Indian Space Programme Discussion

Varoon Shekhar · Post by **Varoon Shekhar** » 18 Aug 2016 02:42

^
Thanks, that's good to know! Monsoon as well as the AN-32 search, is causing the delay. Rather than some technological or political issue.

However, the disappearance of "Resourcesat-2A" from any launch mention, is somewhat puzzling. Originally, it was going up first, followed by "Scatsat" on a different mission. One report even spoke of both Resourcesat and Scatsat being launched on the same mission. That is no longer the case, evidently.

Amber G. · Post by **Amber G.** » 18 Aug 2016 03:53

Wow! Saw some interesting discussion from - Vina, Disha and *many* others.. There were about dozen posts from many people (starting from viewtopic.php?f=3&p=2033119 and lasting for a few PAGES..some one commented ..

Singha wrote:looks like strong JEE kamandu physics fundamentals are needed to move an inch up there.

Don't know if the issues were resolved to everyone's satisfaction but I saw some good science posts and some, let us say "interesting" viewpoints/logic ityadi.. so coming back to technical part of the topic let me add to clarify -- so that everyone is on the same page - first the problem:

Assume a satellite is in circular stable orbit (say 500 Km from surface of earth). Now it fires a thruster (rocket) which gives it a delta-V (change in velocity of say 100 m/s). (Of course Delta-V is in opposite direction of the escaping rocket gas.). Let us consider four cases for the direction of delta-V..

Case A - "UP" (that is outward the line joining the earth and satellite) -- Thrusters are fired straight down --.
Case B - "FORWARD" (In the direction of the orbital velocity) -- Thrusters are fired backwards --
Case C - "BACKWARDS" ( opposite to the direction of the orbital motion)
Case D - "DOWN" (towards earth)

In each case, after the initial jerky motion ends, will the satellite have a new stable orbit. If so, will the new orbit be a) higher or b) lower or c) same? (If new orbit is not circular - (will it be?) use average distance from earth to calculate "higher" or "lower" etc..

Will be interesting to see, if each of previous contributor gives clear answer. Let us see if we have common answer.
(One can skip the reason now if the answer is obvious -- just give the answer).

RoyG · Post by **RoyG** » 18 Aug 2016 07:43

kit wrote:has india any ongoing research into quantum communication tech ? satellites?

prasannasimha · Post by **prasannasimha** » 18 Aug 2016 07:49

If you increase your tangential velocity in line you move up to a higher orbit.(and vice versa) The perpendicular velocity has to be imparted to go back to parking circular orbit.

SSSalvi · Post by **SSSalvi** » 18 Aug 2016 19:58

Here is a overall coverage of IRNSS constellation with 1A through 1G at the time of posting :

The system covers almost entire Earth except Americas.

Amber G. · Post by **Amber G.** » 18 Aug 2016 22:48

^^^ Thanks.

Bheeshma · Post by **Bheeshma** » 18 Aug 2016 23:38

With another two sats in GSO and 4 in the GSO , I am sure we can extend the coverage from Africa to Australasia. That will only leave out the americas.

Amber G. · Post by **Amber G.** » 19 Aug 2016 01:59

^^^
One needs AT LEAST 4 (preferably a few more for required accuracy) satallites visible at the same time to be useful.
The "position" of these seven geosynchronous is more or less "fixed" with respect to us (on the ground). The accuracy for the land mass of India is greatest (~10 m for public, probably better for military) while surrounding area will be less.
One can geometrically "cover" (see) about 80 degrees (50-60 degrees in practice say for weather - much less to be useful accuracy for GPS) from a geosynchronous satellite (easy to calculate, it is about 36,000 Km "up").. it has a large coverage area -- but to extend it further we will need more satallites...

Meanwhile from isro site ..

disha · Post by **disha** » 19 Aug 2016 03:18

Amber G. wrote:
Case A - "UP" (that is outward the line joining the earth and satellite) -- Thrusters are fired straight down --.
Case B - "FORWARD" (In the direction of the orbital velocity) -- Thrusters are fired backwards --
Case C - "BACKWARDS" ( opposite to the direction of the orbital motion)
Case D - "DOWN" (towards earth)

A question., is Case A - "UP" down or "UP" up? Then what is case D - "Down"? "Down" down or "Down" Up?

(ignore this line).

I am assuming Case A -> sat fires the thrusters into the space (and towards an observer hovering over the sat) and on the straight line joining the earth to Sat. Perpendicular to the delta-Direction of the sat at the time of firing.

I am assuming Case D -> sat fires the thrusters towards earth (and away from an observer hovering over the sat) and on the straight line joining the earth to sat (and the observer). Perpendicular to the delta-Direction of the sat at the time of firing.

Correct assumptions?

Amber G. · Post by **Amber G.** » 19 Aug 2016 03:51

"UP" means the direction from center of the earth towards the sat.... ( to achieve this, The thrusters are fired towards earth or in 'down' direction and hence the "reaction" (or delta-V is given in the opposite direction).

(This is, of course perpendicular to direction to the the original orbital motion of the sat)...

****

What I am trying to do is to make the problem clear for understanding. Same problem, say if one fired a gun from a space station and you are asked to find the path (orbit) of the bullet.. or even if some one "jumped" (using one's muscles as "spring") -- tough the "push"/delta-V produced from jumping may be much smaller

..

Basically I am asking, in each of the four cases - will the new orbit will be "higher" or "lower" . (Will the speed (average after the initial jerk) be more or less and the time it takes to go around the orbit is loner or shorter)

It is a straight-forward question, and answer is simple and may help understand the basic physics.

SriKumar · Post by **SriKumar** » 19 Aug 2016 03:55

Amber G. wrote: Assume a satellite is in circular stable orbit[/b] (say 500 Km from surface of earth). Now it fires a thruster (rocket) which gives it a delta-V (change in velocity of say 100 m/s). (Of course Delta-V is in opposite direction of the escaping rocket gas.). Let us consider four cases for the direction of delta-V..

Case A - "UP" (that is outward the line joining the earth and satellite) -- Thrusters are fired straight down --.
Case B - "FORWARD" (In the direction of the orbital velocity) -- Thrusters are fired backwards --
Case C - "BACKWARDS" ( opposite to the direction of the orbital motion)
Case D - "DOWN" (towards earth)

In each case, after the initial jerky motion ends, will the satellite have a new stable orbit. If so, will the new orbit be a) higher or b) lower or c) same? (If new orbit is not circular - (will it be?) use average distance from earth to calculate "higher" or "lower" etc..

Well, the spotlight has been applied. HEre's my 10 paisa.......assuming only one single thrust from a thruster:
Case A: UP- New stable orbit, elliptical. Higher perigee, higher apogee.
Case B: Forward: Stable elliptical orbit. 'Higher' orbit, in the sense... apogee is higher, but perigiee is the same as circular orbit. This orbit is a different orbit from Case A.
Case C: Backwards: Orbit decays and it crashes to earth
Case D: Down: Orbit decays and it crashes to earth.
Using a smaller font in case others want to post their thoughts. DO a 'Quote' to see

Amber G. · Post by **Amber G.** » 19 Aug 2016 07:26

SriKumar wrote:
Amber G. wrote: Assume a satellite is in circular stable orbit ....{ see previous post for full description }
.
Well, the spotlight has been applied. HEre's my 10 paisa.......assuming only one single thrust from a thruster:
Case A: UP- New stable orbit, elliptical. Higher perigee, higher apogee.
Case B: Forward: Stable elliptical orbit. 'Higher' orbit, in the sense... apogee is higher, but perigiee is the same as circular orbit. This orbit is a different orbit from Case A.
Case C: Backwards: Orbit decays and it crashes to earth
Case D: Down: Orbit decays and it crashes to earth.
Using a smaller font in case others want to post their thoughts. DO a 'Quote' to see

Thanks for clear replies. (BTW You are correct in assuming only one single thrust). It will be interesting to see some more posts form others, and to see if all of them in agreement

. For the record I 100% agree with answers given above for ONE of the four cases.

)

SriKumar · Post by **SriKumar** » 19 Aug 2016 07:56

Amber.G,
bummer.....but I'll take what I can get. By the way, when you get a chance, I'd like to draw your attention to a post in the physics thread. I know it is not a place you might want to visit (and I dont blame you), but there is a post made by poster ' sudarshan' who answered a long-standing question I had about the use of rashi-valaya yantra in Jaipur Jantar MAntar. He created a nice image in MATLAB that shows the orientation of earth's axis across seasons relative to ecliptic axis. He later found a (Small) error in the picture he created- constellations were counter-clockwise..., and posted the corrected picture further down.

viewtopic.php?p=1994869#p1994869

Atleast for me, it solved the question for me as to _how_ the rashi-valaya yantra can be used to detect the zodiac constellation. _However_, in looking at the picture it became clear to me that (IMHO) if one knows enough astronomy to design a rashivalaya yantra, one can easily use the night sky to determine the zodiac constellation of the sun (it is the same constellation 6 month away). This begs the question 'why did they need a rashivalaya yantra'. It does not seem like it was needed to determine sun's position in the zodiac constellation- this is the commonly posted function of the RV yantra. If you respond to this anywhere, please let me know. Thanks.

(Mods: Topic not truly OT because it is about space research, as performed in India using indigenous instrumentation, about 300 years ago. The sundial at Jaipur is the largest man-made sundial in the world, as far as I know).

SSSalvi · Post by **SSSalvi** » 19 Aug 2016 12:28

^^^
I had visited the observatory some 30 years ago when I did not have much interest in astronomy. If I had then I would have keenly seen the features.

But if I remember clearly the markings are better than a minute ( of an arc ) graduations. So my gut feeling is it was used to create the Kundali .. I doubt if the printed panchangs were easily available then. So one had to have the knowledge from a proper guru ( only ).

With the observatory it could have been possible for a non-scholar to at least find a rough rashiphalam without indulging in Navmansh table corrections etc.

SriKumar · Post by **SriKumar** » 19 Aug 2016 17:16

^^^ Definitely a place worth visiting...I have not. Thanks to intenet and the posting of pictures, you now have detailed pictures on the web. I posted a few pictures of the markings on the RV yantra in the same BR discussion (or you could google it). In looking for the function of the RV yatra, (i) almost all sites I have seen say it is to determine the constellation of the sun. (ii) Almost nowhere do you find any explanation of _how_ this is be done (other than a very short para). Only sudrashan's detailed drawings actually shed some light on how this is done (there was one other source I found). After seeing the explanations, I now doubt that the purpose was to merely determine the constellation (it can be done simply by looking at the night sky, no RV needed)- but no discussion/descriptions, even on official sites, discusses this. Your explanation of kundali creation might be a possibility. Another related possibility may be that this was the lab from where the panchangs (i.e. astronomical record of position of sun, moon, stars) were calibrated to correct for observational errors* accumulating over time (i.e. over years/decades).
*-astronomical drift might be a better term.

vina · Post by **vina** » 19 Aug 2016 17:30

Amber G. wrote:
Case A - "UP" (that is outward the line joining the earth and satellite) -- Thrusters are fired straight down --.
Case B - "FORWARD" (In the direction of the orbital velocity) -- Thrusters are fired backwards --
Case C - "BACKWARDS" ( opposite to the direction of the orbital motion)
Case D - "DOWN" (towards earth)

Ok. Let me take a stab at this.

Case B -- > Discussed. The orbit goes higher
Case C --> Discussed. The orbit goes lower

Case A -- > This is analogous to throwing a ball straight up from the earth's surface. The earth's gravity is pulling it right down (while it is still in orbit). Suppose, it was a height R (was 500km from the surface the numeric you gave) from the earth's center and the delta V translates into a further radial distance r (r can be worked out given v) , then the force acting on the body will be GMm/(R+r)^2 .
Now if take a first degree approx, the local "g" at that (R+r)this can be reduced to (GM/R^2 )*(1- 2r/R) ..

This looks suspiciously like the F = -Kx kind of Simple Harmonic Motion stuff with the center at a distance R from the earth. So I will stick my neck out and venture that in this case, the body in orbit @ radial distance R, but with a Simple Harmonic motion centered on R superimposed on the orbit.

Case D --> Same as case A, but with signs reversed. So again I would think same answer as A, but maybe with a different phase (if Case A is sin, Case B is probably Cos , if 90 deg phase shift).. Haven't put pen on paper to be sure, but just a quick guess.

Amber G. · Post by **Amber G.** » 19 Aug 2016 21:20

SriKumar wrote:Amber.G,
By the way, when you get a chance, I'd like to draw your attention to a post in the physics thread. I know it is not a place you might want to visit (and I dont blame you), but there is a post made by poster ' sudarshan' who answered a long-standing question I had about the use of rashi-valaya yantra in Jaipur Jantar MAntar. He created a nice image in MATLAB that shows the orientation of earth's axis across seasons relative to ecliptic axis. He later found a (Small) error in the picture he created- constellations were counter-clockwise..., and posted the corrected picture further down.

viewtopic.php?p=1994869#p1994869

<snip>

(Mods: Topic not truly OT because it is about space research, as performed in India using indigenous instrumentation, about 300 years ago. The sundial at Jaipur is the largest man-made sundial in the world, as far as I know).

SriKumar,
I thought Physics Thread went into bit bucket. Now I see it is in GDF which I have not visited. (I had some other priorities so was less active in brf, then I think forum itself was down for a while, now your comment drew my attention to physics dhaga and I must say, when I went through a few posts, one post left me speechless.). Hope physics thread moves back to T&E.

To comment on your question, please see response to your post (<this post> shown here (and may be a few comments near this post for context):
viewtopic.php?p=1594121#p1594121
and this .. viewtopic.php?p=1594962#p1594962

I think if you read it carefully, it will answer your query. I may post more later (or answer specific question if I can)
Basically as said before (please read those posts or look up google "Rashivalaya Yantras")
These are to measure sidereal time. Our daily activities depend on Sun/solar time (24 hours) so the main sundial measures this.
Since the north pole is in "fixed" position wrt to ground, it is easy to construct sundial, you need one fixed, and the slab which gives shadow points to north at an angle=latitude of Jaiput)

Of course, it measures "apparent solar time" and to be accurate with "standard" time one has to add/subtract "equation of time" value.

(Of course, these structures were used to 'calibrate' local clocks which may drift over time so that these clocks remain accurate)

If you are observing stars, you need sidereal time ...If you are documenting positions of the planets, the ecliptic coordinate system may be more convenient. (If you have a calculator with trigonometric functions you may not need separate structures as the calculations can be done faster - but in those days it would have taken more time)

Actually all 12 structures combined is just one instrument. To use it:

First find out which "rashi" the sun is - You can do this by looking a different yantra also a part of the observatory - or just know which month of the year it is)

Now you go to that particular "yantra" from above picture (one of the 12)

And look up the curved scale of that instrument -- lines up with the (1/12) part of the ecliptic that contains the constellation)

You will notice that each of those 12 structures do not point to exact north, or inclination is not the same as latitude of Jaipur

Hope this helps.

Amber G. · Post by **Amber G.** » 19 Aug 2016 22:26

SriKumar, SSS
May be some find this 1954 Hindustan Times picture interesting...(This was designed/constructed by my father).
I liked Jaipur's jantar-mantar a lot, learned a lot about astronomy, spherical trigonometry etc.. and agree with SriKumar that the place is fun and educational to visit. Last time when I took my son and DIL to India, we had a very short time and were able to spend only few hours in Jaipur and spent majority of that time there!
I am partial so I admire the jantar mantar a lot but still have to agree that at the time when this was constructed Europe already had telescope(s) and those observatories - even using smaller structures - were much more precise than Jaipur's. Jaipur had scientists, books (the King had a VERY good collection of all the current work of European astronomers and their methods -- the library still has very valuable books).. yet, IMO they wanted to keep the old tradition so constructed all those structures remaining faithful to Indian designs - and the observatory was less accurate than other best observatories of that era.

Amber G. · Post by **Amber G.** » 19 Aug 2016 22:30

vina wrote:
Amber G. wrote:
Case A - "UP" (that is outward the line joining the earth and satellite) -- Thrusters are fired straight down --.
Case B - "FORWARD" (In the direction of the orbital velocity) -- Thrusters are fired backwards --
Case C - "BACKWARDS" ( opposite to the direction of the orbital motion)
Case D - "DOWN" (towards earth)
Ok. Let me take a stab at this.

Case B -- > Discussed. The orbit goes higher
Case C --> Discussed. The orbit goes lower

Case A -- > This is analogous to throwing a ball straight up from the earth's surface. The earth's gravity is pulling it right down (while it is still in orbit). Suppose, it was a height R (was 500km from the surface the numeric you gave) from the earth's center and the delta V translates into a further radial distance r (r can be worked out given v) , then the force acting on the body will be GMm/(R+r)^2 .
Now if take a first degree approx, the local "g" at that (R+r)this can be reduced to (GM/R^2 )*(1- 2r/R) ..

This looks suspiciously like the F = -Kx kind of Simple Harmonic Motion stuff with the center at a distance R from the earth. So I will stick my neck out and venture that in this case, the body in orbit @ radial distance R, but with a Simple Harmonic motion centered on R superimposed on the orbit.

Case D --> Same as case A, but with signs reversed. So again I would think same answer as A, but maybe with a different phase (if Case A is sin, Case B is probably Cos , if 90 deg phase shift).. Haven't put pen on paper to be sure, but just a quick guess.

Thanks for taking a stab. It seems that your conclusions are different (in 3 cases) than SriKumar's.

(This question is essentially what Shiv and a few others asked/discussed/agreed/disagreed..His two bodies moved in opposite directions so if one piece was case B then the other was case C (or one piece was case A than other piece followed case D)

disha · Post by **disha** » 20 Aug 2016 00:04

AmberG., the answer is simple:

Let us do Case B and Case C first., In

Case B., the sat gains additional delta-V and achieves a new higher orbit. If it is fired at Apogee (like in LAM firing) it gets the most delta-V.
Case C., exactly reverse of Case B., orbit of satellite degrades thus a new low orbit is achieved. The most negative delta-V is achieved this way. This is the way gaganaviharins come back to earth

Now Case A., think discrete. That at the point of the motor kicking sat up, at that delta-point a velocity is imparted to the sat akin to throwing a ball vertically., of course the sat is still in Earth's gravity well and the sat rises up and falls down. The orbital velocity remains the same at the time when the sat reaches the plane as it was before when it was kicked up. If the sat were stationary, it will be akin to throw the sat up vertically. However since the sat is in a orbital flight., the effect to a observer in space is the sat going over a "bump" at the same speed, that is the sat will rise up and then down tracing an arc for the observer in space parallel to sat and at some distance.

A simpler analogy is when sats encounter the gravity weak spot south of S. Lanka - the sat gently goes up and then comes down and continues in its orbit.

Case D., is not reverse of case A. The motor is fired towards earth and the delta-V decreases the distance between two masses., however small this increases the attraction between sat and earth and the sat starts its journey towards earth., slowly accelerating. This is seen in a general degradation in its orbit. If it is not corrected, the sat's orbit eventually degrades enough for it to come back to earth within a short span of time. Though not as quicker as case C.

This is the same principle why Sats have LAM for orbit maintenance. The solar wind acts on the sat slowly pushing it towards the earth. Another reason why Insat-A series had a boom to balance the wind pressure on its other surfaces (other wise the sat starts slowly rotating on its axis - or more work for its gyros to resist the unbalanced pressure).

Amber G. · Post by **Amber G.** » 20 Aug 2016 08:24

disha wrote:AmberG., the answer is simple: ....

Hi Dishaji - Looks like your answers, in few cases, are still different from others..Also, sorry (and certainly no disrespect) but some parts are not even clear (to me at-least).. for example..what does one mean by " If it is fired at Apogee (like in LAM firing) it gets the most delta-V. " -- The initial orbit is circular and hence NO apogee and delta-V is delta-V ( (~100m/s) as given - it depends on the power of thrusters and not position. Perhaps there is confusion with some special transfer orbit maneuvers where selecting delta-V at one particular point is more efficient...

But let us not get sidetracked .. as you said let us keep it simple, from what I can make out (please correct me or add if I am wrong in interpreting you).. I am quoting only parts which are clear to me --.
>>

Case B., the sat [gains additional delta-V] and achieves a new higher orbit.

(This is somewhat consistent with others - but others are saying "higher orbit" means "slower" - so what does it do? gains or loose velocity?)

Case C., exactly reverse of Case B., orbit of satellite degrades thus a new low orbit is achieved. The most negative delta-V is achieved this way. This is the way gaganaviharins come back to earth

Fine, "somewhat/maybe" similar other's answers but again - "lower" orbit means higher velocity and I am still not sure exactly what you mean.. Will the sat come back to earth (in perhaps a spiral ?) or just have a lower *stable* orbit? (Later you compare with case D but put a qualifier "not that quickly")

Case A., think discrete. That at the point of the motor kicking sat up, at that delta-point a velocity is imparted to the sat akin to throwing a ball vertically., of course the sat is still in Earth's gravity well and the sat rises up and falls down. The orbital velocity remains the same at the time when the sat reaches the plane as it was before when it was kicked up. If the sat were stationary, it will be akin to throw the sat up vertically. However since the sat is in a orbital flight., the effect to a observer in space is the sat going over a "bump" at the same speed, that is the sat will rise up and then down tracing an arc for the observer in space parallel to sat and at some distance.

Again I don't exactly know what do you mean by 'observer in space', and what 'plane'??

In simple terms - what the orbit will be? Will it be stable? Will it be higher or lower? (Or does a circular orbit with "bumps" has any standard name?)

Case D., is not reverse of case A. The motor is fired towards earth and the delta-V decreases the distance between two masses., however small this increases the attraction between sat and earth and the sat starts its journey towards earth., slowly accelerating. This is seen in a general degradation in its orbit. If it is not corrected, the sat's orbit eventually degrades enough for it to come back to earth within a short span of time. Though not as quicker as case C.

Again assuming no further 'corrections' what this orbit looks like. What is 'short span' of time to come back to earth - day(s) or month(s) or years or centuries..?

****

Folks, again I am just talking about simple physics and what WILL happen if the experiment is carried out in practice. Basically (in each case)

- Will (new) orbit be stable? If stable orbit - Will it be "higher"/"lower"? Will it take longer or shorter time to go around the earth?

(And we are NOT concerned about atmosphere drag/solar wind or collision with space junk etc..our time frame is say from seconds AFTER the thrusters come to rest to say next 10-20 years -- so you can safely ignore minor effects and consider earth's gravity alone)

I am hoping for one consistent answer which all agree...

vina · Post by **vina** » 20 Aug 2016 09:28

Amber G. wrote: Thanks for taking a stab. It seems that your conclusions are different (in 3 cases) than SriKumar's.

(This question is essentially what Shiv and a few others asked/discussed/agreed/disagreed..His two bodies moved in opposite directions so if one piece was case B then the other was case C (or one piece was case A than other piece followed case D)

I really haven't looked at any other answers, but let me expand on what I wrote for Case A and Case D.

Case A -- > Yes, I think it will be a SHM. For an observer in the same orbit (assuming a viewing from a sister satellite just a few radians separated, it will look like the satellite which fired the thruster in Case A will look like it will be bobbing up and down harmonically , with the mean centered on the orbit. For a ground trace of the orbit, it will look like a sine waveform on an oscilloscope, with the mean however not a straight line , but a circle with radius R (the distance from center of earth) , a rather "circular" sine wave, than "flat" sine wave for want of a better means of describing i.

Case D --> Yes, it does look like the opposite sign of Case A, with a phase difference . Again not putting pen to paper, I think it probably will be 180deg phase difference from Case A, not 90 deg like I wrote earlier.

That said , it is intuitively like this. The case, where you theoretically drill a tunnel from one side of the earth's surface to the opposite side, going through the center, (ie a tunnel through a diameter of the earth) and you drop a ball into it (okay, in Case A , you lob a ball up, it rises against gravity, reaches an apex and drops back, but is really a similar thing as dropping a ball or throwing it down), it becomes a SHM, oscillating between the two apex points with mean at the center of the earth. In this case, the orbit "trace" , if looking from the sun will be like the sine wave looping around to form a circle

In case A and D, the orbit, is like the "center of the earth" equivalent nailed there and the satellite bobs around that like a ball in case of the hypothetical ball dropped in a tunnel through center of earth analogy.

Vayutuvan · Post by **Vayutuvan** » 20 Aug 2016 10:25

AmberG ji, so if we take normal and tangential acceleration (+ and -), that might give some pointers. As disha says, are we looking from earth in a normal direction to the orbit? Because there are infinite number if tangential directions though there is only one normal direction.

Amber G. · Post by **Amber G.** » 20 Aug 2016 17:26

matrimc wrote: Because there are infinite number if tangential directions though there is only one normal direction.

BUT there is ONLY ONE tangential direction in which sat is moving at any single/particular point in time.

vina · Post by **vina** » 20 Aug 2016 19:41

Amber G. wrote: BUT there is ONLY ONE tangential direction in which sat is moving at any single/particular point in time.

I think the questions you posted are clear enough.

Anyways, please look at the Railways thread in "Tech and Economic Forum" . We discussed your question long ago about "How does a trains stay on a track ?" and I pointed out that the "Kinematic" explanation / "coning" does not explain stability , but rather it explains how it follows the curve/steers and pointed the Talgo trainset with independent bogies as an example.

Now that that very same Talgo trainsets are being trialed in India for high speed operations, the question arises how does it go around a curve, since the kinematic explanation/ "coning" doesn't hold ? The guidance and stabilisation mechanisms on that is interesting indeed.

Amber G. · Post by **Amber G.** » 20 Aug 2016 21:59

SriKumar wrote:Amber.G,
....
Atleast for me, it solved the question for me as to _how_ the rashi-valaya yantra can be used to detect the zodiac constellation. _However_, in looking at the picture it became clear to me that (IMHO) if one knows enough astronomy to design a rashivalaya yantra, one can easily use the night sky to determine the zodiac constellation of the sun (it is the same constellation 6 month away). This begs the question 'why did they need a rashivalaya yantra'. It does not seem like it was needed to determine sun's position in the zodiac constellation- this is the commonly posted function of the RV yantra. If you respond to this anywhere, please let me know. Thanks.

(Mods: Topic not truly OT because it is about space research, as performed in India using indigenous instrumentation, about 300 years ago. The sundial at Jaipur is the largest man-made sundial in the world, as far as I know).

I hope my previous post was helpful. Let me know if (and where) there is more discussion. (I can be reached at amber dot shepherd at gmail too)..

(This may be OT for this thread, so this will be last from me -- but allow me to put a post here for convenience - People can safely ignore rest of the message, if not interested)

In short, one can find the rough position of the sun (or other planets for that matter) in zodiac easily enough. It is finding the precise (or as accurate as possible) position in Ecliptic coordinate system where you use this. You need to do this to plot the accurate positions at various time (over years) to make tables/charts (and use data to refine calculations for panchangs etc).

Constructing to use/measure_in Horizontal coordinate system is easy. (You just put a long vertical pole and measure the shadow on the ground) --but to understand motion of sun is quite complex if you use this system. If you want to keep calculations simple for those who use your data this many not be much useful. You need a better observatory to provide data which is easy to use to calculate further astronomical events)

Constructing to use/measure_in Equatorial coordinate system is what most of the yantras in Jaipur does. And measurement is precise. (Big sundial measures hour angle quite accurately and there are other yantras which measure declination.)

Using Ecliptic coordinate system makes calculations/charts (specially for planets/sun/moon's position because latitude is always small, and variation in longitude (wrt to time) in much more uniform ) easier but it is difficult to construct a unit to measure in this system *directly* -- as direction of ecliptic pole is not fixed wrt to ground. Most of the observatories use equatorial system, and convert the data (using calculations/trigonometry) to produce chart in ecliptic system. In Jaipur, one can use one of the 12 yantras , the one which is closest pointing it to the ecliptic pole at that time.

Remember, the observatory was not built JUST for finding time of the day or position of sun at a given point, but to regularly RECORD values over long period of time, and then use the data to calibrate clocks (which were used to record time), charts, and find accurate values of parameters for panchangs to predict/calculate future astronomical events.
Hope this answers your query. Let me know.

disha · Post by **disha** » 20 Aug 2016 22:21

AmberG., forget the LAM firing para - that was orthogonal to the discussion and was just to illustrate a point on how orbit raising maneuvers work. LAM are also sometimes called kick motors!
---
If you are looking for answers:

Case B: Higher stable orbit.
Case C: Lower stable orbit

How:

In Case B., the delta-V imparted increases the PE of the satellite - raising the orbit. That is 'delta-V' is just a kick to the satellite in the direction of its motion to give it a higher PE.

(in my brain) delta-V IS NOT orbital-velocity., (personally I simplify) orbital-velocity as some function of total energy (PE+KE). Orbital velocity will be lower since the object is now higher (delta-v -> PE).

Case C., reverse of Case B.
----
----

Case A: Single "bump or up-down motion" ~~in the orbit~~
Case D: Orbit degenerates slowly, eventually satellite falls down.

Case A., it is a bump in the orbit. After the bump the orbit is the same as before the bump. It will not have any more bumps. Just one single "bump" in the orbit - something like -> ____n____ (the bump is smoother arc and not like 'n')

Initially I also thought that it will "bump-up-down-along-the-orbit" like Vina described. But some more thought leads me to the conclusion that it is bump and that is what I mentioned in my first post earlier.

How: Gravity is the Force (F) which is pulling the sat to the center of the earth, the velocity is 90* perpendicular to force. A 'downward' thruster firing is basically acting against that force - a localized PE is gained against the force - the satellite rises up - since there is no sustained force (f) against the Force (F) which is pulling back the satellite, it loses PE and the satellite "falls" against that force F vector. The original orbit is maintained.

Case D., it is not a trough! Not opposite to Case A.

How: The force works with gravity and the distance between the two masses is reduced, the gravity force is inversely proportional to distance squared., so now the force acting on the body increases further, reducing the distance slightly till the point where the sat falls down. This is seen as degraded orbits. Each orbit is slightly degraded than previous ones.

Amber G. · Post by **Amber G.** » 20 Aug 2016 23:30

^^ and ^^^^ : Thanks for replies.
Matrimc and others - It will be interesting to see some more responses -- even if it just a guess (or intution). Just one line response for each case is ok. (Curious to see the quality of intution of readers of this thread.) Even if you think problem is too easy and answer is obvious, post away the answer too. (As it is I think we have answers/comments from 4 people and *all* responses are not in agreement). (Problem is posted a few posts above <here>

I will post my comments in a short time, so post away.

Amber G. · Post by **Amber G.** » 20 Aug 2016 23:44

prasannasimha wrote:If you increase your tangential velocity in line you move up to a higher orbit.(and vice versa) The perpendicular velocity has to be imparted to go back to parking circular orbit.

This seems to cover case B and C pretty clearly but can you clarify (clear answers) for case A and case D for everyone's benefit. TIA

disha · Post by **disha** » 20 Aug 2016 23:45

I want to correct Case A slightly., I realize that energy has been added to the system and it has to be seen eventually in the rise in PE., so the bump will be a very longish bump in reality mor like a 'r' (ignore the stem) - something like '____r-----', in effect raising the orbit of the satellite to a higher stable orbit.

SriKumar · Post by **SriKumar** » 21 Aug 2016 07:13

Amber G. wrote:...Remember, the observatory was not built JUST for finding time of the day or position of sun at a given point, but to regularly RECORD values over long period of time, and then use the data to calibrate clocks (which were used to record time), charts, and find accurate values of parameters for panchangs to predict/calculate future astronomical events.
Hope this answers your query. Let me know.

THank you for posting details/explanations on this earlier and in this post. I was/am traveling and therefore was not able read it in totality and respond appropriately. I plan to do this shortly. Truly appreciate your mentioning e-khat ka pathaa for follow-up, and I might take you up on it for any future questions(relating to physics that is; and maybe sanskrit ). You may delete it now if you wish.

SriKumar · Post by **SriKumar** » 21 Aug 2016 08:53

I ran some simple (and random) calcs to see what I get. Not sure if they mean anything but here are some (new) thoughts on case A.

In case A where an upward velocity is imparted, I tried to see how that would respond. THe assumption is that a force accelerates the craft upwards to a velocity of 100 m/s and then cuts off. So, what does this do to the orbit? I used 0.5*m*v^2 = m*h*g' (where g' is acc. due to grav. at about 500 km height). Using g = GM/R^2 and R=500 km, I get g= 8.439 m/s. Using this value of g, I get h= 592 m/s. (Actually, I should be using a 'g' at radius (500 km + 0.592 km) but this value is very close, at 8.4383 m/s. So the height estimate is not far off. So, the craft is at a higher orbit by about 592 m.

TO sustain a circular orbit, the centripetal force required should be equal to the force of gravity at that altitude. Once the craft is thrusted upwards, this is no longer the case. Its tangential velocity is still the original 'v' (i.e. orbital velocity for height of 500 km). However, its orbit is now a bit higher. The force of attraction at this new orbit m*GM/(R+h)^2 is not enough to produce the requried centripetal force which is m*v^2/(R+h)....where v= sqrt(GM/R). I am assuming that the tangential velocity 'v' is unchanged from its earlier orbit at 500 km. Required centripetal force at new orbit works out to GM/(R*(R+h)). The required centripetal force is more than the force available for the orbit (R+h) i.e. m*GM/(R+h)^2, so the craft will not be in a purely circular orbit anymore. I think it will go eccentric, albeit very very slightly. I think the orbit will be stable.

SSSalvi · Post by **SSSalvi** » 21 Aug 2016 14:42

You are right Srikumar ji.

The craft will go from a circular to Elliptical orbit.

Since earlier circular orbit was stable the resultant orbit will also be stable unless the impulse is so large that it will escape Earth's field of Influence.

vina · Post by **vina** » 21 Aug 2016 14:51

Okay. I had some time to while away this afternoon and just before taking an afternoon siesta, I thought I will put pen and paper and flesh out my answers to AmberG's questions , Case A and Case D.

Case A :

Consider a frame of reference attached to the satellite in orbit at a distance R from the surface of the earth. Let us assume that the satellite has unit mass for easy of calculations. Now there are there are two "forces" acting on it. They are

1. The Centrifugal force --> w^2 * R (w is the angular velocity)
2. The weight of the satellite -> g(R) (g(R) is the accl due to gravity at height R from center of earth).

Now we have g(R) = w ^2 * R ( equation 1) or g(R) - w^2 *R = 0 (same as equation1)

If due to firing a thruster upwards , the satellite is perturbed by a distance r outwards radially, we have a net force which will be

F = g(R+r) - w^2 *(R+r) ( w is largely const for small displacements) (equation 2)

Now g(R+r) = GM/(R+r)^2) (Newton's law of Gravitation)

This can be re arranged as (GM/R^2) * (1 + r/R)^-2 --> 1st deg approx of --> (GM/R^2) * (1 - 2*r/R)
which can be expanded as GM/R^2 - 2* (GM/R^3) *r (equation 3)

Now g(R) --> GM/(R^2) . So g(R+r) = g(R) - (2GM/R^3) * r

So Equation 2 is now F = g(R) - (2GM/R^3) * r - w^2*R - w^2*r .

Referring to Equation 1, this reduces to F = - (2GM/R^3) *r - w^2 *r --> -1 * ( 2*g(R)/R + g(R)/R) *r --> -(3g(R)/R) *r (equation 4)

Now @ 500 Km.. with surface of earth @ 36500 , the R --> R(e) * 1.014 .. So g(R) -> approx g(surface) / 1.014^2

Equation 4. is of the form F = - K* x , where K is 3g/(R * 1.014^3) and x or r is the distance from the center.

This is the equation of a spring with Force proportional to distance from center .

So the Body is in SHM with a period of 2*pi * sqrt(1/k) .

Case D :

Case D is the same as case A. Now instead of the body going upwards first as in case A, it goes downwards first. So , it out of phase by 180deg .(ie, the starting condition of D is the same as the body in case A which goes up with V, reaches an apex, then descends back to starting point with V downwards and continues).

SriKumar · Post by **SriKumar** » 21 Aug 2016 17:31

vina wrote:Okay. I had some time to while away this afternoon and just before taking an afternoon siesta, I thought I will put pen and paper and flesh out my answers to AmberG's questions , Case A and Case D.

Case A :

Consider a frame of reference attached to the satellite in orbit at a distance R from the surface of the earth. Let us assume that the satellite has unit mass for easy of calculations. Now there are there are two "forces" acting on it. They are

1. The Centrifugal force --> w^2 * R (w is the angular velocity)
2. The weight of the satellite -> g(R) (g(R) is the accl due to gravity at height R from center of earth).

Now we have g(R) = w ^2 * R ( equation 1) or g(R) - w^2 *R = 0 (same as equation1)

If due to firing a thruster upwards , the satellite is perturbed by a distance r outwards radially, we have a net force which will be

F = g(R+r) - w^2 *(R+r) ( w is largely const for small displacements) (equation 2)

vina, I have refrained from commenting on other posts given that amber.G is 'supervising'

the activity, but in this case allow me to jump in.

I am not sure I agree with equation 1 where centrifugal force (you meant centripetal force perhaps or was it centrifugal force- which is fictitious and acts away from the center. We know centripetal force acts towards center) is equated to gravitational force.

As far as I know, there is only ONE force that acts on the orbiting satellite (prior to the thruster getting activated). THis force is the gravitational force, which acts as the centripetal force. THere are no two forces (i.e. gravitational and centripetal) balancing an orbiting satellite; the centripetal force IS the gravitational force. Your earlier post to me several pages ago clarified this point (Thank you for that, actually- I'll confess that the point had not struck me until your post).

So, correct me if I am wrong but there is no force equilibrium in the system (prior to thrusters activating) i..e there are not 2 forces canceling each other out. So, I am not sure that equation 1 (whether assessed at R or R+r) is warranted.

(Still trying to decide between 2 possibilities for case D- and not able to determine which one is right- whether the orbit is stable or unstable- My first guess pointed to the latter, but not sure. There is a reason that it could be stable.).

SriKumar · Post by **SriKumar** » 21 Aug 2016 17:38

SSSalvi wrote:You are right Srikumar ji.

The craft will go from a circular to Elliptical orbit.

Since earlier circular orbit was stable the resultant orbit will also be stable unless the impulse is so large that it will escape Earth's field of Influence.

SSSalvi sahab, for more than one reason, please no 'ji' for me. You have posted many more informative posts than I have (and backed with experience). Just as a posting format, I tend to not use (or accept) 'ji' even though that is polite etiquette. (Possible exception might be posters used their real name on BRF, but then you have Rahul Mehta who uses his real name and if I used ji for him, mods might ban me forever

).

JayS · Post by **JayS** » 21 Aug 2016 17:54

Amber G. wrote: Assume a satellite is in circular stable orbit[/b] (say 500 Km from surface of earth). Now it fires a thruster (rocket) which gives it a delta-V (change in velocity of say 100 m/s). (Of course Delta-V is in opposite direction of the escaping rocket gas.). Let us consider four cases for the direction of delta-V..

Case A - "UP" (that is outward the line joining the earth and satellite) -- Thrusters are fired straight down --.
Case B - "FORWARD" (In the direction of the orbital velocity) -- Thrusters are fired backwards --
Case C - "BACKWARDS" ( opposite to the direction of the orbital motion)
Case D - "DOWN" (towards earth)

In each case, after the initial jerky motion ends, will the satellite have a new stable orbit. If so, will the new orbit be a) higher or b) lower or c) same? (If new orbit is not circular - (will it be?) use average distance from earth to calculate "higher" or "lower" etc..

Will be interesting to see, if each of previous contributor gives clear answer. Let us see if we have common answer.
(One can skip the reason now if the answer is obvious -- just give the answer).

My attempt. Just trying to be intuitive rather than doing rigorous calculations:

Assume a co-ordinate system with vertical axis as Y and horizontal as X. The initial circular orbit is in X-Y plane i.e. plane of the paper if we draw it on paper. Now assume the firing happens at +Ymax location i.e. satellite is topmost position in the diagram.

Easier ones to imagine are:
Case B: Orbit will become elliptical - higher one with the point of firing becoming the Perigee of new orbit. Apogee on diametrically opposite side.
Case C: Orbit will be elliptical - lower one with point of firing becoming the Apogee of new orbit. Perigee on diametrically opposite side.
Both these cases, we will have major axis along the Y axis.

Whether the orbits will be stable depends on deltaV. But will small value like 100m/s I think they will be stable only unless the initial circular orbit is too low that any smaller and the satellite will start facing drag in upper atmosphere.

More complicated ones:
Case A: The orbit will become Elliptical but the Apogee/Perigee will lie on 90deg offset axis. That is the Apogee will come after 90deg of orbit and perigee after 270deg after firing. I think the on an avg the orbit will be higher since the total energy of satellite is increased.

Case D: The orbit will become elliptical like in case A, only now Perigee will come after 90deg and apogee after 270deg. This will be smaller Orbit.

Well when I say 90deg here it means approximately not exactly. Just want to emphasize that the major axis of new orbit will be offset from Y axis and will be slanted. How much slanted will depend on the orbital velocity and delta_V magnitudes since that will decide the velocity vector inclination.

disha · Post by **disha** » 21 Aug 2016 19:40

Regarding eccentricity of the orbit., it depends on how much thrust was provided., but all cases assuming a very short one time firing burst will tend to be elliptical.

PS: saw JayS'jis post. Our posts are the closest in terms of agreement. IMVHO

JayS · Post by **JayS** » 21 Aug 2016 20:05

disha wrote:Regarding eccentricity of the orbit., it depends on how much thrust was provided., but all cases assuming a very short one time firing burst will tend to be elliptical.

PS: saw JayS'jis post. Our posts are the closest in terms of agreement. IMVHO

Plz dnt use "ji". Lets keep it informal.

In your post above you said:
Case A: Single "bump or up-down motion" in the orbit
Case D: Orbit degenerates slowly, eventually satellite falls down.

I am sorry but I do not agree to this bump motion. In my opinion it will be a stable elliptical orbit for case A only the major axis no more aligned with Y-axis (in the co-ordinates I described). My hunch is its along X-axis. But I am not sure, so I mentioned 90deg approximately. Same with Case D, only a smaller ellipse. Of coarse this is true given that the initial orbit was high enough and delta-V was small enough that the perigee of new orbit doesn't come within upper atmosphere where it encounters with drag. Going by the numbers given by AmberG, I think the orbits in most probability stable, but I really do not know LEO region that well to make any firm comments without actual calculations.

And yes, you are right about the eccentricity.

vina · Post by **vina** » 21 Aug 2016 20:22

SriKumar wrote: I am not sure I agree with equation 1 where centrifugal force (you meant centripetal force perhaps or was it centrifugal force- which is fictitious and acts away from the center. We know centripetal force acts towards center) is equated to gravitational force.
.

If you read what I wrote and have done, you will note that I have taken the orbit/satellite as the the frame of reference. Now the satellite is NOT Inertial frame of reference and Newtonian mechanics dont apply correctly to non inertial frames Inertial Frames and Newtonian Mechanics and to apply it correctly, you need to apply the "pseduo forces" , which is what I did and applied the "Centrifugal Force " , acting radially outwards and the weight (due to gravity) radially inwards towards the center.

The math and logic are perfectly correct.

Anyways, AmberG will probably explain this conceptually far better than I can. I will leave that to her.