Chandrayan 3 : Some Frequently Asked Questions

Greetings PAAC Readers!

Here’s another post about some questions on Chandrayan 3 and the Physics behind it, answered by a guest author, Dr. S Rangarajan. Below is the answers to 29 exciting questions about Chandrayan 3, Space and the Physics out there in Space.

It is interesting to note how some of the fundamental STEM concepts learnt in the High School can get reinforced by looking at the recently launched Chandrayan 3 for exploring the Moon. While I was interacting with the school students, they came up with a number of questions and these s may be relevant for a much larger audience and hence this note.

For simplicity, we will take Earth and Moon to be perfectly spherical, so that for points outside the force on any mass we can assume their full mass to be at its centre. The only force we consider is the Gravitational attraction. Mass of Chandrayan 3 is negligible compared to masses of Earth and Moon.

As Earth is 81.3 times heavier than Moon, Moon exerts roughly the same force (magnitude only) as Earth , if it is 9 time nearer (inverse square law). E.g., if Chandrayan 3 is at 360,000 km from Earth, Moon at 40,000 km exerts comparable force. Thus, when distance Moon is far more than this 1/9th factor, we can neglect the effect of Moon and treat it as 2 body problem (Earth-CH3), When Earth is much more than 9 times the Moon distance, it will again reduce to a 2-body problem (Moon-CH3).

1. What is the difference between an orbit and a trajectory?

Total energy (TE) is the sum of Kinetic Energy and Potential Energy. For 2- body problem, per unit mass, TE = v2/2 -μ/r, where v is the velocity and r is the distance from the central body.

If TE < 0, i.e. total energy is negative, it is in a closed path around the central body, called its ORBIT
If TE> 0, it is no longer bound to the central body and it moves off (i.e. escapes) in a trajectory.

2. If a stone is thrown by a man with a horizontal push, do we get a trajectory?

No. TE < 0 and so it would attempt to be in a orbit. But the Earth’s surface blocks it.

3. How do we change the size of an orbit, keeping it in the same plane?

If we want to increase the apogee keeping the perigee same in an Earth’s orbit, we give an increase in velocity at perigee (+ deltaV). If apogee is to be decreased, a decrease in perigee velocity (- deltaV) is to be given.

4. Why was a EBA maneuver carried out at apogee on 16th July?

This was a +deltaV maneuver at apogee to raise the perigee from the originally launched 173km to about 220 km, to prevent a decrease in velocity due to atmospheric drag at low heights (which acts like -deltaV maneuver).

5. Why were several EBN’s carried out? Couldn’t we have carried out a large burn and imparted all the required + deltaV for getting to the Lunar Transfer orbit?

With liquid motors, we can start and stop as we wish. The first burn can be used to precisely calculate the actual deltaV obtained, based on pre and post orbits. These calibrations allow quite precise deltaV to be targeted in all later burns.  Moreover, if we wish to combine all the 5 perigee burns into one, the burn duration will be large, say 90 minutes which will cover a very large arc around perigee. Ideally, we wish to deliver all the deltaV at the perigee point only.

6. What are the required conditions for the choice of the TLI orbit?

The post-TLI orbit is highly eccentric, (218 x 369018 km) , but still has a total energy that is negative and hence is Earth-bound. Its path is so chosen that it comes into a close vicinity of Moon, so that Moon’s gravitational force on Chandrayan 3 has to be taken into account to work out its path.

The rendezvous conditions can be met only if the correct 3 dimensional geometry is considered, as the Moon’s orbit round the Earth is in a different plane compared to the plane in which Chandrayan 3 orbits.

7. What is the importance of the magnitude and timing of the Lunar Orbit Insertion (LOI) firing?

LBN#1 or LOI is most critical as this transfer from an Earth-bound orbit to a Moon-bound orbit. This requires that Chandrayan 3 at the end of this burn has targets like the injection conditions of our SLV’s. At the end of LOI, it should be having a horizontal velocity, whose magnitude should be less than the escape velocity for moon at that altitude. For example, at 118 km altitude to Moon, the escape velocity is 2.3 km/sec, and so we could target a velocity like 2.2 km/sec so that we can get an elliptical orbit of 118 x 18,500 km around the Moon, and bound to the Moon.

LOI burn will be -deltaV maneuver in which we will decrease the pre-burn velocity that could be higher than the escape velocity of Moon , 30 minutes of thruster firing can bring it down from 2.4 km/sec to 2.15 km/sec

8. Why does the same thruster give higher deltaV for the same firing duration?

This is because each burn consumes propellant mass and this, in turn, decreases the mass of the remaining body in orbit.

9. Why do we capture into lunar orbit at the perilune of the new orbit?

The idea of the LOI burn is to bring down the velocity to be less than the escape velocity (or equivalently, make the total energy w.r.t.Moon to be negative) so that it becomes Moon-bound. The post burn is an elliptical orbit around the Moon. In this orbit it is moving the fastest at perilune (Kepler’s second law) and that is what has to be achieved from a higher velocity that the pre LOI path had. At the apolune, the velocity itself is small, say .2 km/sec and if we reduce this value the perilune height will be negative and it would hit the Moon’s surface.

10. Can we explain the concept of a circular orbit from energy considerations?

Yes. PE is energy due to position and so the PE has a value at every point around the central mass. Let us take a spherical Earth.  If we consider small heights compared radius if Earth, we can treat Earth as flat and so draw equi-potentials as horizontal planes. At a height h above the surface PE per unit mass is gh, where g is the assumed constant acceleration due to gravity.

For satellite motion, we have to take the curved Earth, and so equipotential surfaces are concentric spheres around the center of Earth. At an altitude h, this sphere has the radius “Radius of earth + h”. If the horizontal velocity at injection is chosen to be such that its KE is magnitude-wise half of the PE at this height, the TE is negative, implies it is in an orbit. It will be in a circular orbit and so PE does not change. As TE is conserved, KE also cannot change and the speed does not change (Only the direction of velocity changes).

11. Is it correct to say that Chandrayan 3 escaped Earth and was captured by Moon ( LOI of 5 Aug)?

No.

Moon itself is bound to Earth. Prior to the LOI burn, Chandrayan 3 was in a highly elliptical orbit around Earth 218 x 369018 km. If Moon were not close to its path, it would have continued to orbit Earth with a period of around 230 hours.

When Moon gets too close to its path, the lunar gravity influences the path and the dynamics must be studied as a 3 body problem, E-M-CH3.

12. What are the conditions for carrying out a thruster firing for lunar capture?

Chandrayan 3 must be sufficiently close, have the proper orientation, in the proper orbit, and the right amount of -deltav is to be provided at the right time (i.e. the right point in its trajectory).

13. Why is a retro firing (negative deltaV) carried out for LOI?

Prior to the LOI burn, Chandrayan 3 is in a hyperbolic trajectory. Unlike the ellipses, the hyperbola has an eccentricity exceeding 1, does not have a closed path and so left as it is, it would have escaped Moon. This corresponds to the case of total energy (TE) greater than 0. In this hyperbolic trajectory, even when CH3 goes infinitely far, it will have a positive KE and hence a positive velocity (called v∞). By slowing down we reduce the KE and can bring the TE to be less than 0 to be captured by Moon, i.e. bound to Moon.

By retrofiring and reducing the velocity to be less than the escape velocity (w.r.t. Moon) at that altitude, we can make a hyperbolic arrival trajectory to an elliptical orbit around the Moon.

14. Some newspapers report that it takes 15 days for Chandrayan 3 to escape from Earth and then freely fly to Moon?

No. It never escaped Earth. See answer to 11

15. What are slingshot maneuvers? Was it used in Chandrayan 3 mission?

A gravity assist, or a gravitational slingshot is a type of flyby which makes use of a third body to alter the path and/or speed of a spacecraft, to save propellant expense.

Chandrayan 3 did not use any slingshot and the orbit changes were achieved by thruster firings only.

Even LOI was achieved by a thruster firing to reduce the total energy to be less than zero so that an orbit around Moon can be achieved.

16. Was the long duration from launch to LOI due to limitations of the lift off capability of our rockets?

No.

From the injection orbit of apogee 36450 km to the Lunar Transfer orbit of apogee 369018 km, the injection velocity is only marginally higher (Instead of 10. 27 km/s to 10.73 km/s) and well within our capabilities. However, for an unmanned mission the shortening the duration is not a priority. The required increase in velocity was given in instalments, so that the burn durations will cover only a small arc around the perigee and also to use the previous EBNs to calibrate the performance better.

In addition to the slowing down due to LOI maneuver, did Chandrayan 3 experience a large rotation of its velocity vector?

Yes. The hyperbola has two asymptotes and we characterize the angle between them as the turn angle. The lunar arrival hyperbola had an eccentricity of around 1.3 and hence a turn angle of 101 deg.

LOI burn was timed that just after the turn the velocity was brought under the escape velocity at that altitude.

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I may have to add a disclaimer that if something is said wrong, the fault is only mine. But with the diversity of readers, such items can be addressed quickly enough. Please pass on these ’s to younger fold at home so that their STEM concepts will get strengthened and they will also enjoy the Chandrayan 3 events as they unfold.

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18. What is the minimum time needed to land on Moon? Can we cut down the journey time to say one day, by using more powerful rockets?

No.

The mission will consist of two major segments, namely, launch to LOI, and LOI to Landing. The first phase can be completed in about 3 days if the proper rendezvous conditions are met and the second in another 2 days.

More capable rockets can deliver more payload in each launch, but the journey time is decided by orbital geometry.

19. What decides the duration between launch and LOI?

The launch phase ends with Chandrayan 3 injection. Based on the injection position and velocity, the achieved orbit is determined. Assuming the injection is at around 200 km altitude, with the injection velocity horizontal. Then the only remaining parameter is the magnitude of the injection velocity (v).

The shortest duration for LOI would correspond to the initial orbit itself creating the conditions for lunar capture. A necessary but nor sufficient condition for rendezvous with Moon is that the initial orbit should pass through a distance from Earth centre of about 384000km, the typical Earth-Moon distance. That means apogee should exceed 384000 km, and so the injection speed should exceed 10.74 km/sec. At this altitude, the escape velocity is 11 km/sec. If v is more than 11 km/sec, it would get a hyperbolic trajectory and exit Earth.

Thus we have a narrow range for v (10.73 to 11 km/s). Within this range, the larger the v, the period will be higher, but the time at which “r” will meet the

rendezvous condition happens sooner, and with less time duration of contact. E.g., in Apollo 11 mission it was about 76 hours. If we choose just around 10.74 km/sec, the period is around 230 hours and the closest approach to Moon can occur near its apogee, around 115 hours later. We can achieve LOI then within 5days, provided we satisfy several more additional conditions like the direction of approach, orientation, the precision of prediction etc.

20. Why was the first size reduction maneuver done only late night of Aug 6? Was it to give more time for orbit determination?

The orbit transfer from post LOI orbit to 170 x 4313 km orbit, requires a retro firing at the perilune (reducing the speed from around 2.17 km/s to 1.98 km/s). The period of LOI orbit is around 28.2 hours and so the next occurrence over perilune is only after one revolution. At this very first opportunity, the size was reduced. The time lapse between LOI and LBN#1 was not based on time required for orbit determination.

20. What decides the period or the time taken for one revolution in the orbit? Is it the same around Moon as around Earth for the same radius?

In the 170 x 4313 km orbit, the period is 6.25 hours (use Kepler’s III law to derive this value when the SMA got reduced from 10855 to 3977 km).

The laws of Physics are the same. But as the mass of Moon is about 1.23% of Earth’s, the product GM which appears as μ is therefore only 4900 km³/s²,. Around Moon the period of a 100km altitude circular orbit will be around 118 minutes, compared 86 minutes at 100 km altitude of Earth. We will have the Propulsion module, Lander etc. at the 100 km orbit and you will have occasions to benefit from this knowledge. Also, note that unlike a 100 km orbit around Earth that would decay too fast due to atmospheric drag, a 100 km orbit around Moon is quite stable.

21. As per Newton’s second law the same force will produce more acceleration if the mass is less? When we have a succession of thruster firings for orbit correction, mass of the object should decrease due to the ejected mass of the fuel. Can you give us an idea of how this was seen during Chandrayan 3 mission?

In EBN#2, for example, the thruster was fired for 926 sec and it produced a deltaV = 110.98 m/s; The same thruster was fired during LOI for 1834 sec and it produced a deltaV of 272.44 m/s. The doubling of the time gave an extra 22.5% deltaV, attributable to a 22.5% mass depletion from around 3800 to

3100 kg. From these values we can also estimate the thrust level as the rate of change of linear momentum and it works out to around 435N

22. Given that in the EB phase thrusters were fired in the third, 9th, 12th, 17th and 20th perigee transits, besides at Apogee 5, and if the center times of firing are known, can you deduce the apogee and perigee of all the intermediate orbits. Assume that initial orbit is known as 170 x 36450 km?

This is a good exercise for students to get a feel of Kepler’s third law that states that the square of the period is proportional to cube of the size of the orbit (specified by its semi major axis “a”).

• From 170 x 36450, find initial a = 24,688 km.

• Use μ of Earth to be 398600 km³/s², to find Period (T) = 640 minutes
• Since EBN#1 was in 3rd perigee, 2 revolutions in this orbit would take 2x 640 =1280 mts = 21 h 20m; If the injection was at 9.21 UT of 14 Jul, EBN#1 will be centered at 6.41 UT. Indeed it was.
• From the calculated period from successive firings , divided by number of revs, we can compute the new “a” and hence the new orbit size.
• For example, we note that there was firing at 9.01 UT of 25 Jul, followed by a firing at 18.43 Ut of 31 Jul. In the time lapse of 6d 9h 42m, Chandrayan 3 has made 3 revs. Hence period is 3074 minutes. Applying Kepler’s law w.r.t the previous orbit of period 1429 minutes and size 233 x 71351 km (or a = 42176 km), we can find new a = 70572 km , or 236 x 127603 km

23. Why is it that we can see only one side of the Moon always?

The Moon is not perfectly spherical. Gravitational forces stabilize the orientation of the egg-shaped Moon to keep the longer axis pointed always towards Earth as it orbits round the Earth. This gravity-gradient stabilization leads to Moon rotating about its axis at the same rate of its orbital motion. This results in the same face (the near face) towards the Earth. The far side is thus never seen from the Earth.

Furthermore, Earth’s surface has a lot of water. When the moon orbits round the earth, it causes tides that pull up towards the Moon. This is called tidal locking, which reinforces the above effect.

24. Why is it that the day time in Moon is about 15 days?

As the same face of the Moon is locked to the Earth, as the Moon orbits around the Earth, it makes one revolution about its axis as well. The latter results in alternating day and night for the Moon and a full cycle is about a month (around 29.5 Earth days). i.e., if you were standing on the Moon, it would take 29.5 days for Sun to rise, move up on the sky, set, and reappear on the rising horizon. Half of this, namely, around 15 days is then daytime and the other half being the night. Because the same face of Moon always is towards Earth, if you are standing on the Moon, you will see Earth in the same direction always; but the stars and the Sun would move around in the sky.

25. What is the orbital period of the Moon?

The Moon’s orbital period around the Earth is 27.3 Earth days. We can verify this by applying Kepler’s III law w.r.t. any other known satellite like GEO radius of 42,000 km, period 1 day and Moon’s orbital radius of around 384,400 km. Or, we can use μ of Earth to be 398600 km³/s², and period as 2πa^1.5/√μ

Note that 27.3 Earth days is then also the time Moon takes to complete one turn around its axis with respect to the stars, which is called a sidereal day of Moon. However, since Moon is bound to the Earth and Earth goes round the Sun in 365.25 days, the amount of time for Sun to return to the same position in the sky for a Moon’s observer is 29.5 days, which is called the synodic day. Half of this synodic day, of about 15 days is hence the sunlit or “daytime” on the Moon.

Note the similarity to Earth-Sun system where we know that Earth spin period is about 23 h and 56 m, whereas the day is reckoned as 24 hours to account for about 1 deg/day of Earth rotation around the Sun.

26. What makes lunar landing a big challenge that only 3 countries have done it so for?

The Moon’s surface is quite undulated with large craters and rocks, and hence the landing site must be selected carefully. There is no GPS signal to navigate and so one has to depend on absolute inertial sensors and/or on real time observations of the terrain, and matched with maps stored as a reference.

Even as the horizontal component of velocity should be controlled to drive towards the target location (latitude & longitude), the vertical velocity must be brought to near zero for safe and soft landing. The vertical component is subject to the acceleration due to gravity (1.62 m/s²,  near the surface, about 1/6^th  of the value on Earth’s surface). There is no atmosphere and so parachute cannot be used. One has to fire retro thrusters to bring the vertical velocity to zero.

27. What are the special challenges in landing in higher latitudes, particularly nearer to the South pole?

South polar region is ideal for a potential lunar outpost, due to the presence of frozen water and an abundance of resources. However, these regions have dark lighting conditions, besides huge craters. Typical temperatures of ranging from 50 deg c to -230⁰C is tough on the functioning of electronic instruments. The sun lighting exhibits seasonal patterns and so places constraints on the launch window. The Sun is always close to the horizon and hence some of the taller craters produce long shadows. These shadows and the poorer lighting conditions make it difficult to navigate the lander to the chosen site using prestored images of the region. For example, for landing near -70 deg latitude, the Sun is expected to be less than 20 deg above the horizon.

28. Now that the LOI was completed on 5 Aug, can we confirm that the LOI burn decreased the total energy corresponding to the hyperbolic arrival trajectory to that of the achieved elliptical orbit?

The LOI burn was for about 30 minutes, centered at around 7.30 pm IST. The total energy (TE) per unit mass in the arrival trajectory was around 0.373 units.

The positive sign here indicates it is not bound to Moon and would escape if left as it is. The post LOI orbit had a TE of around -.224 units. The decrease of velocity (w.r.t.Moon) from 2.41 km/s to 2.168 km/s assuming an impulsive burn around 7.30 pm IST would result in a change in KE of about 0.55 units. This removal of KE by the -deltaV burn accounts for the lowering of TE from 0.373 to -.224 units.

29. Why Moon and Sun look discs of approximately same size from Earth?

For celestial bodies, the large or small is basically the angle subtended at the eye. This angle is the ratio of the actual size to the distance to the object. The Sun is far away, but bigger in diameter so much compared to Moon. It so happens these two ratios are nearly equal (each about half a degree).

Author Intro:

Dr. S. Rangarajan is a satellite industry veteran with 40+ years of experience in successive leadership roles in the public and private sectors. He was part of the core team that created mission management procedures for APPLE and Bhaskara and successfully implemented them at the control centre. He later became the Director of MCF and ISTRAC and was involved in expanding these facilities, besides building a strong team for satellite operations. Throughout his career he has pioneered various satellite applications, including the space-based Telemedicine and Teleeducation.

Dr. Rangarajan did his Ph.D. degree from Tata Institute of Fundamental Research, Mumbai