This day 20 years back (11^{th} October 1993), a young graduate with a bag full of science books and few pairs of cloths landed here in Bangalore to pursue a career. Born in Andhra Pradesh, studied in Tamilnadu it is the third south Indian state Karnataka, I came to join** Indian Space Research Organization** as ‘Scientific Assistant – B’.

It was a long selection process to get to the job. I had to qualify a written test, an interview with a big panel of ISRO and IISc scientists, a police verification to join central government of India. With the planned Mangalyaan launch on 28^{th} October, would like to give some science behind travelling beyond Earth’s orbit.

If we want to go to moon or mars we can’t aim a rocket towards the target and fire it. As the distance between earth and moon is about 4 million kilometers and to that of mars it is 400 million km. So, we need slightly intelligent way of going there. One way of going there is using Hohmann Transfer Orbit In simple terms, the spacecraft is placed in a highly elliptical orbit around earth and using a ** delta-v** at right point transferred into another elliptical orbit at a suitable time to the target orbit.

There is another low-energy transfer using Lagrange points which will probably take longer transit time. The Interplanetary Transport Network (ITN) formed for deep space missions that travel purely using solar energy or very little fuel to fire the thrusters.

Mangalyaan is taking the first method of Hohmann Transfer Orbit between Earth and Mars to start it journey in this opportunity window to reach mars by November 2014. I wish all the best to my first employer in this mission which is going to prove the technology and ISRO’s ability to apply the science to take the satellite into orbit mars. There are challenges of handling the launch, orbit maneuvers, deep space communication network for both payload data and machine control (it is 20 light minutes distance between earth and mars at the maximum so the two way communication takes 40 minutes making it complex to manage the communications!) It is only unfortunate to see critics criticizing the cost of this mission which is around 450 crores Indian Rupees where as ONE Fodder scam is 950 Crores worth loss to nation; not to mention any other scams of recent past in India.

**How is it related to my career?**Even I ended up using the high energy transfer orbits and low energy transfers during this journey around different companies and quantum leaps to different roles on working on orbits to orbitals, providing management solutions to energy grids and computing grids, optimizing satellite operations to smart metering operations handling data movements in and out of commercial ERP systems and geospatial databases, deriving forecasts of orbits of satellites to insights from big data using analytics in these 20 years.

Apart from the science, math & technology, these 20 years took me around this little globe physically from India to Singapore to various European Countries (Belgium, Germany, France, Netherlands, Luxemburg) to UK then to USA, Japan, Korea; provided an opportunity to work with large enterprises from Australia, China, South Africa etc., to make me meet exceptional personalities from varied cultures, walks of life to interact and learn about the most colorful part of God’s creation.

A majority of the orbiting been with TCS, (which was my second employer between 1996 and 2006 for 10 years) followed by Oracle Corporation and IBM India Pvt. Ltd. to come back to TCS in 2011.

I take this opportunity to thank one and all who helped me through this journey, who have challenged me and those who have been neutral towards me; each of those gestures gave me immense experience and made the journey very colorful and interesting.

With entry and re-entry tested, I hope to go along on the orbit for few more years continuing my learning until the mission retires….

November 9, 2013 at 8:49 pm |

Mangalyaan transfer orbit isn’t a Hohmann transfer. It’s one of those long-path transfers that go more than half-way around the sun.

Earth at 12h UT on 1 December 2013 (departure).

t₁ = JD 2456628.0

x₁ = +0.3242129065 AU

y₁ = +0.9309775284 AU

z₁ = 0 AU

r₁ = 0.9858159905 AU

Mars at 12h UT on 22 September 2014 (arrival).

t₂ = JD 2456923.0

x₂ = +0.3899647804 AU

y₂ = −1.3705690670 AU

z₂ = −0.03825134369 AU

r₂ = 1.4254807128 AU

t₂−t₁= 295 days

d = 2.3028033362 AU

Δθ = 215.113°

Where d is the line-of-sight distance between Earth’s position at departure and Mars’ position at arrival, and Δθ is the arc of true anomaly going from departure to arrival.

We can do a little figuring, in order to guess about the elements of the transfer orbit. Reference all vectors to the heliocentric ecliptic coordinate system.

β denotes the apsidal end of the intended trajectory.

φ denotes the non-apsidal end of the intended trajectory.

For both β and φ, a value of 1 indicates departure, and a value of 2 indicates arrival.

β є [1,2]

φ = 3−β

N = (−1)^φ

If rᵦrᵩ, then Mᵦ = uᵦ = θᵦ = π

The polar equation which relates the heliocentric distance with the true anomaly,

cos θ₂ − cos θ₁ = { [a(1−e²)/r₂ − 1] − [a(1−e²)/r₁ − 1] } / e

The law of cosines,

d² = r₁² + r₂² − 2 r₁ r₂ cos(θ₂−θ₁)

Which, solved simultaneously, subject to the condition that either θ₁ or θ₂ (but not both) must be either 0 or π, leads to

e = 2 (cos θᵦ) rᵦ (rᵦ−rᵩ) / (rᵩ² − rᵦ² − d²)

a = rᵦ / (1 − e cos θᵦ)

Then,

θᵩ = θᵦ + N arccos{(rᵦ² + rᵩ² − d²) / (2rᵦrᵩ)}

The following is true iff 0<e<1,

sin uᵩ = (rᵩ/a) sin θᵩ / √(1−e²)

cos uᵩ = (rᵩ/a) cos θᵩ + e

uᵩ = arctan( sin uᵩ , cos uᵩ )

mᵩ = uᵩ − e sin uᵩ

P = 365.256898326 a^(3/2)

b = 2π/P

If the spaceship's transit has an arc of true anomaly less than π radians,

Δt = (N/b) [mᵩ − π sin(θᵦ/2)]

If the spaceship's transit has an arc of true anomaly greater than π radians,

Δt = P − (N/b) [mᵩ − π sin(θᵦ/2)]

If the transfer orbit has perihelion at Earth departure,

e = 0.204315

a = 1.23895 AU

i = 2.674°

Ω = 70.799°

ω = 0.0°

Δt = 322.443 days

If the transfer orbit has aphelion at Mars arrival,

e = 0.196991

a = 1.19088 AU

i = 2.674°

Ω = 70.799°

ω = 35.084°

Δt = 268.425 days

The inclinations, i, being found by the cross product of the two heliocentric position vectors, and, in both cases, the longitude of the ascending node, Ω, is simply the ecliptic longitude of Earth at the moment of departure.

The argument of perihelion, ω, is zero in the first case because the perihelion is assumed to be at the position of Earth at departure.

In the second case, the perihelion of the transfer orbit is 180° around the transfer orbit from the ecliptic longitude of Mars at arrival, so the argument of the perihelion is the difference in true anomaly along the orbit between the perihelion and the transfer orbit's ascending node.

Now the transfer orbit that Mangalyaan is actually going to use doesn't need to be either of the ones I calculated. Those were just the two transfer orbits that are easiest to discover mathematically. The average in their transit times is 295.434 days, which is near enough to the announced actual transit time (within a reasonable allowance for unknown times of the day) that we can probably estimate Mangalyaan's transfer orbit by averaging the elements of those two. So, just a guess…

e = 0.2007

a = 1.2149

i = 2.674°

Ω = 70.799°

ω = 17.542°

Δt = 295.434 days

November 9, 2013 at 8:52 pm |

Apologies, there is an occasional problem with the blog’s interpretation of “greater than” and “less than” markings as being HTML enclosures instead of mathematical inequality signs. So the rendering of my comment isn’t quite correct.

November 10, 2013 at 2:31 am |

http://www.isro.org/pslv-c25/pdf/pslv-c25-brochure.pdf for ISRO’s document…