Wes Miller 1 Wes Miller Mr. Lancaster SPH 4U1 Jan 31, 2023 Using Our Physics Units to Explain and Analyze the ISS Throughout the semester, each unit touched on a physics concept be it mass, velocity, or distance, that was relatable to the International Space Station. Many questions were asked and equations reviewed, and it was easy to see role they play. For the summative this year, we will review the following units: Motion, Energy and Momentum, Magnetic, Electric and Gravitational Fields, the Wave Nature of Light, and “Modern” Physics to document what was covered in each and how it relates to the ISS. The first unit we covered is motion. Motion is one of the most common physics concepts in our universe, for everything in the universe moves. As we know, the International Space Station is in a constant orbit around the earth. Knowing this, we can calculate many things, such as the velocity of the space station itself.. As long as the time it takes to cover a specific distance is known, we can calculate the velocity with v = βπ βπ‘ , ‘βd’ being the total distance and ‘βt’ being the total time. Another equation relateing to the 2 2 space station would be πΉπ = 4ππ ππ , which is the equation for centripetal force. The reason this equation relates to the ISS is that it is calculating the amount of force pulling the spacecraft into the earth from its movement in orbit. Inputting our units, m = mass of our object, where ~420000kg, r = radius, our distance from the aircraft and the center point of the earth, where ~6800000m, and lastly f = frequency of orbit, or cycles/time. So Wes Miller 2 16/86400, which is 18.5x10^-5. Putting this together we get the following result for centripetal force: 2 −5 2 ,πΉπ = 4(420000)π (6800000)(18. 5 × 10 ) 6 πΉπ = 3. 5π₯10 N. In our Momentum and Energy unit, we learned that there are many applications we can use to calculate elements relating to the International Space Station. The space station is constantly in orbit. A fun thing to know about orbit is that depending on the mass and speed of an object, it would also have to be a certain distance from the earth in order to stay in orbit. Therefore, a certain amount of energy is needed to contain the ISS in orbit. If the ISS was placed a couple of kilometres closer or farther from the earth, it would slowly stray away or start falling toward the earth, falling out of orbit. We can calculate then the required momentum to keep it in orbit with the equation π = ππ£. ‘p’ in this case is momentum, our mass of the ISS is 420000kg, and velocity is 28000km/h or 7888m/s. When substituted, π = 420000 · 7888, which equals 3.3x10^9 kg m/s. We also learned how to calculate the kinetic energy of an object in motion with the equation 2 πΈπ = 1/2(ππ£ ). Using this then in our previously used units, we can create 2 13 πΈπ = 1/2(420000ππ(7888) ), which calculates out to 1.3x10 Joules. Lastly, we learned how to calculate the gravitational energy of an object by using the following equation, πΈπ = ππβ. We do need to consider that due to the distance of the object from the earth, the gravitational force is going to be less than if it was on the surface. In this case, it is 89% of the force that we would feel on earth. Therefore, when substituted in we get πΈπ = 420000ππ(9. 81π · 89%)(408000π) Wes Miller 3 12 πΈπ = 1. 5π₯10 π. Much of this unit covers scenarios and equations around things that collide with each other. Because of that, there are some equations that I can't use accurately enough since the satellite has yet to crash into anything (which is a good thing!). A rule of thumb though is that p = p’. This means that momentum is equal throughout. Even after the collision, the total momentum between all objects that have collided will always be the same. Our third unit is our field unit. This unit contains information on how to calculate gravitational, magnetic, and energetic fields. As we know, the ISS is in a constant orbit around the earth. With this, we can calculate the gravitational potential energy. The gravitational potential energy in simple words is the amount of energy an object has based on its position in a gravitational field. Since the space station is at a very high altitude as well it is high in mass, this means that it would have very high potential energy. We can figure this out by using the πΈππ = 1 − πΊππ( π ) equation. To substitute, we use the −11 Gravitational constant G = 6.67x10 , and the mass of the gravitational object which in this 24 case is earth, M = 5.97x10 , then the mass of the ISS, m = 420000kg, and the radius of the orbit around the earth, r = 6800000m. Then we can substitute the number in to get −11 πΈππ = − 6. 67π₯10 πΈππ = − 2. 5π₯10 25 24 (5. 97π₯10 )(420000)( 1 6 6.8π₯10 ) J Waves is our fourth unit and not applicable to the ISS because it focuses more around how light moves in a wavelike pattern, how it can be manipulated to turn into different colours, or show the ways that light can be dispersed, split, interfered and Wes Miller 4 diffracted. That being said, there is a wave that is used on the space station that doesn't deal with colour but rather with sound. These are known as radio waves and astronauts use them to telecommunicate with Nasa from space. Those massive satellite antennas that can be found around the world are used to send waves into outer space to communicate with telescopes or satellites in space, as well as to look out for objects of interest floating around out there. The antennas range in size depending on the distance that they are needing to communicate from. These antennas are critical as they transmit information from space. Without them, we would not have the knowledge about space or physics that we do since the information we have wouldn’t be supported or proven. Our final unit is “Modern” physics and there we studied the speed at which light travels and how it can create illusions when viewed from different angles and distances. Examples of this are the phenomena called time dilation, length contraction, and mass increase. When viewed from the earth, the space station looks a lot longer than it actually is. This is because of how fast it is travelling and how long light takes to reach the earth to 2 our eyes. An equation that we can use for these phenomena is Ζ = 1 − π£ 2 π . With this equation, we can calculate time dilation, length contraction, and the perceived mass of the object. Starting with time dilation, we will use ππ = 8 plus c = the speed of light, 3x10 , we get ππ = ππ Ζ . When substituting previous values 1π πππππ 2 1− 7888 8 2 (3π₯10 ) ππ = 1 0.89 ππ = 1. 12 π ππππππ Wes Miller 5 This means that for every 1 second on earth, it is comparable to 1.12 seconds on the ISS. Therefore the farther you go from the earth, the quicker things age due to more exposure to space radiation. To solve for length contraction, we use the same equation however we multiply by the length rather than divide. It will instead look like this, 2 πΏπ = 108π • 1 − 7888 8 2 (3π₯10 ) πΏπ = 108(0. 98) πΏπ = 98π This shows that because of the speed the spacecraft is moving, it looks like it is 98 meters long rather than the actual 108-meter length. The final phenomenon, mass increase, can be found using the equation ππ = ππ 2 1− . 7888 8 2 (3π₯10 ) When substituting the actual mass we get ππ = 420000ππ 0.98 ππ = 471910ππ This shows that from the earth, the spacecraft's perceived mass is 470 thousand kgs when comparing to its actual mass of 420 thousand kgs. With all of this information and learning how physics is involved in everything, one can calculate the motion, energy, momentum, magnetic, gravitational fields, and the nature of light of almost anything in the entire universe. Physics is an extremely powerful tool to have, especially if you’re NASA and venture into space. Having the wrong calculation could ruin an entire mission, putting time, money, and lives at risk. Having the knowledge and power of physics in our world has allowed us to advance to where we are today. I can’t imagine where we’d be without it.
