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Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: petelamana on 20/02/2018 17:27:34

Title: Would a Hohmann Transfer be more fuel efficient to attain lunar orbit?
Post by: petelamana on 20/02/2018 17:27:34
To achieve lunar orbit would it be more fuel efficient to utilize a Hohmann Transfer, rather than a "direct" approach, as the Apollo program used?

And, for that matter, are HTs used to translate satellites into higher orbits?
Title: Re: Would a Hohmann Transfer be more fuel efficient to attain lunar orbit?
Post by: evan_au on 20/02/2018 20:37:24
Quote from: petelamana
To achieve lunar orbit would it be more fuel efficient to utilize a Hohmann Transfer, rather than a "direct" approach, as the Apollo program used?
Yes, the Hohmann transfer orbit is the most fuel-efficient, if all other things were unchanged - but they aren't.

The Hohmann transfer orbit uses just two (quite intense) engine burns.
But it takes about half an orbit between the two engine burns - in the case of an Earth to Moon transfer, that is 2 weeks, compared to about 3 days for the Apollo approach.

That means 4 times as much food, oxygen and muscle cramps, which means more mass, and more fuel.

In the case of Apollo 13, when their oxygen tank exploded, depriving them of electrical power and heating, they continued on their orbit around the Moon, and back to Earth after about 5 days. If they had been on a Hohmann transfer orbit, it would have taken about 4 weeks to return to Earth orbit - and they would have died.

See: https://en.wikipedia.org/wiki/Hohmann_transfer_orbit

Quote
And, for that matter, are HTs used to translate satellites into higher orbits?
With a chemical rocket, you can achieve the very intense burn to kick a satellite into a Hohmann transfer orbit - and the second intense burn to keep it there once it reaches the farthest point in the orbit. However, these chemical rockets are not very efficient.

See, for example: https://en.wikipedia.org/wiki/Inertial_Upper_Stage

Many of today's geosynchronous satellites have ion thrusters, which are very efficient, but can't achieve a very intense burn. So these satellites reach geosynchronous orbit by doing a number of low-thrust burns on successive orbits. This takes longer to reach geosynchronous orbit, but by carrying less fuel, they can carry more payload (communications antennas, transmitter channels, etc).

See: https://en.wikipedia.org/wiki/Hohmann_transfer_orbit#Low-thrust_transfer
Title: Re: Would a Hohmann Transfer be more fuel efficient to attain lunar orbit?
Post by: Janus on 20/02/2018 21:27:44
Quote from: petelamana
To achieve lunar orbit would it be more fuel efficient to utilize a Hohmann Transfer, rather than a "direct" approach, as the Apollo program used?
Yes, the Hohmann transfer orbit is the most fuel-efficient, if all other things were unchanged - but they aren't.

The Hohmann transfer orbit uses just two (quite intense) engine burns.
But it takes about half an orbit between the two engine burns - in the case of an Earth to Moon transfer, that is 2 weeks, compared to about 3 days for the Apollo approach.

That means 4 times as much food, oxygen and muscle cramps, which means more mass, and more fuel.

In the case of Apollo 13, when their oxygen tank exploded, depriving them of electrical power and heating, they continued on their orbit around the Moon, and back to Earth after about 5 days. If they had been on a Hohmann transfer orbit, it would have taken about 4 weeks to return to Earth orbit - and they would have died.
A Hohmann transfer orbit from LEO to Moon orbit only takes 10 days total.  What Apollo 13 used was a  circumlunar free return trajectory.   This basically means that you can whip around the Moon and use its gravity to  return you to LEO.  However, in this case it required a corrective burn using the LEM to accomplish.  Unlike Apollos 8, 10 and 11 which were put directly into circumlunar free return trajectories , Apollo 13 was initially put into a highly elliptical orbit  which fell well short of the Moon, which would return them to the Earth. Then after a docking with the LEM and a system check out, this was changed to  a non-return trans-lunar trajectory.  The accident occurred after this course correction, which required the LEM burn to put them into that circumlunar return trajectory (they could have done a direct abort by using the Service Module engines, but were worried that the explosion may have compromised the SM's structural integrity).


Title: Re: Would a Hohmann Transfer be more fuel efficient to attain lunar orbit?
Post by: wolfekeeper on 20/02/2018 22:49:50
OK, hardcore orbital mechanics follows, you have been warned.

First, it's not true that Hohmann is necessary minimal (bielliptic transfers can be lower delta-v). To be accurate you can't even use a Hohmann transfer to get to the moon, because a Hohmann transfer only works between two circular orbits around a single central gravitating body. But with the moon there, there's a second body. It turns out though that that's a good thing.

The Apollo transfer uses less fuel than a Hohmann transfer would because it uses Oberth effect:

https://en.wikipedia.org/wiki/Oberth_effect

In other words as the vehicle falls into the lunar gravitational field it speeds up. But rockets work better at high speed/lower altitude, so it can do a small burn at the closest point and enter an orbit around the moon. That also means that the closer the orbit is to the moon when you do the braking burn, the more efficient the trip is. If you did a normal Hohmann at lunar orbital radius, but away from the moon, the vehicle would be at very high altitude above the Earth and the vehicle would be moving very slowly, and so during the circularising burn (to oversimplify slightly) the exhaust would end up moving very fast, and waste lots of energy and fuel.
Title: Re: Would a Hohmann Transfer be more fuel efficient to attain lunar orbit?
Post by: evan_au on 23/02/2018 09:09:13
Quote from: OP
Would a Hohmann Transfer be more fuel efficient...
One case where a direct Hohmann Transfer orbit is not used is where a very large change in velocity is needed (very high delta-v). An example of this is NASA's upcoming Parker Solar Probe, which is intended to study the Sun's corona from the inside.

We have no rocket that is capable of providing the intense burn necessary to put this spacecraft within 6 million km of the Sun's surface using a Hohmann Transfer from Earth's orbit. Instead, a series of 7 gravitational assists from Venus will progressively reduce the perihelion (closest approach to the Sun) while leaving the aphelion (farthest point from the Sun) at around the orbit of Venus, ready for the next gravitational assist.

This is very fuel efficient, since most of the acceleration (or in this case, deceleration) is done by the gravitational field of Venus.

Previous space probes to the outer Solar System like Voyager 2 have used gravity assists from Jupiter and Saturn to accelerate outward; in these cases, they have not attempted to insert the space probe into a circular orbit, so they did not need the second intense burn (and huge amount of fuel) required by a Hohmann Transfer orbit.

See: https://en.wikipedia.org/wiki/Parker_Solar_Probe#Trajectory
Title: Re: Would a Hohmann Transfer be more fuel efficient to attain lunar orbit?
Post by: wolfekeeper on 23/02/2018 16:45:04
I should also mention one other thing, when it arrives near the moon, the vehicle is, relative to the moon on a parabolic trajectory:

https://en.wikipedia.org/wiki/Parabolic_trajectory

In principle an arbitrarily small burn at periapsis (the closest point) can push the orbit into a stable orbit (in practice you need a bit more of a burn because Earth's gravity tends to destabilise very high apoapsis (the highest point) orbits.

The answer to the second question, are hohmann orbits ever used? Yes, quite often- or approximations anyway. Many GEO orbital transfers have to start in circular orbits in LEO, and have to move up to Geostationary orbits. In this case, the only significant gravity is that of Earth. However, many launches are not from the equator, and so they end up in non equatorial orbits, and have to perform a plane change to reach the geostationary orbit. I haven't checked, but I'm pretty sure they do the plane change at GEO altitude, because it would take far less fuel and they combine that with the circularisation burn.
Title: Re: Would a Hohmann Transfer be more fuel efficient to attain lunar orbit?
Post by: Janus on 23/02/2018 17:46:43
However, many launches are not from the equator, and so they end up in non equatorial orbits, and have to perform a plane change to reach the geostationary orbit. I haven't checked, but I'm pretty sure they do the plane change at GEO altitude, because it would take far less fuel and they combine that with the circularisation burn.

Such a inclination change is known as a "broken plane" maneuver
If beginning and ending orbit are both circular, then the delta v can be found by

dV = 2v sin (di/2)
where v is the orbital velocity and di is the change of  inclination.
This is done at a point where the two orbital planes cross.   When going from LEO to GEO, doing this at GEO is more efficient as v is lower.   Doing this at the same time as the circularizing burn would require that the line of the two planes crossing passes through the point where you want your Geosynchronous satellite to be positioned.

The broken plane maneuver is also needed whenever we launch a probe to another planet as the planets don't orbit is exactly the same plane.    This is done where the transfer orbit crosses the intersecting line of the two orbits.  Because of the timing requirements for such a transfer, we usually don't have the option of doing this at the destination orbit, but at somewhere else along the trajectory.
The equation for changing the inclination of a non-circular orbit without changing its other properties is
dV = (2 sin (di/2) sqrt(1-e^2) cos (w+f) na /(1-e cos(f))
Where e is the eccentricity of the orbit
w is the argument of periapis
f is the true anomaly
n is the mean motion*
a is the semi major axis.

* the mean motion is equal to 2 pi/P if measured in rad/sec and 360/P if measured in deg/sec
where P is the orbital period.