In the mean time, the Cornell team has begun machining structural parts for the cubesat that will be part of the final unit to be sent into space. Additionally, GNURadio install scripts for various flavors of Linux have been released on the KickSat GitHub. A wiki describes the install process for those interested in building a local ground station.

]]>Parts one and two of modeling the atmosphere derived a way to describe atmospheric pressure as a function of altitude, assuming a constant temperature and acceleration due to gravity throughout the ideal gas. In this installment of Modeling the Atmosphere, the same function will be derived, this time using the Boltzmann distribution.

The Boltzmann distribution states that the fraction of particles in an energy state is given by the equation

where represents the degeneracy of the energy level and the partition function

This distribution works for discrete energy levels, so a continuum approximation must be made for classical systems. By using a density of states , which supposes that there are states in the energy interval to , the probability distribution for the energy

The first goal in when using the Boltzmann distribution to model an atmosphere is to find an expression for the energy .

The equipartition theorem states that each degree of freedom for a particle receives of energy. So, in an isothermal atmosphere of ideal gas, each particle receives the same amount of kinetic energy . The potential energy of a particle depends on its altitude and is given by the equation , where is the mass of a particle, is the acceleration due to gravity (once again assumed to be constant), and represents altitude. Thus, the total energy of an ideal gas particle in an isothermal atmosphere is given by

Assuming that the density of states is a constant , then the continuous distribution

Calculating the integral:

Evaluate at zero:

And at infinity:

Thus:

Therefore

For a large system, this equation describes the fraction of particles at a height between and :

Notice that the coefficient has units , which cancels out to be unitless.

How can this expression for the factional number of particles at a height be transformed to describe pressure as a function of altitude? The ideal gas law allows pressure to be expressed in terms of the number of particles :

The number of particles between the height and is given by the fraction of particles between and times the total number of particles in the system:

So the pressure as a function of altitude is given by:

Notice that the coefficient now has units of pressure:

Using the boundary condition at , this coefficient is equal to the initial pressure :

This equivalence

makes sense conceptually. To see why, note that , or the mass of all of the particles in the atmosphere, and , where is the surface area of the ground. Then

which describes the force of the entire atmosphere across the surface.

Thus, the exact same expression for pressure as a function of altitude (previously derived using a different method) has been found, this time starting from the Boltzmann distribution!

]]>This equation is short to write, but it is not particularly useful. Before applying it to real-life situations, it must be transformed into the barometric equation.

First, recall that density represents mass per unit volume. In Modeling the Atmosphere, pt. 1, was used to replace the expression , where represents the mass of a thin slice of air with volume . Now,

Air is not actually composed of slices, so it would be more convenient to represent the mass in terms of particles. By volume, Earth’s atmosphere is composed of about 78% , 21% , and 1% . Approximating Earth air as an ideal gas and using some information from the periodic table, the mass of one mole () of air particles can be calculated:

So on average, air particles (a.k.a. one mole of air particles) masses approximately . Using this knowledge, the mass of one slice of air may be re-expressed as , where represents the number of moles of air particles in the slice of atmosphere. The same process holds for any other collection of particles that can be modeled as an ideal gas. For a general gas, each mole might mass instead of . Generally:

and for Earth’s atmosphere in particular,

Now that the mass of the slice of atmosphere has been re-expressed as , let’s re-express the volume of the atmosphere slice using the ideal gas law:

Solving for volume yields:

Substitute:

and simplify:

At this point, we’ve almost transformed the starting equation into the barometric equation. To review, represents the mass per mole of the ideal gas (for Earth’s atmosphere, ), is the acceleration due to gravity, is the Boltzmann constant, and is the temperature in Kelvin (assumed to be constant – regardless of height – in this idealized atmosphere).

The next step requires calculus. First, “multiply both sides” by the differential :

and divide both sides by :

Using the classic technique for solving differential equations, integrate both sides:

Exponentiate both sides:

Using the boundary condition when , it’s clear that the constant corresponds to the pressure at zero altitude: . Thus,

Finally, we have derived the barometric equation. Using this equation, the pressure of an idealized atmosphere can be determined as a function of altitude . Interestingly, this derivation is not the only method for obtaining the barometric equation. Check back for an alternate derivation and an application to satellite testing with weather balloons.

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During the first CRS mission, the Falcon 9 rocket experienced an anomaly in one of its first stage engines. A sudden loss of pressure caused a small explosion before the engine was shut down and a new ascent trajectory was calculated. This time, no anomalies with the Falcon rocket were reported.

However, once the Dragon capsule achieved orbit, ground control discovered that three of the capsule’s four thruster pods were being inhibited from initializing. Deciding to delay solar panel deployment until at least two of the thrusters were operational, the team prepared to signal an override to the inhibit command as Dragon passed over an Australia ground station.

Finding the thruster pod three tank pressure to be positive, the capsule’s solar panels were deployed. The team then proceeded to bring up thruster pods two and four. With the help of the U.S. Air Force’s long range communication system, all four thruster pods were engaged, allowing Dragon to transition from free drift to active control. The capsule entered an orbit-raising burn, putting it on track to rendezvous with the ISS a day later than originally planned.

The Dragon capsule docked with the ISS at 4:31 AM CDT on Sunday, March 3, and remained attached to the station for about three weeks. During its stay, the space station crew unloaded supplies and loaded a return payload with research results, experiments, and equipment. Over 3000 pounds of equipment and packing material are loaded onto the craft.

At 5:56 AM CDT on Tuesday, March 26, Dragon departed the ISS for Earth. The departure took place one day later than originally planned do due sea weather at the landing site. The capsule then conducted three departure burns, followed by a ten minute deorbit burn. As the capsule reentered the armosphere, it ejected its trunk and solar arrays to burn up in the atmosphere.

The pod thrusters performed nominally during the return approach. After blasting through the atmosphere, Dragon successfully deployed its drogue parachute, followed soon after by its three main chutes. Splashdown occurred 200 miles out in the Pacific at 11:34 AM CDT. Less than two days later, Dragon arrived at the Port of LA, cargo intact.

An interactive panoramic view inside the Dragon capsule can be viewed here.

SpaceX’s third commercial resupply mission, CRS-3, will launch no earlier than 2 October 2013, with KickSat onboard.

]]>As explained previously, KickSat and its payload of Sprites launches on the SpaceX CRS-3/ELaNa-5 mission no earlier than 2 October 2013. The final launch date depends on several factors, one in particular being the success of the SpaceX CRS-2 mission. The CRS-1 mission, summarized in an earlier post, started shaky but ended strong – an overall success.

Barring setbacks, CRS-2 launches tomorrow, March 1. A SpaceX Dragon capsule will be carried into orbit by one of SpaceX’s Falcon 9 rockets. The Dragon capsule is expected to reach the International Space Station the next day. Read on for NASA’s official mission overview and links to live internet streams of the launch.

SpaceX CRS-2 press kit mission overview:

After SpaceX’s successful first official resupply mission to the International Space Station last October, SpaceX is set to launch its second Commercial Resupply (CRS) mission to the orbiting lab. The SpaceX CRS-2 mission is targeting launch at 10:10AM EST Friday, March 1 from Launch Complex 40 at the Cape Canaveral Air Force Station, Florida.

If all goes as planned, Dragon will arrive at station on Saturday, March 2, where it will be grappled and berthed to the complex for an expected three-week visit. Dragon is scheduled to return to Earth on March 25 for a parachute-assisted splashdown off the coast of Baja California. Dragon is the only cargo craft capable of returning a significant amount of supplies from station to Earth, including experiments.

Information for viewing the launch live:

The launch will be webcast live, with commentary from SpaceX corporate headquarters in Hawthorne, California, at spacex.com/webcast, and NASA’s Kennedy Space Center at www.nasa.gov/nasatv.

NASA TV and web will begin pre-launch coverage at 8:30AM EDT.

The official SpaceX webcast will begin approximately 40 minutes before launch.

Good luck to SpaceX on the CRS-2 mission!

Apologies for the color balance of the images, usually I wait to take photos in natural light so that the temperature comes out a little cooler. I arrived home tonight to this kit in the mailbox and couldn’t help but post something right away!

*For an overview of the project, visit the About page.*

It’s a horrible joke. Non-physicists don’t find it funny, and physicists just nod and think, “yes, that’s exactly how we work.” But as an anecdote it proves useful, particularly for physicists about to present what will look like an incredibly reductionist model. Let’s take a look at one of physics’ most simplistic but useful representations of an atmosphere.

The exponential atmosphere model, also called the isothermal or barotropic atmosphere model, rests on three assumptions: the atmosphere is composed of an ideal gas, temperature is constant regardless of height, and acceleration due to gravity remains the same at all altitudes. These assumptions allow for a model in which pressure is a function of height:

which means that atmospheric pressure depends only on the altitude .

By definition, an ideal gas is composed of a very large number (at least one mole, i.e. ) of identical, spherical particles. These particles follow Newton’s laws of motion and don’t interact with each other or the system’s boundaries except for perfectly elastic collisions (i.e. collisions in which no kinetic energy is lost). Particle size measures very small (at least ten times as small) compared to the average distance between particles. The particles distribute uniformly and randomly throughout an ideal gas, and in equilibrium conditions some density (mass per unit volume) characterizes it. These assumptions taken together lead to the ideal gas law, stated mathematically as

where represents pressure in Pascals , represents volume in , equals the number of particles (not to be confused with the for units of force in Newtons), is the Boltzmann constant , and represents the temperature in Kelvin .

Working from the ideal gas model, we can begin to construct a decently useful atmospheric model. To start, consider a thin slice of atmosphere parallel to the ground. We want this thin slice of atmosphere to be general, so instead of specifying a particular altitude for to occupy we say that it resides at some arbitrary height from the ground. The slice measures very thin, only some small amount that we will call from bottom to top.

What can we say about our slice of atmosphere? Although is paper-thin, it extends very far in the and directions. We can visualize as a large plane with some surface area (Technically, gradually curves, extending all the way around the Earth to form a thin spherical shell around the planet, but this model ignores that detail). Our slice of atmosphere may be thin in , but its great extent in and means that, in total, contains many particles of air. Each particle contains some mass, and the sum of all these masses totals a surprisingly substantial aggregate mass for our slice of atmosphere.

We now know that is a large object of mass a height above the ground. Working from these facts, a reasonable question to ask is whether is falling downward. After all, most large, massive objects standing in midair do. But the answer is that it can’t be! Remember, our slice of atmosphere is general; it occupies an arbitrary altitude and represents any slice of atmosphere we might consider. Whatever holds true about our slice of atmosphere must hold true for any and all slices of atmosphere, regardless of altitude, otherwise wouldn’t be general. So, if is falling downward, then so must all slices of atmosphere. Under these circumstances, the entire atmosphere collapses into the ground. None of the atmospheres we observe display this behavior, so our arbitrary slice of atmosphere must not be falling.

We can ask the same question but in the opposite direction: is our slice of atmosphere rising? Again, it cannot be. If any given slice of atmosphere rose away from the ground, then the entire atmosphere would fly off into space. We do not observe this phenomenon, so it seems that an arbitrary slice of atmosphere does not rise. Therefore, because an arbitrary slice of atmosphere neither falls nor rises, we must conclude that is stationary – it’s suspended in midair.

How is it possible that an object of mass stands stationary in midair? The object must feel a downward force from to gravity in accordance with Newton’s second law of motion:

In this system, the acceleration vector would be the acceleration due to gravity , and the mass would be the mass of our slice of atmosphere . Therefore, the force due to gravity on our slice of atmosphere is given by:

(Remember, this atmospheric model assumes a constant acceleration due to gravity, i.e. remains the same at all heights).

Clearly, our slice of atmosphere experiences a downward force. So why doesn’t it fall? The answer lies with pressure.

Pressure is a force per unit area (units ); an ideal gas exerts pressure at its boundaries. In our model, the slice forms a two-dimensional boundary of surface area for all of the air beneath it. Therefore, it experiences a force upwards due to pressure from the air underneath it:

Air resides above our slice of atmosphere as well, exerting a force due to pressure downwards. Since has a thickness of , the pressure from above differs slightly from the pressure from below:

An object stands motionless only when all forces acting on that object sum to zero. All three forces must sum to zero in order to ensure that our slice of air remains in place. Since the vectors for all of our forces are in the same direction , we can work with their scalars:

Rearranging gives

If we divide both sides by , some interesting things start to happen

The left side, , may be replaced by using the definition of the derivative (see equation 6). The change on the right side is less opaque. In the denominator, the product of the surface area of our slice of atmosphere with its height yields the volume of :

Now our equation has the form

There remains one final simplification. On the right side of the equation, , the fraction represents a particular quantity. Let’s examine the units in order to decide the nature of this quantity. Mass has units , and volume has units . What quantity represents mass per unit volume ? Density! Using density , the equation becomes

A small, single equation that describes the entire atmosphere! At least, to a reasonable approximation. Remember, this model assumes an ideal gas, constant temperature, and uniform acceleration due to gravity – none of which are true for any atmosphere! That reminds me of a physics joke…

At this point, we can describe the change in pressure with respect to altitude in terms of density and acceleration due to gravity . Although this relation is short and sweet, its form is not particularly useful. Future posts will delve into what exactly this equation means and explore the applications of its modified (more user-friendly) form, the barometric equation.

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Dear KickSat Backer,

Enclosed is your souvenir Sprite spacecraft. This is an actual prototype Sprite, exactly the same as the ones we are using in our pre-flight testing, and just like the ones that will fly in space!

Thanks to your support, we’re now on track for launch in Fall 2013 on the CRS-3/ELaNa-5 mission. You can keep track of our progress as we build KickSat by reading our Kickstarter blog updates. You can also get involved yourself by putting together a ground station to listen for the Sprites as they fly over your town.

Thank you for sharing my dream of making spaceflight accessible to everyone. Together we are making this dream a reality!

Sincerely,

Zac Manchester

The letter drops a major piece of news: the SpaceX CRS-3 mission has been pushed from summer 2013 to fall 2013. This change follows a shift of the second NASA Commercial Resupply Services mission to the ISS from December of this year to no earlier than 1 March 2013. As a result, the July launch date of the CRS-3 mission has been rescheduled to no earlier than 2 October 2013.

The souvenir boards are for display only. The printed circuit board is genuine, but the solar cell is not. The development kits will be shipping during the first or second week of January. Instructions for constructing a ground station have not yet been published.

]]>October was a busy month for commercial spaceflight! Not only did SpaceX complete its first commercial resupply mission to the International Space Station, but Nanosatisfi completed its first successful high-altitude test of the ArduSat payload. Since the first high-altitude test didn’t get off the ground, this mission was a big moment for the Nanosatisfi team.

Nanosatisfi partnered with Colorado Springs based Edge Research Lab to send its sensor payload to the edge of space on Saturday, October 27. The mission launched from Monument, Colorado, a town already 2.1 km above sea level, and flew towards space underneath a 1600 g high-altitude Hwoyee meteorological balloon.

The entire flight apparatus massed approximately 4.3 kg, about the mass of two freshman physics textbooks. Once filled with hydrogen gas, the balloon provided 56 N of lift at the neck. After making sure that tracking information and flight telemetry were being received, the group called the local air traffic controller and waited for FAA clearance. The balloon was released at 12:00 noon CDT (10:00 am local time) and proceeded to rise through the atmosphere for close to an hour.

The balloon climbed to an altitude of nearly 26 km before bursting at 12:56 CDT. Slowed only by a small drag parachute, the ArduSat sensor payload fell back to Earth. About 40 minutes later, at 1:33 PM CDT, the payload landed approximately 129 km from the launch site. The payload was recovered about 25 km south of the intersection of State Highway 94 and State Highway 71 (aka Punkin Center). A summary video of the entire mission from the perspective of the sensor payload can be viewed below:

The video was shot using two cameras, one pointed at the horizon and one at the balloon. At least one of these cameras was a HackHD camera, a small camera advertised to shoot 1080p video at 30 fps. Both cameras are owned and operated by Edge Research Lab, and neither are a part of the ArduSat payload.

Pictured above is a three-dimensional representation of the ArduSat payload flight path. The relatively heavy sensor payload meant that the balloon needed to be filled with more gas (you can see in the video that the ArduSat balloon is much more inflated), which in turn led to a low bursting altitude. Balloons with lighter payloads can be filled with less gas and achieve a higher altitude before bursting, between 40 and 50 kilometers. The ArduSat payload reached a maximum speed of about 54 m/s, or 193 km/hr, over the course of its flight.

After recovery, the ArduSat team announced happily that the sensor payload was intact and functional. They have released a picture of what appears to be Pike’s Peak, taken from the sensor payload’s camera, and confirmed the collection of payload sensor data. The team promised a more in-depth analysis of the payload sensor data “coming soon,” and followed up with a Kickstarter update on November 5. Unfortunately, the update provided little in the way of details concerning the outcome of the flight’s sample programs and sensor tests.

]]>To answer this question, I’ll start with the most simplistic model of signal power reduction as it travels to Earth: free space path loss. This model describes the power of a signal along a path ray pointing outward from the source as it spreads through vacuum, and it gives the best case scenario for a Sprite signal arriving at a ground station on Earth. Free space path loss takes two phenomena into account.

The first phenomenon this model takes into account is the inverse square law. Mathematically,

,

where is the power per unit area (spatial power density) as a function of distance from the power source and is the total power transmitted from the power source in watts. For the Sprites, the distance and the total power transmitted . Thus,

.

The second phenomenon this model takes into account has to do with how well the receiving antenna absorbs power from an incoming signal. Mathematically, the power received

for an isotropic (ideal) receiver, where is the wavelength of the incoming signal. Sprites transmit at a frequency of , so

.

Using a wavelength of , then

.

So, in a best case scenario, Sprite signals will reach a ground station antenna with a power of only 119 attowatts. That’s incredibly small! Surely there is no hope of picking up such a weak signal?

Before abandoning the project, let’s examine the signal strength of a different group of satellites: GPS. Satellites that are part of the GPS constellation orbit the Earth at an altitude of approximately and transmit signals at a power of . A signal has a wavelength

.

Thus the received power

.

In other words, GPS signals received by smartphones have a power no greater than 46 attowatts!

It seems as if these incredibly small values are not so unusual. In fact, they are common enough that a different unit of measurement, the dBm, is often used instead of watts so as to avoid calculations with overly small numbers. By defining one dBm to equal the power ratio in decibels of some measured power referenced against one milliwatt, the range of values tightens considerably. Mathematically, for some power in watts the equivalent value in dBm is given by the expression

.

So the Sprites’ 119 attowatt signals can be alternately expressed as

.

If a smartphone can receive 46 attowatt signals from GPS satellites, then a ground station can receive 119 attowatt signals from a Sprite. However, the extremely low power of these signals means that they can be easily corrupted by background noise in the electromagnetic spectrum. Signals of any power can be received and amplified without a problem. But unless some sort of filtering is applied for weak signals, noise will drown out their message. In order to ensure that received signals are readable, both the sender and receiver must make use of some clever tricks. I will cover some of these tricks – such as forward error correction and CDMA (code division multiple access) – in follow-up posts.

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