ISS is 100 meters away or less), until it drifts toward a position for station-keeping and eventually, docking.

-RODION

*nosebleeds on prograde and retrograde* but i'm learning.

[rodsky]

*nosebleeds on prograde and retrograde* but i'm learning.

A prograde orientation is one that has the ship facing in the same direction as it is travelling in its orbit. A prograde burn is a burn made in that direction, i.e. that adds to the ship's orbital speed.

A retrograde orientation is one that has the ship facing in the opposite direction from its direction of travel in its orbit. A retrograde burn is a burn made in that direction, i.e. that subtracts from the ship's orbital speed.



Thus, when the shuttle is facing into the direction of its orbit, it's in prograde orientation. If it's facing away from the direction of orbit (i.e. "lubot first" hehehe), then it's in retrograde orientation.

-RODION

[randzg]

hey rodsky. sorry i just got online now.

Anyway, i will try:

1. elliptical orbit is easier because if you are trying to achieve a circular orbit, the velocity and fuel burning you will need has to be very accurate. If you burn fuel a little to much, elliptical na dayon, a little to less, elliptical na pud dayon. if this is the case, aw, mas sayon cguro mag achieve nalang elliptical orbit.

2. Earth is not a perfect sphere to begin with.. naa cguro changes in gravitational pull (small maybe).

3. Earth is also rotating around the sun and has velocity - assuming ari ta sa equator dapit mo orbit.

4. Gravity from other bodies like moon might also affect orbit.

How did i do?

[cptn_star]

geeeshh...these stuff could be learned in a small while but even the simulation of doing the whole duration of the shuttle's mission could take me years to master....astronauts do deserve their big paychecks...