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Old 04-07-2012, 07:40 PM   #1
RGClark
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Question Gravity assists for Mars mission?

This is a question involving the orbital mechanics of spaceflight. There are a few different components to the question. First, if our Mars rocket departed from the Moon or a Lagrange point propellant depot fully fueled towards Earth at, say, 11 km/s or more, so it's moving at speeds beyond Earth's escape velocity, then in just passing by the Earth it should pick up additional speed equal to Earth's escape velocity about 11 km/s. So at least temporarily it should have a speed of 22 km/s. But the problem is that it still will be slowed down by the Earth as it proceeds to Mars, so it will lose some of this speed. How much speed will it lose?

What I want to do is leave Earth's vicinity at such high speed so that you don't have the long travel times of the Hohmann orbit, and in fact so that the trajectory approximates a straight-line path and if you do it at closest approach of Mars then the travel time could be say 60,000,000 km/22 km/s = 2,700,000 s, about 31 days. (You would have the problem of aerocapture at Mars at such highly elevated speeds but I'll leave that to another discussion.) So another question I have is at what high speed would you need so that the path is approximately straight-line?

This is just using Earth flyby. Could we in addition also use a Venus flyby? You would need an orbital arrangement where both Venus and Mars are near the Earth at the same time. Say you are now traveling at 22 km/s towards Venus, minus the amount you're slowed by leaving the Earth. You can likewise pick up about 11 km/s additional speed by just passing by Venus on the way to Mars, perhaps arranging it so that the path is bent by Venus to aim the craft towards Mars. So you could conceivably be traveling now at 33 km/s, again though I need to know how much speed you would lose in leaving Venus. You would also have to factor in the additional time it takes to get to Venus and the longer straight-line distance to Mars from Venus. Also, in being within Venus's orbit around the Sun, the greater gravitational effects of the Sun will have a greater effect to curve the trajectory.

Finally, could we use repeatedly the gravitational boosts of Earth and Venus? Suppose we are now at 33 km/s, more or less, after leaving Venus but we arrange it so our path is bent completely around to head back towards Earth. Could we once more get an additional 11 km/s to bring our velocity to 44 km/s after the Earth boost? Could we do this repeatedly to get arbitrarily high speeds?


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Old 04-07-2012, 07:52 PM   #2
MaverickSawyer
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"Venus flyby"? Don''t you mean "Venus FRYby"?
Radiation concerns aside, why would you want such high speeds? You'll have to shed the velocity somehow, and you can either aerobrake or use engines. In either case, the faster you go, the more mass you need to stop. However, if you are talking about mutli-year missions, you're going to need a BIG ship, and it all goes downhill form there. A simple transfer is the best option: it's fast, it's fuel efficient, and you can do aerocapture with a relatively light aeroshield. Speed is not your friend in this case.
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Old 04-07-2012, 08:03 PM   #3
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Some information about gravity assists: The DV you gain by it, depends on the angle between entering/leaving velocity vector and periapsis radius vector. The bigger this angle is, the more impulse is exchanged.

This means: The more you change the direction of your velocity vector relative to the planet, the more DV you get - or in the other direction: If you try a gravity assist with minimal change in direction, you get nothing.

That puts some limits on what you can get by a gravity assist: Without a change in direction, you get nothing, thus just letting Earth pull your apoapsis higher for example, would not work out as a classic gravity assist.

(There you can use some more complex voodoo, the interplanetary highway system, traveling slowly from lagrange point to lagrange point: Takes a lot of time, but works with minimal fuel)

During an Earth-Earth-Mars trajectory, you could thus get most DV if you leave Earth with already some non-tangential velocity, which makes it not really effective. A deep space maneuver would be needed to line up with Earth properly for having some serious gains.

An Earth-Venus-Earth-Mars trajectory could work better with the gravity assists, but would require likely more DV and take longer than the classic Earth-Venus-Mars transfer.

Depends of course on the launch window that you use, there is no perfect option for all situations.
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Old 04-09-2012, 03:41 PM   #4
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I'm having trouble disentangling the
gravitational slingshot effect gravitational slingshot effect
and the
Oberth effect Oberth effect
.
By the Oberth effect I can get greater velocity if I apply my rocket burn when I'm closest to the planet. Plugging in some speeds into the equation on the Wikipedia page I am able to get an additional boost about that of Earth's or Venus' escape speed if I make the rocket burn high, say, 10 km/s or above. The problem is this page seems to be suggesting to get the gravity boost, I need to apply a rocket burn but I wanted to get the gravity boost without having to apply an additional rocket burn.
On the other hand the Wikipage on the gravitational slingshot effect suggests I can get an additional boost without having to supply an additional burn. But the problem here is I want to get these additional boosts while my craft is already moving at high speed, to boost even higher, but if I'm going too fast I won't swing around the planet but instead go right by it without getting the slingshot effect.
But this slingshot effect is potentially quite large though according to the Wikipedia page. It can be as high as twice the speed of the planet around the Sun. Since for the Earth this is about 30 km/s, this means you can get a boost of about 60 km/s (!) How fast can you be going beforehand and still get swung around by the planet to still get the boost?


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Old 04-09-2012, 03:46 PM   #5
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Yes, twice the speed of the planet without propulsion makes sense. But there is a cave-eat. You need to fly through a suitable tunnel below the surface of the planet and do it in a away that you try to invert your direction of travel.

In practical spaceflight, such extremes are unlikely. But Jupiter gets pretty close to the maximum, since it is very dense at its high mass.
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Old 04-10-2012, 12:52 AM   #6
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You are misunderstanding the Oberth Effect. This has nothing to do with getting a gravity assist - it is simple mechanics. It's not really about gravity at all, but velocity - which is usually higher when you are closer to a gravity source.

It all has to do with work. Work is a function of force, mass, and distance over time. Increasing the force, mass or the distance increases the amount of work done, and reducing the time increases the work done.

So, lets say I make a ten second burn. The force produced by the engine, and the mass of the vessel both remain the same regardless of our velocity at the beginning of the burn. However, the higher our initial velocity, the further we will travel during that 10 seconds - we have increased the distance so more work was done. That means I have increased the energy level more - resulting in a higher velocity gain.

There are two basic rules for efficient space flight. The First rule is "Make changes to the amount of your velocity when your velocity is high". The Second is "Make changes to the direction of your velocity when your velocity is low". The first rule is the result of the Oberth Effect, the second is the result of inertia.
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Old 04-10-2012, 08:29 PM   #7
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Quote:
Originally Posted by Tommy View Post
 You are misunderstanding the Oberth Effect. This has nothing to do with getting a gravity assist - it is simple mechanics. It's not really about gravity at all, but velocity - which is usually higher when you are closer to a gravity source.
It all has to do with work. Work is a function of force, mass, and distance over time. Increasing the force, mass or the distance increases the amount of work done, and reducing the time increases the work done.
So, lets say I make a ten second burn. The force produced by the engine, and the mass of the vessel both remain the same regardless of our velocity at the beginning of the burn. However, the higher our initial velocity, the further we will travel during that 10 seconds - we have increased the distance so more work was done. That means I have increased the energy level more - resulting in a higher velocity gain.
Thanks for that. If they are different effects then we may be able to apply them separately to get an even bigger velocity boost. A problem though is that the Oberth effect requires you to do the rocket burn while you are closest to the planet but this will have the effect of pushing you further away from the planet which will decrease the effect of the gravitational slingshot.


Bob Clark

---------- Post added at 04:29 PM ---------- Previous post was at 07:40 AM ----------

Quote:
Originally Posted by RGClark View Post
 ...
But this slingshot effect is potentially quite large though according to the Wikipedia page. It can be as high as twice the speed of the planet around the Sun. Since for the Earth this is about 30 km/s, this means you can get a boost of about 60 km/s (!) How fast can you be going beforehand and still get swung around by the planet to still get the boost?
The gravitational slingshot won't work from Earth since the spacecraft even if you launch from the Moon is still moving in the same direction as Earth with respect to the Sun.
It might work from Venus. I remember reading some of the plans to reduce the return time from a Mars mission is to do a swingby of Venus. As I recall though, the reduction in time was not that dramatic as to reduce the trip time to days instead of months, so likely the same would be true for using a Venus swingby for the outbound trip.
I think we could use the Oberth effect though.


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Old 04-11-2012, 12:19 AM   #8
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There are a couple problems with using the Oberth Effect for this mission. First, if the ship and fuel are produced on Earth, the cost of getting them to the Moon or L-point more than offsets the savings gained by the Oberth Effect.

Even if the fuel or ship is produced on the Moon, the velocity you carry over from escaping the Moon is enough to provide Earth escape velocity after the swingby - without any additional burn. It is, in fact, enough to get you to Mars.

You would have to be going at least to Jupiter before the Oberth Effect will help.
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Old 04-11-2012, 05:59 PM   #9
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Quote:
Originally Posted by Tommy View Post
 There are a couple problems with using the Oberth Effect for this mission. First, if the ship and fuel are produced on Earth, the cost of getting them to the Moon or L-point more than offsets the savings gained by the Oberth Effect.
Even if the fuel or ship is produced on the Moon, the velocity you carry over from escaping the Moon is enough to provide Earth escape velocity after the swingby - without any additional burn. It is, in fact, enough to get you to Mars.
You would have to be going at least to Jupiter before the Oberth Effect will help.
Recall I wanted to get a speed beyond just escape velocity. I wanted the speed to be high enough to make the trajectory approximately straight-line from the Earth to Mars at its closest approach. I wanted then to have the trip time cut down to be counted in days rather than in months.
A speed of about 20 km/s is possible at Earth by Oberth if you apply a delta-V burn at perigee of about 11.2 km/s, escape velocity. This is well within current capabilities since lunar and planetary probes are routinely given this speed to escape Earth. Plugging this into the last formula on the Oberth effect wikipage, you see this burn gets boosted to about 19.4 km/s.
This method presupposes you have propellant depots at a Lagrange point or on the Moon.


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Old 04-15-2012, 05:46 PM   #10
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Quote:
Originally Posted by RGClark View Post
 Recall I wanted to get a speed beyond just escape velocity. I wanted the speed to be high enough to make the trajectory approximately straight-line from the Earth to Mars at its closest approach. I wanted then to have the trip time cut down to be counted in days rather than in months.
Hello Bob.

The first requirement is a side effect of the second: instead of focussing on the "approximately stright-line", you should focus on how long you want the transfer time to be.

Define the transfer time, you can then compute the Sun-centered transfer orbit that will bring you from Earth to Mars. Once you have the transfer orbit, you can compute the velocity difference between Earth orbit and your transfer orbit. The modulus of this velocity is the hyperbolic excess velocity V_\infty, which basically is the velocity that your spacecraft has once it is at infinite distance from the central body (Earth, in this case).

The hyperbolic excess velocity V_\infty is also directly related to your orbital energy as
V_\infty^2 = 2 (\epsilon_k + \epsilon_p)
in every point of your orbit. \epsilon stands for specific energy (energy per unit of mass) and the subscripts _k and _p stand for kinetic and potential, respectively.

In order to go from the Moon to Mars, assuming that you are starting from a parking orbit around the Moon, you basically have two options:
  1. Perform a Hohmann transfer from Moon orbit to near Earth orbit. Once you reach perigee hit the gas to reach the proper orbital energy that will launch you towards Mars.
    This method is quite efficient, but it will add the journey from the Moon to Earth to your travel time.
  2. Hit the gas directly on the Moon: it will require a bigger delta-V than the first option, but it saves the time that is needed to reach Earth from the moon.
These two options arise from the usual trade-off of mission planning: the one between delta-v and travel time.

Quote:
Originally Posted by RGClark View Post
 A speed of about 20 km/s is possible at Earth by Oberth if you apply a delta-V burn at perigee of about 11.2 km/s, escape velocity.
You should forget about the concept of "escape velocity" here. Escape velocity is the velocity needed for an object on ballistic trajectory on a planet surface to reach parabolic escape. You have a specific target escape velocity (given by your mission parameters). Consequently, you have a target orbital energy. All you have to do is reach that energy.

Quote:
Originally Posted by RGClark View Post
 This method presupposes you have propellant depots at a Lagrange point or on the Moon.
As others before me noted, why start on the Moon when you can start from LEO? You are already deep in the Earth gravitational field, where burns give the most kinetic energy. Having depots on an L-point means that you have to bring it there. Why not just move it from Earth to LEO and start your journey from there?
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Old 04-15-2012, 06:35 PM   #11
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also, another aspect of travel times: a few days longer travel means a lot less fuel needed, until you reach the Hohmann transfer window.

There are a lot of other reasons why a straight line trajectory is pretty bad actually: You leave perpendicular to Earths velocity vector, which means you have much higher fuel requirements already for leaving Earths orbit in the desired trajectory.

Also, a perfect straight line requires the speed of light as velocity, otherwise you are stronger subject to gravity and actually traveling a more or less curved path by the gravity of the planets, especially Earth and Mars.

The affordable optimum is currently a ~90 day travel to Mars, there you can get with current technology options without an extreme technological risk. ~60 days would be theoretically possible with some experimental propulsion systems (like Orbiters Deltagliders), ~39 days would be the minimum thinkable with even very hypothetical propulsion systems that still obey the laws of thermodynamics.
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Old 04-15-2012, 06:46 PM   #12
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It also occurs to me that Mars is so relatively close to Earth (comparatively speaking, of course) that a Venus flyby would have limited ability to reduce the total transit time between Earth and Mars, given that you'd first have to get to Venus to fly by it, which takes time. Best to just use an Hohmann-esque trajectory immediately.
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