Challenges cover a variety of topics
Algebra is necessary to solve this challenge.
We left an unsolved system of equations in Fascinating Frequencies 2. In this challenge, you must solve it.
\vec v=\frac{\vec d\pi}{72000s}\\[8pt] \frac{4\pi\vec v\vec d}{sm}=\frac{Gm_1}{\vec d^2}\\[8pt] G=\frac{6.67408m^3}{10^{11}s^2kg}\\[8pt]The challenge is to find the instantaneous acceleration of a point on a sphere when it touches the larger sphere it is rolling inside of. Before finding the acceleration of a point on the circumference of the circle, let’s find that of the centre. The radius of the larger circle is 1 metre. The centre of the small circle is not on the circumference though. It is 25cm away because the small circle has radius 25cm. This means the centre of the smaller centre is 100-75 = 25 centimetres away from that of the larger one. The smaller circle completes a rotation whose circumference is 2π*75cm = 150πcm every 4 seconds. It’s velocity is thus 375π millimetres per second sideways. This velocity changes, and it’s vector’s endpoint will naturally also complete a full circular rotation each second. This circle has radius 1.5π m/s and thus circumference 2π*375π mm/s = 75π2 centimetres per second inward. To complete this circle every second, the acceleration must be 75π2 cm/s2. The same rule applies to smaller circle. The acceleration equals 4π2r/s2 inward times the rate of rotation. For the smaller circle, this gives us 1π2 m/s2 toward it’s centre. When the point touches the rim, these vectors will align, and the point will feel their cumulative acceleration. It’s counterintuitive that the point experiences the most force when it stops, but it’s true. Add 1π2 m/s2 inward and 75π2 cm/s2 inward together to get our answer, 175π2 cm/s2 inward.
Before trying to find the apoapsis of a geostationary orbit, first remember what a geostationary orbit and apoapsis are. Can a path always at the same height have a highest point? No. They are all equal! The speed necessary to complete a revolution every 20 hours is dπ/ 20 per hour, where d is the distance from the object. The orbital speed necessary at any point for a geostationary orbit is that at which centrifugal force equals gravity. Centrifugal force’s acceleration equals 4π2dv/sm outward. And that of gravity equals Gm/d2 inward. We are left with a system of equations that has too many units. We will solve it in Ubiquitous Units.
\vec v=\frac{\vec d\pi}{72000s}\\[8pt] \frac{4\pi\vec v\vec d}{sm}=\frac{Gm_1}{\vec d^2}\\[8pt] G=\frac{6.67408m^3}{10^{11}s^2kg}\\[8pt]In this challenge, you will need to use geometry, calculus, algebra, and angle theorems.
A roulette is a shape created by rolling things. If you roll circles, you have a trochoid. If you draw the path of a point on the circumference, you have a cycloid. And if you roll that circle inside another circle, you have a hypocycloid. Simple enough. Say you have a circle with radius 25 centimetres and a circle with radius 1 metre. When you rotate the smaller circle inside the bigger one, the path traveled by a point on the circumference is a hypocycloid. It will touch the bigger circle 4 times every rotation. At these points, the instantaneous velocity of the point is zero. The challenge is to find the instantaneous acceleration if the small circle completes 1 revolution inside the bigger one each second. This is different from rolling completely over the inside, which it will do every 4.
This challenge involves orbits.
A geostationary orbit is one where the satellite stays in the same position relative to the ground. This works because the earth spins; the velocity that keeps it over the same spot increases while the velocity needed to orbit decreases. At some point, they intersect. All geostationary orbits must be at the equator, and have the same altitude althroughout. If you wanted to get a satellite into geostationary orbit around a planet with mass of 5,000 yottagrams that completes 1 revolution every 20 hours, how fast would your orbital speed be at apoapsis? The apoapsis is the highest point in an orbit.
We wanted to find the frequency of a gamma radiation particle. We know that if it had undergone beta decay, it would have released the same amount of energy, but by releasing 2 beta radiation particles, both traveling at half of c. The equation to find the energy of these particles is shown below, where m0 is the particle’s rest mass energy, ≈0.511 meV. It gives ≈8.5 microjoules of energy per particle, or ≈17 microjoules in all. We then use the second equation to find the frequency of a photon with ≈17 microjoules of energy. Our answer is thus that the frequency of the photon is ≈2.566×1024 rotations per second.
E\approx\sqrt{\frac{(m_0\times0.5c)^2}{1-\frac{v^2}{c^2}}c^2+(m_0c^2)^2}\approx 8.5\times10^{-6}J\\[32pt] r=\frac Eh\approx\frac{17\times10^{-6}J}{6.626\times10^{-34}Js}\approx\frac{2.566\times10^{28}}sWe are trying to find the value of the expression below. Note than the tan stuff approaches infinity as k increases. This is because 360/k approaches 0, and so 90- 360/k approaches 90. And as x approaches 90, tan(x) approaches infinity. So the reciprocal of the tan stuff gets smaller and smaller. So the side lengths approach zero while the number of sides increases. The shapes with more sides approximate circles. Their circumference equals the number of sides times the length of each side. K approaches 10 times more than the tan stuff. And so this will reach infinity like Exquisite Equations did.
\prod_{k=1}^\infin\Bigg(A\bigg(k,\,\frac1{tan (90\degree-\frac{360\degree}k)}\bigg)\Bigg)\\[16pt]This challenge is the first of a new science challenge series, Fascinating Frequencies. You will need to use physics to solve this challenge.
Some radioactive particles are decaying. They are decaying in different ways, but each type of decay releases the same amount of energy. If the particle undergoes beta decay, it will release an electron and positron, both traveling at half the speed of light in a vacuum. If it undergoes gamma decay, it will release a photon. The challenge is to find the frequency of this photon.
Trigonometry, geometry, ad algebra are required to solve todays challenge.
Pi notation is to Sigma notation as plus is to minus. Sigma notation is the sum of a bunch of stuff, Pi is the product. Pi is also ≈3.1416, but we will note that as π. In Strange Sums, we evaluated an expression with 2 Sigmas. Here will will find the product of the outputs of an equation. Specifically, the equation for the area of a regular polygon. Better go find that one out right now if you don’t know it. We will express this function as A(n, s), where n is the number of sides and s the length of them. Now solve the expression below.
\prod_{k=1}^\infin\Bigg(A\bigg(k,\,\frac1{tan (90\degree)-\frac{360\degree}k}\bigg)\Bigg)We want to find the solution to the Exquisite Equation. However, many of you may notice that there are 3 variables and only 2 equations. This means that x can be any amount of protons! Tough luck, you solve the equation so can’t do your experiment. Well, you could build a new LHC in fifteen minutes.
x=6\,000\,000+E^+\frac{2-e^{\frac{-1}{N^+}}-e^\frac{-3}{N^+}}{N^+}-3E^+-1\\[8pt] \frac{E^+-E^+e^\frac{-3}{N^+}}{N^+e^\frac1{N^+}}= 1\,000\,000\\[32pt] x\in\Bbb Z