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]Tag: solution
Solution to Fascinating Frequencies 1
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}}sSolution to Peculiar Products
We 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]Solution to Exquisite Equation
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 ZSolution to Conounding Crystals
Let’s find the angles of the triclinic crystal! A triclinic crystal has all sides and angles different. The triclinic crystal we’re dealing with is a prism. All of its sides are parallelograms. With faces with areas 30cm2, 35cm2, and 42cm2, it could be orthorhombic, with all right angles. That would give it sides 5cm, 6cm, and 7cm, but we know that those aren’t the side lengths. It could also have 1 right angle, and then it could have sides 3cm, 11cm, and 14cm. 3*10 = 33, and 3*14 = 42. And 11*14 = 154. 90 degree angles give the biggest area, so 2 bigger areas are fine. If you didn’t know about triclinic crystals, you’d probably guess that. It’s the obvious solution. But for no angles to be 90, all the products have to be higher than necessary. The 33 must go to 30. The 42 can’t go to 42, the 154 must. And so the 42 goes to 35. 33>30, 42>35, and 154>42. No problem. The portion of the area is the sin of the angle. Simple trig. So our answers are arcsin(30/33), arcsin(35/42), and arcsin(42/154), which gives us 65.38°, 56.44°, and 15.83°.
Solution to Bouncy Ball
The new instantaneous velocity of the ball becomes the negative square root. This bouncy ball defies physics, but we can still calculate it’s velocity after 100 jumps. Most importantly, The ball’s velocity at the end of a bounce is -1 times it’s velocity at the beginning of it. So the velocity is square-rooted every bounce. If a non bouncy object is dropped from a height of A metres, and it’s terminal velocity is B metres per second, then B = √(2gA)/2. g is the force of gravity, 9.807 metres per second squared. The initial height, A, is 1 metre. 2gA is 19.614 m2/s2. Thus B is 4.429 m/s. Square-rooting 99 times is the equivalent of raising it to the power of 2-99. We get approximately 1 metre per second upwards. It’s slightly more, but very little.
Solution to Strange Sums
Let’s solve an equation! The sum of a decreasing geometric series is described by the second equation. If you would like to know why, see Peculiar Pets. Knowing this, the full equation is clearly the harmonic series, 1/1 + 1/2 + 1/3 + … . The harmonic series adds to infinity, our answer.
\sum_{k=2}^{\infin}\Bigg(\sum_{l=0}^{\infin} \Big(\frac1{k^l}\Big)\Bigg)\\[8pt] \sum_{l=0}^{\infin}\Big(\frac1{k^l}\Big)= \frac1{k-1}\\[8pt] \sum_{k=1}^{\infin}\Big(\frac1{k}\Big)= \infin\\[8pt]Solution to Perplexing Protons
We want to find the number of protons in the particle collider. We know that 1 Kj of energy was released. 1 electron capture releases 511 electron volts. There are ≈6.242 * 1021 electron volts in a kilojoule. Thus there were 1.221 * 1019 reactions. The rate of reactions is proportional to the number of electrons. Let the number of neutrons be N and the electrons E. The increase in electrons is E+, and that of neutrons, N+. E is increased linearly and decreased proportionally to itself. The distance from equilibrium will decrease by a factor of e in every length of time over which it would become zero if it were only being decreased, and at a constant rate equivalent to the instantaneous force of decrease at t=0. We know this from Battle of the Blobs and Spinning Stick. This duration of time equals 1/n. The equilibrium is at nE=E+, so the number of electrons will approach E+/N+. E, at any time, equals E+/N+ * (1-e-t/N+). The number of neutrons is E+t minus all that stuff. We make the equations below for t=1, t=3, t=3.25 and t=3.5 to find our answer. The final equation to get E+ and N+ is rather difficult, and it is necessary to find them to solve the problem. This equation will be a challenge of it’s own, Exquisite Equation, coming next week.
P_{t=1}=5\,000\,000+\frac{E^+}{N^+} (1-e^{\frac{-1}{N^+}})\\[8pt] E_{t=3}=\frac{E^+}{N^+}1-e^{\frac{-3}{N^+}} \\[8pt] E_{t=3.25}=\frac{E_{t=3}}{e^\frac1{N^+}} =1\,000\,000\\[8pt] P_{t=3.5}=P_{t=1}+\frac {E^+}{N^+}(1-e^{\frac {-3}{N^+}})+E_{t=3.25}-3E^+-1\\[8pt] \frac{E^+-E^+e^\frac{-3}{N^+}} {N^+e^\frac1{N^+}}=1\,000\,000Solution to Complex Coins
We wanted to find the probability that you beat John at your 3 sided coin game. John has 3 points and you have only one. In 2 more flips, John has a 4/9 probability of winning. In any given roll, either round or tails gives him a point. There is a 1/9 probability you will tie it up. The chances are then fifty fifty. The remaining 4/9 have John ending at 4 points. 2 of them give you 1 point, 2 give you two. When the score is 4-2, there is a 1/9 chance of getting 2 heads in a row to tie is up. With 4-3, it’s 1/3. When it’s tied up, you have a 50% chance of winning. 50% of a ninth of 2/9 is 1/81 or 2/162. 50% of a third of 2/9 is 1/27 or 6/162. And 50% of 1/9 is 1/18 or 9/162. (2+6+9) / 162 = 17/162. Not very likely at all.
Table \space of \space Probabilities\\[16pt] \begin{matrix} Points&0&1&2&3&4&5\\[8pt] 0&1\\[4pt] 1&\frac12&\frac12\\[4pt] 2&\frac16&\frac23&\frac16\\[4pt] 3&\frac1{18}&\frac49&\frac49&\frac1{18}\\[4pt] 4&\frac1{54}&\frac29&\frac{14}{27}&\frac29& \frac1{54}\\[4pt] &&&… \end{matrix}Solution to Vile Voting
We wanted to find the standings and satisfaction of Brutus and Craw’s vote. Brutus will vote Red, Orange, Yellow, Purple, Blue, Green. Then Craw moves his extremes to the middle to vote Blue, Purple, Green, Red, Yellow, Orange. Now Brutus moves his extremes to the middle to get Orange, Yellow, Red, Green, Purple, Blue. It will never end! Serves them right.
Table \space of \space votes\\[8pt] \begin{matrix} &\#1&\#2&\#3&\#4&\#5&\#6\\ B_1&R&O&Y&P&B&G\\ C_1&B&G&P&R&Y&O\\ B_2&Y&R&O&G&P&B\\ C_1&G&P&B&O&R&Y\\ \end{matrix}