The Chandrayan take 5 earth orbit modifications to reach a translunar trajectory. I thought less orbit modifications should be enough:

  1. orbit inclination modification
  2. burns to go into a synchronization orbit
  3. an Hohmann transfer to intercept the moon orbit

The description on wikipedia looks like modifying the orbit eccentricity several times until it reaches the correct Hohmann transfer orbit. Why not only going into the last elliptic orbit earlier?


2 Answers 2


It didn't have enough thrust. Small rocket engines are easier to build than large ones, and weigh a lot less. That probe (I assume you mean Chandrayaan 1) had a 440 N main thruster that it used to get to Lunar orbit. It weighed initially about 1350 kg all from wiki. That amounts to an acceleration of approximately 0.3 m/s/s. You need around 4000 m/s for a trans lunar injection using basic Hohmann transfer, although there are far more efficient trajectories called low-energy transfers.

An extra 4000 m/s with that little thruster would require several hours of thrusting time, even with the rocket equation helping you as you lose mass. Hohmann transfers assume that you basically make an instantaneous impulse change at one of your apsides, which is impossible if you are thrusting for that long near Earth. So, multiple burns are required to take full advantage of the efficiencies gained by such a transfer orbit.

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    $\begingroup$ I'm not entirely sure if it matters what the specific impulse of the engine is, other than dealing with efficiency of fuel use. Whether the engine had great efficiency or not isn't relevant - it had low thrust, so required multiple burns. $\endgroup$ Commented Jul 12, 2019 at 16:27
  • $\begingroup$ The total impulse of an engine refers to the total amount that engine can change the momentum of the spacecraft before it runs out of fuel. It's the integral of the force applied by the engine with respect to the time the engine is on. Specific impulse is a measure of "fuel efficiency" in the sense of how much impulse a kilogram of fuel will give you when burned in the engine. What you're looking for is delta V, or total change in velocity, which is not dependent on the engine but on orbital mechanics $\endgroup$ Commented Jul 12, 2019 at 17:05
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    $\begingroup$ Haha well it's not strictly just thrust if you are thinking about it in general, since different spacecraft have different masses, and mass changes as fuel is expended, and different specific impulses for the same thrust will impart delta V at different rates since fuel is used faster or slower. However, the term you are looking for is just "acceleration", which is itself dependent on both thrust and Isp, plus spacecraft mass properties $\endgroup$ Commented Jul 12, 2019 at 17:28
  • $\begingroup$ Link to discussion between Octopus and Michael - migrated for space concerns. Migrated, great answer which prompted me to way overthink it :). $\endgroup$ Commented Jul 12, 2019 at 17:33
  • $\begingroup$ A 3.1 km/s burn from LEO is enough to get an apogee at lunar height. And the so called low energy transfers invest this 3.1 km/s. Something Belbruno usually neglects to mention when he's boasting about Hiten. For achieving a loosely bound lunar orbit, Belbruno's route saves maybe .2 km/s over the Farquhar route. Am downvoting this answer for that reason. $\endgroup$
    – HopDavid
    Commented Jul 25, 2019 at 16:26

The multi-revolution transfer orbits can be called as 'Phasing elliptical orbits'. In the mission planning, it gives us some advantage such as reducing the finite burn arc losses which may depending on the particular cases to about 80 m/s that can translate directly to savings in mass. Such transfers also give us good time to characterize the propulsion system and operational flexibility.

Such orbits are achieved mainly utilizing the positive effects of the natural forces that disturb the orbits around the Earth such as gravitational forces of the Sun and the Moon, Earth's atmospheric drag, Earth's non-spherical gravity field, etc. (http://adsabs.harvard.edu/abs/2002cosp...34E1372S).

  • $\begingroup$ Good additional point! $\endgroup$
    – uhoh
    Commented Sep 12, 2019 at 8:36

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