Good books about the moon landings hoax?

[Glom]

And there's the essential failing in 99.99999% of the guff that HB spout. They seem to think that just because they imagine that something is possible then it follows that is how it happened.
I regularly imagine what I would do if I won the Lottery. History shows though that it has not happened!

That's just fine. I imagine 6 lunar landings between 1969 and 1972 are possible therefore they happened.

[smartcooky]

Is this type of orbit even possible?...or is it just a question of power. Wouldn't such a craft have to be under constant thrust to maintain such an orbit?

A spacecraft orbits around the center of mass of the primary (the body that it is orbiting).  To orbit in a manner described by Jarrah would require constant thrust, so much so that it is prohibitive.  In every practical sense such an orbit is impossible.

.. in fact, I'm not even sure such a path can even be classed as an orbit, by definition

"an orbit is the gravitationally curved path of an object around a point in space"

Even though gravity will cause some of the curvature, surely the bulk of it would be caused by the thrust applied?


Speaking of orbits, and slightly off topic, I watched a fascinating programme on "BBC Knowledge" last night called "The Truth about Voyager". It had a detailed explanation of how the "Grand Tour" concept came about for the Voyager missions to the outer planets, including the solving of the "three body problem" using a computer the size of a large room. It also had the most elegant description of how the gravitational slingshot works, including a something that hadn't previously occurred to me; its the speed through space of the planet being approached that is the key factor in the acceleration of the spacecraft as it is literally flung in its new direction. It was a truly interesting programme that goes right back the the 1960s to the very birth of the concept.   

[smartcooky]

Again, I cannot find the post, but the C rock photo also appeared on the front of a magazine within weeks of the Apollo 16 landings. Again, no C was visible on the rock. I'm sure the magazine cover was posted when on the proboards.

It's here:
https://www.flickr.com/photos/83874396@N06/13325730833/

Edit: I see I was beaten to it, should have read more of the thread.

Sheesh... AW&ST for a dollar a copy!!!

[Chew]

It also had the most elegant description of how the gravitational slingshot works...

This article at the Planetary Society helped me wrap my mind around it a lot better: http://www.planetary.org/blogs/guest-blogs/2013/20130926-gravity-assist.html

The Wikipedia is pretty good too.

[Luke Pemberton]

It also had the most elegant description of how the gravitational slingshot works...

This article at the Planetary Society helped me wrap my mind around it a lot better: http://www.planetary.org/blogs/guest-blogs/2013/20130926-gravity-assist.html

The Wikipedia is pretty good too.

Interesting links. Have you ever looked at the Phet physics simulation of various orbits? I quite like the visual impact.

[Chew]

Yes! I have provided that link many times to people making grand claims but who obviously didn't understand celestial mechanics.

[JayUtah]

...impact.

No pun intended.

Reminds me of the Kuiper object plotting some people were doing with supercomputers.  N-body gravitation, where n is around 800.

[Luke Pemberton]

...impact.

No pun intended.

Not that time. Well spotted.

Reminds me of the Kuiper object plotting some people were doing with supercomputers.  N-body gravitation, where n is around 800.

Three bodies is bloody hard enough, but n -> 800, I can understand why you needed a supercomputer. You've mentioned the Kuiper object problem before, why were they being plotted (If you can't explain, I understand)? Is it related to the New Horizons project?

[JayUtah]

...why were they being plotted (If you can't explain, I understand)? Is it related to the New Horizons project?

No, it's not related to New Horizons.  Some university was investigating various models of the Kuiper belt formation.  I can't explain in detail because I wasn't that intimately involved.

[Bob B.]

Well, it should be possible to calculate the needed thrust as a function of vehicle mass and the latitude (can't remember what's latitude or longtitude - bear over with me if I'm wrong) required.

In an equatorial orbit, the thrust should be zero, in a "polar Jarrah orbit" the thrust should be equal to the gravity exerted on the craft.

I decided to put pencil to paper and came up with some crazy numbers.  I assumed we're going to "orbit" about the artic circle.  To get there I figure we'd launch into a normal orbit with an inclination of 66.5 degrees.  When arriving at the northernmost part of our orbit we'd be over the artic circle.  We'd then burn our engine to keep us circling over the latitude of 66.5^o N.  I assumed we'd retain our original speed, thus allowing us to return to the original orbit just be shutting off the engine.  I assumed an orbital radius of 6,600 km (an altitude of about 230 km).

Any object moving in a curved path has a centripetal acceleration that points in toward the center of curvature.  For a body in a orbit around Earth, that acceleration is provided by gravity and is directed toward the center of Earth.  For our initial orbit, the centripetal acceleration is about 9.15 m/s².  This is less than the surface value of 9.8 m/s² because we're further away.  The velocity of our spacecraft is 7,771 m/s.

To constantly orbit above the artic circle we must follow a curve of much smaller radius, which is 6600*cos(66.5) = 2,632 km.  The centripetal acceleration needed to maintain this curved path is calculated using the equation v²/r.  Since we're maintaining our initial velocity, the required centripetal acceleration is 7771²/2632000 = 22.95 m/s².  The center of curvature, toward which the acceleration is directed, is located at the intersection of Earth's polar axis with the plane containing the artic circle.

The acceleration that our engine must provide to maintain this orbit is the difference between the acceleration vector of the "Jarrah orbit" and the acceleration vector of gravity.  Note the use of the word vector.  We must subtract one vector from the other, thus we must account for both magnitude and direction.  We calculate the required acceleration as follows:

SQRT[ (22.95 - 9.15*cos(66.5))² + (0 - 9.15*sin(66.5))² ] = 21.05 m/s²

That's 2.15 g that must be continuously applied to maintain the Jarrah orbit.  The Apollo SPS engine produced only about 0.3 to 0.8 g, depending on the mass of fuel remaining.  It could also only burn for about 10 minutes before the fuel was gone.

(ETA)  Let's say we placed a rocket stage on the artic circle fueled with LOX and LH2.  The best mass ratio we could expect from this would be about 10.  Given this configuration we'd get a delta-V of about 10,000 m/s.  Providing an acceleration of 21.05 m/s², we'd last about 8 minutes before going kaput.

[Allan F]

Brilliant - just shows how little the hoaxers know and WANT to know.

And the total dV you have calculated is inside one order of magnitude of the total dV of the Saturn V launching an Apollo mission into LEO - and the burn time is also close. So you'd need a fully fueled Saturn V for roughly each 10-minute part of a Jarrah Orbit (TM).

edit: Total dV is 30% from a fully fuelled Saturn V stack - which burned for 11 ½ minute before TLI.

[Bob B.]

Damn, I just noticed that I kept spelling arctic as artic.  I hate that I did that and it's too late to edit.

[smartcooky]

What happens when the vehicle in the "Jarrah" orbit runs out of fuel? Does it carry on in an earth-centric orbit at the inclination (in the direction) it happened to be going in when the fuel ran out?

[JayUtah]

What happens when the vehicle in the "Jarrah" orbit runs out of fuel? Does it carry on in an earth-centric orbit at the inclination (in the direction) it happened to be going in when the fuel ran out?

Altitude, direction, and velocity at cutoff determine what orbit follows.  Generally if you're describing a constant-latitude, constant altitude orbit, it seems to me you'll go into a circular orbit at that altitude with an inclination equivalent to your latitude and the ascended antinode aligned with the longitude at cutoff.

[smartcooky]

What happens when the vehicle in the "Jarrah" orbit runs out of fuel? Does it carry on in an earth-centric orbit at the inclination (in the direction) it happened to be going in when the fuel ran out?

Altitude, direction, and velocity at cutoff determine what orbit follows.  Generally if you're describing a constant-latitude, constant altitude orbit, it seems to me you'll go into a circular orbit at that altitude with an inclination equivalent to your latitude and the ascended antinode aligned with the longitude at cutoff.

OK, so I'm just trying to get my head around what that means.

If you were to draw a circle, centred on the centre of the earth, that intersected the vehicle in the "Jarrah" orbit at the point at which it ran out of fuel with the tangents of the circle and the "Jarrah" orbit coinciding exactly, then the circle will be the new orbit. True?