Why is there no dust on the Lunar Lander's footpads?

[ka9q]

What is pressure thrust?What does the balance of the thrust consist of?

I think I can answer this. The purpose of a chemical rocket engine is to convert the heat of combustion into kinetic energy of the exhaust, as that's how you generate thrust. While all heat consists of kinetic energy of the individual gas molecules, they're going in random directions; the purpose of the engine is to convert as much of that energy as possible into smooth linear motion out the nozzle. This makes a rocket a heat engine, just like the one in a gasoline-powered car, that converts heat into useful work. As in any real heat engine, this can only be done with less than 100% efficiency.

The combustion products start very hot and under high pressure in the combustion chamber, but with essentially no linear velocity. It's the nozzle's job to convert this heat to linear motion as efficiently as possible.

The standard nozzle design is a De Laval or converging-diverging nozzle. It brings the gases up to the local speed of sound in the throat (the narrowest part) and then further accelerates them to supersonic velocity as they expand and cool in the divergent part. If the pressure at the mouth of the nozzle is exactly equal to the ambient pressure, then the nozzle is optimally expanded and there is no pressure thrust; all the thrust comes from the momentum of the exhaust gases leaving the nozzle.

But if the nozzle is under-expanded, then the exhaust gases leave the nozzle with greater than ambient pressure. The difference between this local pressure and ambient, times the exit area of the nozzle, is the pressure thrust. That thrust is summed with the momentum thrust. There may or may not be pressure thrust in an atmosphere, and it can even be negative if the nozzle is too long and over-expands the exhaust gases so that they exit at less than ambient pressure. But in a vacuum there is always at least a little pressure thrust.

I think of momentum thrust as that produced by the combustion gases pushing on the forward end of the combustion chamber, while pressure thrust is that component produced by the expanding exhaust gases pushing forward on the inside of the nozzle. Because  pressure thrust results from underexpansion, it takes away even more from the momentum thrust so zero pressure thrust is the optimum condition.

This is why rockets designed for operation at sea level are designed with shorter nozzles than those designed for vacuum operation. This is most easily seen in the 9-SRB versions of the Delta II launch vehicle. Six of the strap-on solid boosters are lit on the ground, and they have short nozzles. Three more strap-on solids are lit at T+60 seconds, and they have noticeably longer nozzles since they operate at higher altitude and lower ambient air pressure.

[Andromeda]

This is something that I think a lot of people still don't get their heads around...an object on a ballistic trajectory that starts at 'h' height above the ground, a with an initial motion parallel to the ground, will fall to the ground at the same rate and in the same time as an object that fall directly from 'h' height to the ground.

I remember watching a demonstration of this in about Year 9 physics. The teacher had a simple spring-loaded device which popped one ball-bearing out sideways, while simultaneously dropping another ball bearing vertically onto the ground. The experiment worked really well in the lab with its lino-on-concrete floor, as the ball bearings hit the floor with a very audible click. And only one click, because they hit the ground simultaneously.

Yes, the "monkey and the hunter" setup.  What your teacher didnt tell you was that the whole rig was so finely balanced it could take an hour to set up so it actually worked!  I don't miss doing that 

[JayUtah]

I think I can answer this.

That's an answer worthy of an introductory textbook:  concise and correct.

If the pressure at the mouth of the nozzle is exactly equal to the ambient pressure, then the nozzle is optimally expanded and there is no pressure thrust...

True.  And since it is impossible in practice to achieve zero ambient pressure with any achievable nozzle design, there is always pressure thrust in a vacuum.

There may or may not be pressure thrust in an atmosphere, and it can even be negative if the nozzle is too long and over-expands the exhaust gases so that they exit at less than ambient pressure.

Indeed the design for launch vehicle nozzles is the most demanding because you need knowledge of mission requirements in order to know where to set the sweet spot.  The SSME nozzle is an overexpansion nozzle, hence the shock diamonds and Mach cone seen when the engine reaches steady-state operation at launch.

The cone is actually just a degenerate diamond.  The plume, with its low static pressure, is "squeezed" by the ambient air pressure as it exits the nozzle.  This takes a few milliseconds for a response in the form of increased incandescence due to higher density and temperature, hence its distance from the exit plane.  The cone shape occurs because the center of the plume moves faster than the periphery, and because the elasticity of the gas under compression forces the center of plume to travel farther before undergoing its Mach transition (i.e., to subsonic flow).

That design was chosen to maximize the LOX/LH2 propulsion at high altitude after SRB staging.  For launch vehicles without SRBs, optimal nozzle expansion for the first stage is tuned for sea level.

I think of momentum thrust as that produced by the combustion gases pushing on the forward end of the combustion chamber, while pressure thrust is that component produced by the expanding exhaust gases pushing forward on the inside of the nozzle.

That's often the way it's depicted in diagrams.  Unfortunately for the momentum case, the "little arrows pressing against all the walls of the thrust chamber and also out the exit plane" is misleading because it conveys the wrong impression that momentum thrust is simply chamber pressure multiplied by exit-plane area.  Momentum thrust is a function of the fluid dynamics in the nozzle.

For practical purposes, the center of thrust is at the throat of the nozzle.

[JayUtah]

Apologies for the thread derail, but could you expand on this, please, Jay.

Sorry, just noticed this request after having been directed to it.  KA9Q's answer is as good as you're going to get.

What is pressure thrust?

Static pressure of the exhaust gas at the exit plane.  It exerts pressure upward (well, in all directions, but only in the upward case is there anything to expand against and thereby convey mechanical force).  You can think about it in terms similar to air-cushion vehicles.

What does the balance of the thrust consist of?

Depends on operating environment and engine design.  In atmospheric boost operations, only a small percentage of total thrust is contributed by pressure thrust.  For the throttled DPS in a vacuum, up to 40% of total thrust is pressure thrust.

In what ways do they differ?

In practically all ways, except in their ability to convey mechanical force to the spacecraft.  Momentum thrust is a pure application of Newton's third law.  Pressure thrust is more governed by classical gas laws.

Do the proportions change at different levels of thrust? Or for different types of engines?

Yes, and more.  In engines that can be throttled, plume expansion in a fixed nozzle and fixed ambient pressure varies, leading to variations in exit-plane pressure.  Since the definition of pressure thrust is the difference between exit pressure and ambient pressure, clearly a change in the ambient changes the proportion.

[Tanalia]

In a vacuum, even a microscopic dust particle that weighs 0.000000000000000000001 of a gramme will fall at exactly the same rate as a 10lb concrete breeze-block; on the moon that is 1.63 m/s

Minor nitpick -- they will accelerate at 1.63 m/s².

[smartcooky]

In a vacuum, even a microscopic dust particle that weighs 0.000000000000000000001 of a gramme will fall at exactly the same rate as a 10lb concrete breeze-block; on the moon that is 1.63 m/s

Minor nitpick -- they will accelerate at 1.63 m/s².

Very true

However, when you start talking about metres per second squared to most HBs, the hand waving and glazed expressions begin.

[gillianren]

Honestly, I don't myself entirely understand how you square a second.

[JayUtah]

Honestly, I don't myself entirely understand how you square a second.

The same way you turn a phrase. 

It's a calculus thing.  The "squared" second is in the denominator, which designates a rate.  In calculus there's always a rate at which the rate-of-change changes -- that is, an endless chain of differentials.  When you unravel the algebra, it's m/s² or m/s/s, which may be easier thought of as (m/s)/s.  (m/s) is a velocity -- change in position over time.  Acceleration is a change in velocity over time, so (m/s)/s.

It's best not to try to visualize the behavior too concretely from the arithmetic of the units.  Not even we who do this for a living manage to get that right.

[smartcooky]

.....when you start talking about metres per second squared to most HBs, the hand waving and glazed expressions begin.

It's a calculus thing.  The "squared" second is in the denominator, which designates a rate.  In calculus there's always a rate at which the rate-of-change changes -- that is, an endless chain of differentials.  When you unravel the algebra, it's m/s² or m/s/s, which may be easier thought of as (m/s)/s.  (m/s) is a velocity -- change in position over time.  Acceleration is a change in velocity over time, so (m/s)/s

I couldn't have come up with a better demonstration of my point than that!!

All I recall about this what what my physics teacher told me, more years ago that I care to admit.

There is a height above which, on the Moon, an object dropped will impact the surface at a higher speed than one dropped from the same height on the Earth, even though the moon has a much lower gravity.

An object dropped on the Earth accelerates at 9.8 m/s² until it reaches its "terminal velocity", the speed at which it can no longer fall any faster because of air resistance - i.e. drag. (for a human skydiver, in a standard free-fall attitude, this is about 120 mph)

On the Moon, while the same object only accelerates much more slowly (1.63 m/s²) there is no air, therefore no terminal velocity, it just keeps accelerating until "wham", it hits the lunar surface.

[ka9q]

Honestly, I don't myself entirely understand how you square a second.

It's even easier than Jay made it out to be.

On earth, ignoring atmospheric drag, for every second that some object free-falls, it picks up 9.8 m/s of downward velocity. If it falls for 1 second, it picks up 9.8 m/s. If it falls for 2 seconds, it picks up 19.6 m/s, and so on. This is true regardless of starting speed, even upward. If you toss something straight up at 9.8 m/s, then after 1 second it will come to a dead stop, and after 2 seconds it will be coming down at 9.8 m/s (and quite likely hit you in the head). So we say that the acceleration of gravity is 9.8 meters per second, per second. Or just 9.8 m/s² for short.

On the moon, simply substitute 1.622 everywhere you see 9.8 in the above paragraph.

[Zakalwe]

On the Moon, while the same object only accelerates much more slowly (1.63 m/s²) there is no air, therefore no terminal velocity, it just keeps accelerating until "wham", it hits the lunar surface.

I'm no mathematician, so please accept my apologies if this is incorrect.
The effects of gravity decreases with the square of the distance, so an object twice as high would "feel" 1/4 of the gravitational attraction when compared to an object at half the height. So it would accelerate more slowly.

With this in mind, does the Moon then have a "terminal" velocity? For example, if a body (with no movement relative to the Moon's surface) appeared at a certain altitude, would it impact at a higher velocity than a similar object appearing at a different altitude? And is there a point where increasing the altitude does not increase the impact velocity?

[Glom]

On the Moon, while the same object only accelerates much more slowly (1.63 m/s²) there is no air, therefore no terminal velocity, it just keeps accelerating until "wham", it hits the lunar surface.

I'm no mathematician, so please accept my apologies if this is incorrect.
The effects of gravity decreases with the square of the distance, so an object twice as high would "feel" 1/4 of the gravitational attraction when compared to an object at half the height. So it would accelerate more slowly.

Yes, but be careful.  That distance is from the centre of the attractor's mass, not the surface.  The difference in weight  between an object 100m up above the surface and an object 200m up is only seen deep down in the jillionth decimal place.  For practical purposes, when we're talking about acceleration due to gravity on the surface, it is a constant.

With this in mind, does the Moon then have a "terminal" velocity? For example, if a body (with no movement relative to the Moon's surface) appeared at a certain altitude, would it impact at a higher velocity than a similar object appearing at a different altitude? And is there a point where increasing the altitude does not increase the impact velocity?

No.  Terminal velocity is due to air resistance, which increases with the square of airspeed.  When falling in an atmosphere, you accelerate due to gravity minus the air resistance until you speed reaches the terminal speed where the air resistance is now equal but opposite to the weight so there is no longer a net force to cause further acceleration.  On the Moon, without an atmosphere, you keep accelerating so the higher you start, the faster you'll be going when you hit the surface.  It's also worth pointing out that terminal speed differs from object to object due to drag coefficient.

[Glom]

On the Moon, while the same object only accelerates much more slowly (1.63 m/s²) there is no air, therefore no terminal velocity, it just keeps accelerating until "wham", it hits the lunar surface.

With this in mind, does the Moon then have a "terminal" velocity? For example, if a body (with no movement relative to the Moon's surface) appeared at a certain altitude, would it impact at a higher velocity than a similar object appearing at a different altitude? And is there a point where increasing the altitude does not increase the impact velocity?

Rereading, I wondered if you meant a different kind of terminal speed due to the diminishing gravitational force as distance increases.  The answer is no.  The greater the height at the start of the freefall, the faster the speed at impact.  You can apply conservation of energy to see how this is the case.  The higher an object is above the surface, the more gravitational potential energy it has.  In freefall, this energy becomes kinetic energy.  More GPH at the beginning mean more kE at the end hence higher speed.

[Valis]

With this in mind, does the Moon then have a "terminal" velocity? For example, if a body (with no movement relative to the Moon's surface) appeared at a certain altitude, would it impact at a higher velocity than a similar object appearing at a different altitude? And is there a point where increasing the altitude does not increase the impact velocity?

If an object is falling towards the Moon, its impact velocity increases with increasing altitude. Even though the accelerating force decreases with increasing distance, it still accelerates the object. Think of an example of two objects falling from different heights. Hold the lower object in place until the higher object has reached it. Now, the initially higher object has already gained some velocity (let's say v₀), while the lower object starts at zero. As they experience the same gravitational attraction with the Moon from the lower object's initial position onwards, from that point on they gain the same amount of velocity until the impact. Let's call this velocity gain  v.  As v is the only velocity gained by the lower object, it's its impact velocity. However, the higher object's impact velocity is  v₀+v.

Terminal velocity comes from a medium's (like air) resistance of the falling motion. In a vacuum, the velocity increases without a limit in Newtonian mechanics (in real life, relativity would kick in at some point).

[cjameshuff]

Rereading, I wondered if you meant a different kind of terminal speed due to the diminishing gravitational force as distance increases.  The answer is no.  The greater the height at the start of the freefall, the faster the speed at impact.  You can apply conservation of energy to see how this is the case.  The higher an object is above the surface, the more gravitational potential energy it has.  In freefall, this energy becomes kinetic energy.  More GPH at the beginning mean more kE at the end hence higher speed.

Actually, there is a limit. An object dropped from an infinite distance would (disregarding the infinite time taken to fall) hit the surface at surface escape velocity. This is quite different from the concept of a terminal velocity, however.