This morning, July 20, 2021, Blue Origin’s New Shepard rocket launched in to space for its first 11-minute passenger flight.

Congrats to Bezos and team!

HOWEVER, the commentator on one of the science channels made this statement:  “The rocket needs to go high enough to reach orbit, even briefly”

A popular misconception is that orbit and outer space are the same thing – actually, I’ve written about this previously.  However, I would expect a science commentator on a science-related channel to know the difference!

Just to recap:

• The Kármán Line, at 100Km (about 62 miles) above the Earth’s surface is considered for treaty purposes to be the delineator between the Earth and “outer space”.  From what I can tell, New Shepard will briefly surpass this, making all of its passengers “official” astronauts.
• For reference, the International Space Station’s orbit fluctuates, but the average is about 240 miles above Earth’s surface.
• Reaching outer space doesn’t automatically mean that you are in orbit, nor that you somehow magically become weightless.

Even hundreds of thousands of miles away from the surface, any object is still within the Earth’s gravitational influence.  So, if you fire a rocket straight up (perpendicular to the Earth’s surface), even hundreds or even thousands of miles in to space, eventually, the rocket will run out of fuel, and the Earth’s gravity will pull it back down.

In orbit, however, an object’s velocity parallel to the Earth’s surface creates centrifugal force which balances the pull gravity.

E. A rocket launched perpendicular to the Earth’s surface eventually runs out of fuel and is pulled back down by gravity. F. To achieve orbit, a rocket must attain both height and velocity parallel to the surface.

My understanding of New Shepard is that it will basically travel straight up, give the passengers a few minutes of weightlessness, and then return to Earth.

Although gravity will be acting on the spaceship and its passengers during the descent, the passengers will experience weightlessness (which is different than outer space and also different than orbit) because gravity acts on both nearly-equally.  The state or sensation of weightlessness occurs when there are no net forces acting on the passengers, which is only true inside the ship, and only relative to the ship itself.

Although the exact height of the New Shepard’s mission isn’t listed, it is expected to go “well above” the Kármán Line.  So if we were to guess 65 miles, we would be in the right ball park.

This means that New Shepard’s velocity at 65 miles above the Earth’s surface will be zero, after its rockets turn off, and gravity bleeds away all of its upward velocity.  At that exact point, with gravity acting constantly, it will begin to accelerate back down to Earth.  Until the ship hits turbulence in the atmosphere, or they fire their re-entry rocket, the passengers will continue to fall at an ever faster rate toward Earth, pulled by gravity, yet experiencing weightlessness within the ship.

In contrast, to attain orbit at 65 miles above the Earth’s surface, the ship would have to be travelling almost 17,780 miles per hour parallel to the Earth’s surface in order to have enough centrifugal force to exactly counter the force of gravity.

How do we go about calculating this?  Glad you asked…

Force of gravity:

F = m ⋅ g

• m = mass of the object
• g = acceleration due to gravity

Centrifugal force (same as centripetal force, but in the opposite direction):

F =	m ⋅ v2
	r

• m = mass of the object
• v = linear velocity
• r = radius of the orbit (from the center of the Earth)

In orbit, the net force is zero, so:

m ⋅ g =	m ⋅ v2
	r

We quickly see that mass is irrelevant, and we can solve for v:

v = √ g ⋅ r 

If we measure gravity and orbital radius in feet, we get orbital velocity in feet per second.  To simplify things, we can use 78,545 miles per hour² as the acceleration due to gravity:

v = √ 78545 ⋅ r 

The radius of the Earth is about 3960 miles.  To find orbital velocity, we have to have the radius from the center of the Earth, which include’s the Earth’s radius plus the height above the surface:

v = √ 78545 ⋅ ( 3960 + r ) 

If we plug in 65 miles, we get about 17,780 miles per hour required for orbit.

So again, good luck to Bezos and team – although you won’t be “in orbit”, you will definitely be in outer space.

SIMulation vs EMulation

This question came up in the context of another topic, and I immediately envisioned an excellent interview question:

Explain the difference between simulation and emulation

As someone who has written thousands of simulations, this is how I would answer that question…

SIMulation

Simulation is a closed system where the state “S” contains all relevant variables and their values, and S(i) describes those values at a specific iteration within the simulation.

S = { V0..Vk }

S(i) = {V0,i; V1,i; .. Vk,i}

Based on the variable values in S(i) and a system of rules A{R0… Rm}, the simulation calculates a new state, S(i+1).

In the simulation, we have a starting state, S(0) and an eventual end-state S(n), which is created after n iterations of the simulation.  We start with S(0) and apply the rules of A{} in order to obtain S(1), and this continues until we reach S(n).

So explicitly, a simulation requires feedback from the current state in order to create a subsequent state, given a specific set of variable values and a specific set of rules that operate on those values.

EMulation

Emulation is a simulation whose rules are copied from another system.

Given a set of rules that describe system B: {R0..Rm}, if system A is an emulation of B then it simply contains the same rules:

A = B = {R0..Rm}

Given any state S(i) and a set of rules A{}, if we apply those rules to the current state, we get a new state S(i+1):

S(i+1) = A{} -> S(i)

Given the same state S(i) and another set of rules B{}, if we apply the rules in B{} to the current state S(i), we get S'(i+1), which is an “alternate-reality copy” of S(i+1):

S'(i+1) = B{} -> S(i)

If two different sets of rules A{} and B{} applied to any possible state S(i) produce the same next state, then:

S'(i+1) = S(i+1)

And therefore:

A is an emulation of B or vice-versa.

The important detail is that both systems A and B are equivalent, but they might not be implemented the same way – for example, the rules of B might be implemented in hardware, while A might be a software emulation of B.

Comparison

Emulation is a specific simulation that answers the question: “how would some other specific system respond, given a state of S”.

Conversely, in a simulation, you can have a set of independent rules that aren’t tied to any other system – the simulation is simply answering the question: “what would happen, if…”

I’ve seen definitions of both simulation and emulation that tie to copying something from the real world, but I disagree.   For example, I could easily write a simulation of a 4D solid traversing 3D space, which is a completely hypothetical model that has no tie to the real world.

Likewise, if you copy the rules of some arbitrary simulation over to another system, then you’ve emulated the original simulation, with no ties to the real world – for example, you could emulate our clearly-contrived 4D solid simulation by copying its rules from a state diagram or a set of software commands to some other equivalent logical system – for example, the logical rules within a game simulation.

Which brings us to a final point:  Simulations (and by extension, emulations) can contain other simulations.  For example, you could have a simulation of a digital pet within a simulation of a digital world.  The state logic and iteration of the digital pet could be driven inside the world simulation, or it could be driven externally and simply rendered based on its current state.  If the digital pet is driven inside the world simulation, it could be driven within the code of the simulation (created at compile time), or it could be implemented as part of the logical rules of the world – for example, as a script or state machine that’s created using in-world tools during runtime.

Why the Sun does NOT track East-West

(Most of the time)

We were brought up learning “the sun rises in the east, and sets in the west”, which is absolutely true.

Anyone with experience in the outdoors, including scouting, will probably have been told:

In the morning, follow your shadow to travel west.

In the afternoon, follow your shadow to travel east.

And… moss grows on the north sides of trees, etc…

HOWEVER….

That’s not always EXACTLY true…

About a year ago, I had a discussion about a road that runs north-south, and why the sun doesn’t track east-west.

It went like this:

Him: East is THAT way (points southwest)

Me: Um… No.  The road there runs north-south, so east is perpendicular to that road.

Him:  Well, I was always taught that the sun points east-west.

Me: That’s not exactly true, because the Earth is tilted on its access, causing certain parts of the Earth’s surface to be closer to the sun…

Him: I don’t believe all that…

Me:  It’s spring.  Go stand at the fence pole and walk from the fence to the house, following the sun’s path.  Then, do it again, same time of day, 6 months from now.

(To this day, we disagree on due east)

When I instinctively mentioned the Earth’s orbit, and the 23 degree tilt of Earth’s axis, I knew that to be correct, but I ended up dwelling on the actual mechanics.

So, here we go…

The Earth is tilted on its rotational axis by 23 degrees, relative to its orbital orientation around the sun.

As a result, over the course of a year, which is the Earth’s orbital period, that tilt causes the Earth’s orientation relative to its orbital plane to change.

In the winter and summer, if you were to slice through the center of the sun with a big knife, along the axis of the Earth’s rotation, the sun would be sliced vertically – perpendicular to the orbital plane, which is the path that the Earth follows as it orbits the sun.

However, in the spring and fall (and all other times), the Earth is canted with respect to the orbital plane, and thus if you were to slice through the sun’s center with a knife aligned with the Earth’s rotational axis, the sun would be cut at an angle with respect to the orbital plane.

In the summer, the Earth’s north pole is canted toward the sun.  From the sun’s perspective, its track is perfectly parallel to the Earth’s equator, which is perpendicular to the Earth’s rotational axis.

Likewise, in the winter, although the Earth’s north pole is canted away from the sun, the sun’s track is still parallel to the Earth’s equator.

In the spring and fall, the plane of the Earth’s rotational axis is tilted from the sun’s perspective.

In the spring, from the sun’s perspective, the Earth appears tilted to the right, causing the sun’s track to pass along a path that’s rotated counter-clockwise relative to the Earth’s equator and rotational axis.

Likewise, in the fall, the Earth is tilted in the opposite direction due to its orbit, and the sun’s track appears rotated clockwise relative to the Earth’s equator and rotational axis.

Thus, in winter or summer, the sun’s track follows a true east-west path.  At every other time, the Earth appears slightly tilted to the sun, causing the sun’s track to follow a northeast-southwest path in the spring, or a southwest-northeast path in the fall.

So, to all of you scouts out there…

• In the winter or summer, your shadow points east-west.
• In the fall, put the sun over your right shoulder to follow an east-west path.
• Likewise, in the spring, put the sun over your left shoulder to follow an east-west path
• When a map shows that a road runs north-south or east-west, it runs north-south or east-west, regardless of where the sun points.