**Physicist**: If you have plenty of chalkboard space and absolutely nothing better to do, you can write down numbers, letters (Greek if you’re δυσάρεστος), and mathematical operators and eventually you’ll have the longest equation ever written down. So if you really want to see the most complicated equation ever, call in sick for a few weeks. A better question might be “what is the most (but not needlessly) complicated equation?”.

There are plenty of equations that are infinitely long, but often they’re simple enough that we can write them compactly. For example, This equation goes on forever, but it’s fairly straight forward: every term you flip the sign and increase the denominator from one odd number to the next. You can write it in mathspeak as . Like π itself, this sum goes on forever, but it isn’t complicated. You can describe it simply and in such a way that anyone (with sufficient time and chalk) can find as many digits of π as they like. This is the basic idea behind “Kolmogorov Complexity“; the length of the shortest possible set of written instructions that can produce a given result (never mind how long it takes to actually compute it).

If you’re looking for an equation that needs to be complicated, a good place to look is physics (I mean, what else do you really *need* math for?). If you want to describe the behavior of a ball flying through the air it’s not enough to say “it goes up then down”; there’s a minimum amount of math that goes into accurately calculating the path of falling objects, and it’s more complicated than that.

Arguably, the universe is pretty complicated. But like π, it is deceptively so (we hope). If you want to do something like, say, describe the gravitational interactions of every star in the galaxy, you’d do it by numbering the stars (take your time: star 1, star 2, …, star n), determine the position and mass of each, and , and then find the force on each star produced by all the others. In practice this is absurd (there are a few hundred billion stars in the Milky Way, but we can’t see most of them because there’s a galaxy in the way), but the equation you would use is pretty straight forward. The force on star k is: . This is just Newton’s law of gravitation, , repeated for every possible pair of stars and added up. So while the situation itself is complicated, the equation describing it isn’t. Evidently, if you want an equation that genuinely needs to be complicated, you don’t need a complicated situation, you need complicated dynamics.

The equation for the gravity between many objects is just the equation for the gravity between every pair added up. Not much more complicated.

The whole point of physics, aside from understanding things, is to describe the rules of the universe as simply as possible. To that end, physicists love to talk about “Lagrangians”. Once you’ve got the Lagrangian of a system, you can describe the behavior of that system by applying the “principle of least action“, which says that the behavior of a system (the orbit a planet takes, the path of light through materials, etc.) will always be such that the “chosen” path will have the minimum total Lagrangian along that path. It’s a cute recipe for succinctly and simultaneously describing a lot of dynamics that would make Kolmogorov proud.

The Lagrangian gives every point on this picture a value and the total along an entire path is the “action”. The principle of least action says that the path a system will actually take has the least action. With this principle, a single Lagrangian can be used to derive many physical laws at once, so it’s a good candidate for equations that aren’t *needlessly*complex.

For example, you can sum up Newton’s physics almost instantly. Rather than talking about kinetic energy and momentum and falling, you can just say “Dudes and dudettes, if I may, the Lagrangian for an object flying through the air near the surface of the Earth is , where m is mass, v is velocity, and z is height”. From this *single*formula, you get the conservation of energy, conservation of momentum (when moving sideways), as well as the acceleration due to gravity. There are also Lagrangians for everything from orbiting planets to electromagnetic fields.

Generally, when you look at the same dynamics applied over and over, the equations involved don’t get much more complicated (although their solutions definitely do). And if you want to describe the dynamics of a system, Lagrangians are an extremely compact way to do it. So what’s the most (but not needlessly) complicated equation in the universe? Arguably, it’s the Standard Model Lagrangian, which covers the dynamics of every kind of particle and all of their interactions. Notably, it doesn’t cover gravity, but be cool. It’s a work in progress.

The equation of everything (except gravity).

In some sense this equation is compressed data. All the relevant dynamics are there, but there’s a lot of unpacking to do before that becomes remotely obvious. All equations must have some context before they do anything or even mean anything. That’s why math books are mostly words. “2+2=4” means nothing to an alien until after you tell them what each of those symbols mean and how they’re being used. In the case of the Standard Model Lagrangian, each of these symbols mean a lot, the equation itself is uses cute short hand tricks, and it doesn’t even describe dynamics on its own without tying in the Principle of Least Action. But given all that, it’s describing the most complicated thing we can describe, which is nearly everything, without being needlessly verbose (“mathbose”?) about it.

**Answer Gravy**: This isn’t part of the question, but if you’ve taken intro physics, you’ve probably seen the equations for kinetic energy, momentum, and acceleration in a uniform gravitational field (like the one you’re experiencing right now). But unless you’re actually a physicist, you’ve probably never been freaked out by seeing a Lagrangian work. This gravy is full of calculus and intro physics.

The “action”, , is a function of the path a system takes, . More specifically, it’s the integral of the Lagrangian between any two given times:

where t_{1} and t_{2} are the start and stop times, is a path, is the time derivative (velocity) of that path, and is some given function of and . If you want to chose a path that extremizes (either minimizes or maximizes) S, then you can do it by solving the Euler-Lagrange equations:

This is called the Euler-Lagrange equations (plural) because this is actually several equations. Each different variable (x_{1}=x, x_{2}=y, x_{3}=z) tells you something different. In regular ol’ calculus, if you want to find the value of x that extremizes a function f(x), you solve for the value x. Using the Euler-Lagrange equations is philosophically similar: to find the path that extremizes S, you solve for the path .

The Lagrangian from earlier, for a free-falling object near the surface of the Earth, is:

For z:

So the E-L equation says:

or

In other words, “everything accelerates downward at the same rate”. Doing the same thing for x or y, you get , which says “things don’t accelerate sideways”. Both good things to know.

You wanna be even slicker, note that this Lagrangian is independent of time. That means that . Therefore, applying the chain rule:

But we have the E-L equations! Plugging those in:

And therefore:

This thing in the parentheses is constant (since it never changes in time). In the case of we find that this constant thing is:

Astute students of physics 1 will recognize the sum of kinetic energy plus gravitational potential. In other words: this is a derivation of the conservation of energy for free-falling objects. A more general treatment can be done using Noether’s Theorem, which says that every symmetry produces a conserved quantity. For example, a time symmetry ( doesn’t change in time) leads to conservation of energy and a space symmetry ( doesn’t change in some direction) leads to conservation of momentum in that direction.

]]>Static friction is a force that keeps an object at rest. It must be overcome to start moving the object. Once an object is in motion, it experiences kinetic friction. If a small amount of force is applied to an object, the static friction has an equal magnitude in the opposite direction. If the force is increased, at some point the value of the maximum static friction will be reached, and the object will move. The coefficient of static friction is assigned the Greek letter “mu” (μ), with a subscript “s”. The maximum force of static friction is μ_{s} times the normal force on an object.

*force of static friction ≤ (coefficient of static friction)(normal force) maximum force of static friction = (coefficient of static friction)(normal force)*

*F _{s} ≤ μ_{s} η , and F_{s}^{max} = μ_{s} η*

*F _{s} *= force of static friction

*μ _{s} *= coefficient of static friction

*η* = normal force (Greek letter “eta”)

≤ means “less than or equal to”

*F _{s}^{max}* = maximum force of static friction

**Static Friction Formula Questions:**

1) A 5500 N force is applied to a sled full of firewood in a snow-covered forest. The skis of the sled have a coefficient of static friction *μ _{s}* = 0.75 with the snow. If the fully-loaded sled has a mass of 700 kg, what is the maximum force of static friction, and is the force applied enough to overcome it?

**Answer:** On a flat surface, the normal force on an object is *η = mg*. Using this, the maximum force of static friction can be found:

*F _{s}^{max} = μ_{s} η*

*F _{s}^{max} = μ_{s} mg*

*F _{s}^{max}* = (0.75)(700 kg)(9.8 m/(s

*F _{s}^{max}* = 5145 kg∙m/s

*F _{s}^{max}* = 5145 N

The maximum force of static friction is 5145 N, and therefore the applied force of 5500 N is enough to overcome it, and start moving the sled.

2) A person building a brick-making machine wants to measure the coefficient of static friction between brick and wood. To do this, she places a 2.00 kg brick on a flat piece of wood, and gradually applies more and more force until the brick moves. She finds that the brick moves when exactly 11.8 N of force is applied. What is the coefficient of static friction?

**Answer:** The force that was applied was exactly the right amount to overcome static friction, so it is equal to *F _{s}^{max}*. On a flat surface, the normal force on an object is

*μ _{s}* = 0.6020…

Give the answer with three significant digits, to match the other numbers in the equation:

*μ _{s}* ≈ 0.602

We find the coefficient of static friction between the brick and wood to be 0.602.

]]>If the state of an electron in a hydrogen atom is slightly perturbed, then the electron can make a transition to another stationary. The transition will emit a photon with a certain wavelength. If the electron state is characterized by the quantum number n the wavelength is given by the Rydberg formula.

**(1/wavelength of the emitted photon) = (Rydberg constant)(1/(integer 1) ^{2} – 1/(integer 2)^{2})**

The equation is:

1/λ = R(1/(n_{1})^{2} -1/(n_{2})^{2})

with n_{1} < n_{2}

Where:

R: Rydberg’s constant (R=1.097 * 10^{7} m^{(−1)})

λ: Wavelength of the emitted photon

n_{1}: integer 1

n_{2}: integer 2

**Rydberg Formula Questions:**

1) Assume an electron transition occurs from the n_{1}=2 to the n_{2}=3, what is the wavelength of the emitted photon?

**Answer:**

Substituting the data in the Rydberg formula

1/λ = (1.097 * 10^{7} m^{(−1)})*(1/2 – 1/3)

1/λ = (1.097 * 10^{7} m^{(−1)})*0.1666 = 0.182 *10^{7} (1/m)

λ = 5.47 * 10^{(-7)} m

2) Assume an electron transition occurs from the n_{1}=1and the wavelength of the emitted photon is 1.7 * 10^{(-7)} m, what is the integer number associated with the transition?

**Answer:**

Substituting the data in the Rydberg formula

1/1.7 * 10^{(-7)} m = (1.097 * 10^{7} m^{(−1)})*(1- 1/n)

From this formula we find n

0.64 = 1 – 1/n

=1 – 0.64

=0.46

=1/0.46

=2.15

n ≈ 2

]]>The Archimedes principle states that the upward buoyancy force exerted on a body partially or completely immersed in a fluid is equal to the weight of the fluid that the body displaces and acts in an upward direction in the center of the mass of the displaced fluid. The Archimedes principle is a fundamental law of physics for fluid mechanics. It was formulated by Archimedes of Syracuse.

**push = density of fluid * gravity acceleration * volume of object.**

The equation is:

p=ρ_{f}*g*V

we have,

p = push

ρ_{f} = density of fluid.

g = gravity acceleration.

V = volume of object.

**Archimedes Principle Questions:**

1)A steel ball of 5cm radius is immersed in water. It calculates the thrust and the resulting force (lead density = 7900kg/m^{3}).

**Answer:** To calculate the resulting push, to calculate the push we must calculate the volume of the ball.

V = 4/3πr^{3}= 4/3π (0,05 m)^{3} = 5.24*10^{-4} m^{3}

and know the density of water (1000 kg/m^{3}).

p = Pf*g*v= 1000 kg/m^{3} * 9.8 m/s² *5,23*10^{-4} m^{3} = 5.1254 N.

p = 5.1254 N.

Let’s go with the resulting force. Here two forces act: the push of the water upwards and the weight of the ball downwards. We still have to calculate this last one:

the density of lead is 7900kg/m^{3}, then the mass of lead is

m_{l} = 7900kg/m m*5.24*10^{-4} m^{3} = 4.1396 Kg

The weight of the ball is

w = m_{l}*g = 4.1396 Kg* 9.8m/s^{2} = 40.568 N

The resulting apparent weight is:

W_{a} = w-p = 40.568 N – 5.1254 N = 35.443 N

W_{a} = 35.443 N.

2) Weigh a cube of 10cm edge into the air resulting in 19 N and then weighed immersed in water giving a value of 17 N. Calculate the apparent weight and the push.

**Answer:** The apparent weight is the weight of the object immersed in a fluid, or in other words, the result of the actual weight and thrust. Therefore, the apparent weight is 17 N .

W_{a} = 17 N.

The push is:

w = 19 N.

W_{a} = w-p-→

p = w-W_{a} = 19 N- 17 N = 2 N

p = 2 N.

]]>Bayes’ theorem is a way to figure out conditional probability. Conditional probability is the probability of an event happening, given that it has some relationship to one or more other events. For example, your probability of getting a parking space is connected to the time of day you park, where you park, and what conventions are going on at any time. Bayes’ theorem is slightly more nuanced. In a nutshell, it gives you the actual probability of an **event **given information about **tests**.

- “Events” Are different from “tests.” For example, there is a
**test**for liver disease, but that’s separate from the**event**of actually having liver disease. **Tests are flawed**: just because you have a positive test does not mean you actually have the disease. Many tests have a high false positive rate.**Rare events tend to have higher false positive rates**than more common events. We’re not just talking about medical tests here. For example, spam filtering can have high false positive rates. Bayes’ theorem takes the test results and calculates your*real probability*that the test has identified the event.

Bayes’ Theorem (also known as Bayes’ rule) is a deceptively simple formula used to calculate conditional probability. The Theorem was named after English mathematician Thomas Bayes (1701-1761). The formal definition for the rule is:

In most cases, you can’t just plug numbers into an equation; You have to figure out what your “tests” and “events” are first. For two events, A and B, Bayes’ theorem allows you to figure out p(A|B) (the probability that event A happened, given that test B was positive) from p(B|A) (the probability that test B happened, given that event A happened). It can be a little tricky to wrap your head around as technically you’re working backwards; you may have to switch your tests and events around, which can get confusing. An example should clarify what I mean by “switch the tests and events around.”

You might be interested in finding out a patient’s probability of having liver disease if they are an alcoholic. “Being an alcoholic” is the **test** (kind of like a litmus test) for liver disease.

**A**could mean the event “Patient has liver disease.” Past data tells you that 10% of patients entering your clinic have liver disease. P(A) = 0.10.**B**could mean the litmus test that “Patient is an alcoholic.” Five percent of the clinic’s patients are alcoholics. P(B) = 0.05.- You might also know that among those patients diagnosed with liver disease, 7% are alcoholics. This is your
**B|A:**the probability that a patient is alcoholic, given that they have liver disease, is 7%.

Bayes’ theorem tells you:**P(A|B) = (0.07 * 0.1)/0.05 = 0.14**

In other words, if the patient is an alcoholic, their chances of having liver disease is 0.14 (14%). This is a large increase from the 10% suggested by past data. But it’s still unlikely that any particular patient has liver disease.

Another way to look at the theorem is to say that one event follows another. Above I said “tests” and “events”, but it’s also legitimate to think of it as the “first event” that leads to the “second event.” There’s no one right way to do this: use the terminology that makes most sense to you.

In a particular pain clinic, 10% of patients are prescribed narcotic pain killers. Overall, five percent of the clinic’s patients are addicted to narcotics (including pain killers and illegal substances). Out of all the people prescribed pain pills, 8% are addicts. *If a patient is an addict, what is the probability that they will be prescribed pain pills?*

Step 1: **Figure out what your event “A” is from the question.** That information is in the italicized part of this particular question. The event that happens first (A) is being prescribed pain pills. That’s given as 10%.

Step 2: **Figure out what your event “B” is from the question.** That information is also in the italicized part of this particular question. Event B is being an addict. That’s given as 5%.

Step 3: **Figure out what the probability of event B (Step 2) given event A (Step 1)**. In other words, find what (B|A) is. We want to know “Given that people are prescribed pain pills, what’s the probability they are an addict?” That is given in the question as 8%, or .8.

Step 4: **Insert your answers from Steps 1, 2 and 3 into the formula and solve.**

P(A|B) = P(B|A) * P(A) / P(B) = (0.08 * 0.1)/0.05 = 0.16

The probability of an addict being prescribed pain pills is 0.16 (16%).

A slightly more complicated example involves a medical test (in this case, a genetic test):

There are **several forms of Bayes’ Theorem **out there, and they are all equivalent (they are just written in slightly different ways). In this next equation, “X” is used in place of “B.” In addition, you’ll see some changes in the denominator. The proof of why we can rearrange the equation like this is beyond the scope of this article (otherwise it would be 5,000 words instead of 2,000!). However, if you come across a question involving medical tests, you’ll likely be using this alternative formula to find the answer:

Watch the video for a quick solution or read two solved Bayes’ Theorem examples below:

1% of people have a certain genetic defect.

90% of tests for the gene detect the defect (true positives).

9.6% of the tests are false positives.

If a person gets a positive test result, **what are the odds they actually have the genetic defect?**

The first step into solving Bayes’ theorem problems is to assign letters to events:

- A = chance of having the faulty gene. That was given in the question as 1%. That also means the probability of
*not*having the gene (~A) is 99%. - X = A positive test result.

So:

- P(A|X) = Probability of having the gene given a positive test result.
- P(X|A) = Chance of a positive test result given that the person actually has the gene. That was given in the question as 90%.
- p(X|~A) = Chance of a positive test if the person
*doesn’t*have the gene. That was given in the question as 9.6%

Now we have all of the information we need to put into the equation:

P(A|X) = (.9 * .01) / (.9 * .01 + .096 * .99) = 0.0865 (8.65%).

The probability of having the faulty gene on the test is 8.65%.

I wrote about how challenging physicians find probability and statistics in my post on reading mammogram results wrong. It’s not surprising that physicians are way off with their interpretation of results, given that some tricky probabilities are at play. Here’s a second example of how Bayes’ Theorem works. I’ve used similar numbers, but the question is worded differently to give you another opportunity to wrap your mind around how you decide which is event A and which is event X.

**Q. Given the following statistics, what is the probability that a woman has cancer if she has a positive mammogram result?**

- One percent of women over 50 have breast cancer.
- Ninety percent of women who have breast cancer test positive on mammograms.
- Eight percent of women will have false positives.

Step 1: Assign events to A or X. You want to know what a woman’s probability of having cancer is, given a positive mammogram. For this problem, actually having cancer is A and a positive test result is X.

Step 2: List out the parts of the equation (this makes it easier to work the actual equation):

P(A)=0.01

P(~A)=0.99

P(X|A)=0.9

P(X|~A)=0.08

Step 3: Insert the parts into the equation and solve. Note that as this is a medical test, we’re using the form of the equation from example #2:

(0.9 * 0.01) / ((0.9 * 0.01) + (0.08 * 0.99) = 0.10.

The probability of a woman having cancer, given a positive test result, is 10%.

**Remember when (up there ^^) I said that there are many equivalent ways to write Bayes Theorem?** Here is another equation, that you can use to figure out the above problem. You’ll get exactly the same result:

The main difference with this form of the equation is that it uses the probability terms *intersection*(∩) and *compliment *(^{c}). Think of it as shorthand: it’s the same equation, written in a different way.

In order to find the probabilities on the right side of this equation, use the multiplication rule:

P(B ∩ A) = P(B) * P(A|B)

The two sides of the equation are equivalent, and P(B) * P(A|B) is what we were using when we solved the numerator in the problem above.

P(B) * P(A|B) = 0.01 * 0.9 = 0.009

For the denominator, we have P(B^{c} ∩ A) as part of the equation. This can be (equivalently) rewritten as P(B^{c}*P(A|B^{c}). This gives us:

P(B^{c}*P(A|B^{c}) = 0.99 * 0.08 = 0.0792.

Inserting those two solutions into the formula, we get:

0.009 / (0.009 + 0.0792) = 10%.

Bayes’ theorem problems can be figured out *without* using the equation (although using the equation is probably simpler). But if you can’t wrap your head around why the equation works (or what it’s doing), here’s the non-equation solution for the same problem in #1 (the genetic test problem) above.

Step 1: Find the probability of a true positive on the test. That equals people who actually have the defect (1%) * true positive results (90%) = .009.

Step 2: Find the probability of a false positive on the test. That equals people who don’t have the defect (99%) * false positive results (9.6%) = .09504.

Step 3: Figure out the probability of getting a positive result on the test. That equals the chance of a true positive (Step 1) plus a false positive (Step 2) = .009 + .09504 = .0.10404.

Step 4: Find the probability of actually having the gene, given a positive result. Divide the chance of having a real, positive result (Step 1) by the chance of getting any kind of positive result (Step 3) = .009/.10404 = 0.0865 (8.65%).

Bayes’ Theorem has several forms. You probably won’t encounter any of these other forms in an elementary stats class. The different forms can be used for different purposes. For example, one version uses what Rudolf Carnap called the “**probability ratio**“. The probability ratio rule states that any event (like a patient having liver disease) must be multiplied by this factor PR(H,E)=P_{E}(H)/P(H). That gives the event’s probability conditional on E. The **Odds Ratio Rule** is very similar to the probability ratio, but the likelihood ratio divides a test’s true positive rate divided by its false positive rate. The formal definition of the Odds Ratio rule is OR(H,E)=P_{H,}(E)/P_{~H}(E).

Although Bayes’ Theorem is used extensively in the medical sciences, there are other applications. For example, it’s used to filter spam. The **event **in this case is that the message is spam. The **test **for spam is that the message contains some flagged words (like “viagra” or “you have won”). Here’s the equation set up (from Wikipedia), read as “The probability a message is spam given that it contains certain flagged words”:

The actual equations used for spam filtering are a little more complex; they contain more flags than just content. For example, the timing of the message, or how often the filter has seen the same content before, are two other spam tests.

In this screenshot, you can see what may be an “anomalous aerial vehicle.”(Image: © To the Stars Academy of Arts & Science )

U.S. Navy pilots and sailors won’t be considered crazy for reporting unidentified flying objects, under new rules meant to encourage them to keep track of what they see. Yet just a few years ago, the Pentagon reportedly shut down another official program that investigated UFO sightings. What has changed? Is the U.S. military finally coming around to the idea that alien spacecraft are visiting our planet?

The answer to that question is almost certainly no. Humans’ misinterpretation of observations of natural phenomena are as old as time and include examples such as manatees being seen as mermaids and driftwood in a Scottish lochbeing interpreted as a monster. A more recent and relevant example is the strange luminescent structure in the sky caused by a SpaceX rocket launch. In these types of cases, incorrect interpretations occur because people have incomplete information or misunderstand what they’re seeing.

Based on my prior experience as a science advisor to the Air Force, I believe that the Pentagon wants to avoid this type of confusion, so it needs to better understand flying objects that it can’t now identify. During a military mission, whether in peace or in war, if a pilot or soldier can’t identify an object, they have a serious problem: How should they react, without knowing if it is neutral, friendly or threatening? Fortunately, the military can use advanced technologies to try to identify strange things in the sky.

“Situational awareness” is the military term for having complete understanding of the environment in which you are operating. A UFO represents a gap in situational awareness. At the moment, when a Navy pilot sees something strange during flight, just about the only thing he or she can do is ask other pilots and air traffic control what they saw in that place at that time. Globally, the number of UFO reportings in a year has peaked at more than 8,000. It’s not known how many the military experiences.

Even the most heavily documented incidents end up unresolved, despite interviewing dozens of witnesses and reviewing many written documents, as well as lots of audio and video recordings.

UFOs represent an opportunity for the military to improve its identification processes. At least some of that work could be done in the future by automated systems, and potentially in real time as an incident unfolds. Military vehicles — Humvees, battleships, airplanes and satellites alike — are covered in sensors. It’s not just passive devices like radio receivers, video cameras and infrared imagers, but active systems like radar, sonar and lidar. In addition, a military vehicle is rarely alone — vehicles travel in convoys, sail in fleets and fly in formations. Above them all are satellites watching from overhead.

Sensors can provide a wealth of information on UFOs including range, speed, heading, shape, size and temperature. With so many sensors and so much data, though, it is a challenge to merge the information into something useful. However, the military is stepping up its work on autonomy and artificial intelligence. One possible use of these new technologies could be to combine them to analyze all the many signals as they come in from sensors, separating any observations that it can’t identify. In those cases, the system could even assign sensors on nearby vehicles or orbiting satellites to collect additional information in real time. Then it could assemble an even more complete picture.

For the moment, though, people will need to weigh in on what all the data reveal. That’s because a key challenge for any successful use of artificial intelligence is building trust or confidence in the system. For example, in a famous experiment by Google scientists, an advanced image recognition algorithm based on artificial intelligence was fooled into wrongly identifying a photo of a panda as a gibbon simply by distorting a small number of the original pixels.

So, until humans understand UFOs better, we won’t be able to teach computers about them. In my view, the Navy’s new approach to reporting UFO encounters is a good first step. This may eventually lead to a comprehensive, fully integrated approach for object identification involving the fusion of data from many sensors through the application of artificial intelligence and autonomy. Only then will there be fewer and fewer UFOs in the sky — because they won’t be unidentified anymore.

Iain Boyd, Professor of Aerospace Engineering, *University of Michigan*

*This article is republished from **The Conversation** under a Creative Commons license. Read the **original article**.*

The power and propulsion element provides a communications relay capability for NASA’s Gateway, enabling it to serve as a mobile command and service module for human and robotic expeditions to the lunar surface.Credits: NASA

In one of the first steps of the agency’s Artemis lunar exploration plans, NASA announced on Thursday the selection of Maxar Technologies, formerly SSL, in Westminster, Colorado, to develop and demonstrate power, propulsion and communications capabilities for NASA’s lunar Gateway.

“The power and propulsion element is the foundation of Gateway and a fine example of how partnerships with U.S. companies can help expedite NASA’s return to the Moon with the first woman and next man by 2024,” said NASA Administrator Jim Bridenstine. “It will be the key component upon which we will build our lunar Gateway outpost, the cornerstone of NASA’s sustainable and reusable Artemis exploration architecture on and around the Moon.”NASA announces the first partnership of its kind with MAXAR Technologies to power the future lunar orbiting station.

The power and propulsion element is a high-power, 50-kilowatt solar electric propulsion spacecraft – three times more powerful than current capabilities. As a mobile command and service module, the Gateway provides a communications relay for human and robotic expeditions to the lunar surface, starting at the Moon’s South Pole.

The power and propulsion element of NASA’s Gateway is a high-power, 50-kilowatt solar electric propulsion spacecraft – three times more powerful than current capabilities.Credits: NASA

This firm-fixed price award includes an indefinite-delivery/indefinite-quantity portion and carries a maximum total value of $375 million. The contract begins with a 12-month base period of performance and is followed by a 26-month option, a 14-month option and two 12-month options.

Spacecraft design will be completed during the base period, after which the exercise of options will provide for the development, launch, and in-space flight demonstration. The flight demonstration will last as long as one year, during which the spacecraft will be fully owned and operated by Maxar. Following a successful demonstration, NASA will have the option to acquire the spacecraft for use as the first element of the Gateway. NASA is targeting launch of the power and propulsion element on a commercial rocket in late 2022.

“We’re excited to demonstrate our newest technology on the power and propulsion element. Solar electric propulsion is extremely efficient, making it perfect for the Gateway,” said Mike Barrett, power and propulsion element project manager at NASA’s Glenn Research Center in Cleveland. “This system requires much less propellant than traditional chemical systems, which will allow the Gateway to move more mass around the Moon, like a human landing system and large modules for living and working in orbit.”

Charged with returning to the Moon within five years, NASA’s lunar exploration plans are based on a two-phase approach: the first is focused on speed – landing on the Moon by 2024 – while the second will establish a sustained human presence on and around the Moon by 2028. We then will use what we learn on the Moon to prepare to send astronauts to Mars.

For more information about NASA’s Moon to Mars exploration plans, visit:

-end-

]]>A new study using data from NASA’s Chandra X-ray Observatory and ESA’s XMM-Newton suggests that dark energy may have varied over cosmic time, as reported in our latest press release. This artist’s illustration helps explain how astronomers tracked the effects of dark energy to about one billion years after the Big Bang by determining the distances to quasars, rapidly growing black holes that shine extremely brightly.

First discovered about 20 years ago by measuring the distances to exploded stars called supernovas, dark energy is a proposed type of force, or energy, that permeates all space and causes the expansion of the Universe to accelerate. Using this method, scientists tracked the effects of dark energy out to about 9 billion years ago.

The latest result stems from the development of a new method to determine distances to about 1,598 quasars, which allows the researchers to measure dark energy’s effects from the early Universe through to the present day. Two of the most distant quasars studied are shown in Chandra images in the insets.

The new technique uses ultraviolet (UV) and X-ray data to estimate the quasar distances. In quasars, a disk of matter around the supermassive black hole in the center of a galaxy produces UV light (shown in the illustration in blue). Some of the UV photons collide with electrons in a cloud of hot gas (shown in yellow) above and below the disk, and these collisions can boost the energy of the UV light up to X-ray energies. This interaction causes a correlation between the amounts of observed UV and X-ray radiation. This correlation depends on the luminosity of the quasar, which is the amount of radiation it produces.

Using this technique the quasars become standard candles, as implied by the artist’s illustration. Once the luminosity is known, the distance to the quasars can be calculated from the observed amount of radiation.

The researchers compiled UV data for 1,598 quasars to derive a relationship between UV and X-ray fluxes, and the distances to the quasars. They then used this information to study the expansion rate of the universe back to very early times, and found evidence that the amount of dark energy is growing with time.

Since this is a new technique, the astronomers took extra steps to show that this method gives reliable results. They showed that results from their technique match up with those from supernova measurements over the last 9 billion years, giving them confidence that their results are reliable at even earlier times. The researchers also took great care in how their quasars were selected, to minimize statistical errors and to avoid systematic errors that might depend on the distance from Earth to the object.

A paper on these results appears in Nature Astronomy on January 28, 2019, by Guido Risaliti (University of Florence, Italy) and Elisabeta Lusso (Durham University, United Kingdom). It is available online at https://arxiv.org/abs/1811.02590

NASA’s Marshall Space Flight Center in Huntsville, Alabama, manages the Chandra program for NASA’s Science Mission Directorate in Washington. The Smithsonian Astrophysical Observatory in Cambridge, Massachusetts, controls Chandra’s science and flight operations.

Image credit: NASA/CXC/Univ. of Florence/G.Risaliti & E.Lusso

**Read more from NASA’s Chandra X-ray Observatory.**

For more Chandra images, multimedia and related materials, visit:

]]>Press Trust of India | London Last Updated at May 1, 2019 18:10 IST

Last Updated at May 1, 2019 18:10 IST

Photo: Wikimedia Commons

- ALSO READUniverse’s first molecule, formed 14 billion years ago, detected in spaceCold morning in Delhi today, minimum temp at 5.2 degree C; air remains poorCold morning in Delhi: Clear sky, but minimum temperature at 5.5 degrees C‘Living fossil from early universe’: Hubble spots new galaxy near Milky WayPleasant morning in Delhi today, sky to remain clear with moderate fog: IMD

Scientists say they have removed the recent doubts cast on the presence of the elusive dark matter within galaxies, disproving the empirical relations in support of alternative theories.

Dark matter is one of the greatest enigmas of astrophysics and cosmology, and is thought to account for 90 per cent of the matter in the universe.

However, its existence has been demonstrated only indirectly, and has recently been called into question.

The study, published in the Astrophysical Journal, offers new insights into understanding the nature of dark matter and its relationship with ordinary matter.

From the expansion of the universe to the movement of stars in the galaxies, there are many phenomena that cannot be explained by the presence of baryonic matter alone.

The attractive force generated by matter is insufficient to explain observable gravitational effects.

This had led to the theory of the existence of undetectable dark matter, and the idea that galaxies are embedded in its spherical halo.

“Three years ago, a few colleagues at Case Western Reserve University strongly questioned our understanding of the universe and the in-depth work of many researchers, casting doubt on the existence of dark matter in the galaxies,” said Chiara Di Paolo, a doctoral student at International School of Advanced Studies (SISSA) in Italy.

“Analysing the rotation curves of 153 galaxies, principally the ‘classical’ spiral kind, they obtained an empirical relationship between total gravitational acceleration of the stars (observed) and the component which we would observe in the presence of only ordinary matter in the classical Newtonian theory,” Di Paolo said.

“This empirical relationship, which seemed valid in all the galaxies they analysed and at any galactic radius, motivated the explanation of gravitational acceleration without necessarily calling into question dark matter, but involving, for example, theories of modified gravity such as modified Newtonian dynamics (MOND),” she said.

Di Paolo and her collaborators wanted to verify this relationship, analysing the rotation curves of galaxies other than the classical spiral kind — 72 galaxies with low surface brightness (LSB) and 34 dwarf disc galaxies.

They produced more extended results, finding a relationship, which, besides total gravitational acceleration and its ordinary component, also involves the galactic radius and the morphology of the galaxies.

“We have studied the relationship between total acceleration and its ordinary component in 106 galaxies, obtaining different results from those that had been previously observed,” said Paolo Salucci, professor of astrophysics at SISSA and one of the research authors.

“This not only demonstrates the inexactness of the empirical relationship previously described but removes doubts on the existence of dark matter in the galaxies.

“Furthermore, the new relationship found could provide crucial information on the understanding of the nature of this indefinite component,” Salucci said.

]]>
(China Daily) 08:09, May 23, 2019

*Sun Jiadong, former chief designer of the Beidou navigation system, tries out a device combining Beidou and augmented reality technologies at an exhibition in Beijing on Wednesday. WANG ZHUANGFEI / CHINA DAILY*

Chinese researchers are developing a satellite-based monitoring system capable of tracking and reporting the location of airliners in flight in nearly real time, according to project leaders.

Sky Mirror, designed by China Electronics Technology Group Corp’s 54th Research Institute in Shijia-zhuang, capital of Hebei province, will enable air traffic controllers to avoid another tragedy similar to that involving Malaysia Airlines Flight 370 in which ground control lost the tracking of the jetliner due to suspected human factors, and will also help rescuers quickly and accurately locate airliners in distress.

With these advantages, the system will extensively improve civil aviation industry safety around the world, said Wei Haitao, deputy head of the institute’s satellite navigation operation.

MH370, which carried 239 passengers and crew, disappeared from radar screens during a flight from Kuala Lumpur to Beijing on March 8, 2014. “After the MH370 incident, the need for time-sensitive air traffic surveillance has rapidly increased,” he said.

“We were asked by our civil aviation authorities to design and construct a space-based network that can monitor and track the real-time location and flight trajectory of civil airliner.”

Currently, air traffic controllers mainly rely on ground surveillance radar to track flights, but this approach is subject to many restrictions such as the radars’ detection range and the absence of such equipment on the sea.

The improved method, involving the use of a satellite-enabled device called automatic dependent surveillance-broadcast, or ADS-B, in which a plane obtains its position via satellite navigation and periodically broadcasts the information to ground control, also requires a certain number of ground facilities along the flight route.

In contrast, Sky Mirror will consist of a number of satellites placed in low-Earth orbit and will receive signals emitted by the ADS-B from airborne airliners, no matter where they are, and then transfer the location to ground control, Wei explained.

He said new airliners to be delivered to domestic carriers in the future will all be equipped with ADS-B capable of receiving navigation and positioning signals from China’s Beidou Navigation Satellite System, and will benefit from Beidou’s high-accuracy services.

Ye Hongjun, project manager of Sky Mirror, said the system’s first satellite will soon be launched on a Long March 11 carrier rocket to perform a two-year technological demonstration. He added that eight such satellites are scheduled for launching in 2021 to form a regional system covering the Asia-Pacific region and will then offer commercial services to airlines.

China is the second nation, after the United States, that is developing and building a space-based air traffic surveillance network. The researchers made the remarks on the sidelines of the 10th China Satellite Navigation Conference which opened in Beijing on Wednesday.

]]>