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All parameters of this problem given in SR units_The following question examines an alien space craft traveling towards the Earth, according toanobserver on the Ear...

Question

All parameters of this problem given in SR units_The following question examines an alien space craft traveling towards the Earth, according toanobserver on the Earth_ Forthe purposes ofthis problem, you may assume thatthe Earth is in an inertial reference frame which is at rest i.e_ the Home frame You may assume the satellite is also at rest in the Earth's rest frame From the alien's inertial reference frame, i.e: the Other Frame; they measure the Earth to travel distance 0f L = %Is i

All parameters of this problem given in SR units_ The following question examines an alien space craft traveling towards the Earth, according toanobserver on the Earth_ Forthe purposes ofthis problem, you may assume thatthe Earth is in an inertial reference frame which is at rest i.e_ the Home frame You may assume the satellite is also at rest in the Earth's rest frame From the alien's inertial reference frame, i.e: the Other Frame; they measure the Earth to travel distance 0f L = %Is in a time of Al' 5 Is after the satellite passes them. The alien spacecraft measures velocity for the Earth and satellite of & J1 Actb 0istv+ } 0apt = ttvt Sata Il Vata F0 (a) (5 points) According to the Alien, how far did they travel from the time the satellite passed them to the time the Earth arrived at them, |Ad'|? Briefly explain your answer: (b) (3 points) What velocity does the Earth observer measure for the Alien's spaceship, 0? (c) (12 points) What time interval does the Earth observer measure for the Alien's travel towards the Earth, 4t? What distance does the Earth observer measure for the Alien's displacement magnitude_ Adj? How could the Earth observer calculate how much time elapsed in the Alien's frame from measurements in their own frame? (d) (5 points) What drove Einstein to develop his theory of special relativity? What are the two fundamental postulates of Einstein's theory of special relativity?



Answers

An alien spaceship traveling at 0.600c toward the Earth launches a landing craft with an advance guard of purchasing agents and environmental educators. The landing craft travels in the same direction with a speed of 0.800c relative to the mother ship. As observed on the Earth, the spaceship is 0.200 1y from the Earth when the landing craft is launched. (a) What speed do the Earth-based observers measure for the approaching landing craft? (b) What is the distance to the Earth at the moment of the landing craft’s launch as observed by the aliens? (c) What travel time is required for the landing craft to reach the Earth as observed by the aliens on the mother ship? (d) If the landing craft has a mass of $4.00 \times 10^{5} \mathrm{kg}$ , what is its kinetic energy as observed in the Earth reference frame?

Hi, everyone. This is the problem. Based on general theory off relativity Here it is given on Earth. Satellite used in the global Positioning System moves in a circular orbit with time period off. 11 hour, 58 minutes. We have to calculate in the first part radius of the orbit, we have to calculate its speed. Right? If frequency off oscillator in the satellite is 1500 75.42 megahertz, what is the fractional changing? Free considered toe then fractional change in frequency Do toe time relation and gravitational Belousov toe the frequency. He overall change in frequency. We have to find overall changing frequency. Let us start solving it first part gravitational pull on the satellite Do toe earth will provide necessary centripetal force required to revolve in the orbit so it can be written as g m e are upon We sweat on Dhere. Speed off the satellite is two pi r variety. So by using a question Bannon toe we can write so are can be return. It's G m e the square by full pi squared Hold the power one by three Substituting the value G is 6.67 March of the art, 5.98 10 to the power 24 time period is in seconds. We have to use 43,000 square upon four by square Hold the power one by three. So on solving it you will get the orbital radius. Toby 2.66 and toe 10 to the power. 7 m. Okay. Be part the orbital speed off the satellite by our upon t to fight are is 2.66 10 to the power seven upon 43,000. You know a zero seconds. So it will be 3.87 and tow 10 with the power 3 m per second. See part. The small fractional decrease in frequency received is equal in magnitude to fractional. Increasing period up. Moving oscillated. Do toe time delay and it's available B minus gamma minus one one minus B squared. The square means are bitterly speed. So we will get minus 834 in tow, 10 to the power minus 11. Mhm deeper. The orbit altitude is a large compared toa radios off earth. So gravitational potential energy G m e m. Upon art and changing gravitational potential energy we confined 6.67 10 to the power 11. 5.98 in tow, 10 to the Power 24 and to mask Opta Satellite upon 2.66 and toe 10 to the power seven minus minus plus 6.67 10 to the power 11. 5.98 10 to the power 24 into mass upon 6.37 and to turn to the about 6 m. So it will be 4.76 10 to the power seven Jewell per k G and changing frequency will be fractional change in frequency due to gravitational effect. Bill B did Yuji upon I am into C squared so 4.76 10 to the power seven upon and three into 10 to the power eight Holy square. So it will be 5.29 in tow, 10 to the power minus 10. So effective change you can find minus 8.34 5.29 10 to the power minus 10. It will be 4.46 and tow 10 to the Power Minister. That's all for this problem. Thanks for watching it

Well, we lost your lander with respect to the earth is equal to We lost your lander with respect to space shitless the lot steals a spaceship with respect to the earth divided by one plus the lost you of lander. With respect to SpaceIy less sorry, it's multiplied by it's multiply by we lost of space. You've been respect to the Earth Divided by C Square. Let's plug into values. Um, 0.800 times the speed of light less are 0.600 times the speed of light, divided by one less. Uh, Zito find hate little zero times the speed of light multiplied by 0.600 times the speed of light divided by C square. And therefore we lost your lander with respect to the earth is zero point 946 times the speed of light. Oh, this was a party. Now let's to part the well, we know it contracted land l A sickle to lp multiply by square Rudolph one minus B squared, divided by C Square. Let's plug in the values with proper length in this case is properly and is 0.200 light years into square root off one minus Zito find 600 hold square. And therefore ah, contract id Land physical to 0.16 theater light years. Okay. No, it's two parts. See? Well, we know it d He is equal to how divided by we the time is distance divided by speed and speed. In this case is a speed off lander with respected e space shitless speed off space, You with respect to the earth and putting the values we have 0.160 lying tears divided by aiding both lost easily him 1.4 words in the times the speed of light. And this is equal to see the point 114 Uh, is there a 0.114 ah, years years. All right. And finally you need to do part D. Well, we know that kinetic energy, uh, kinetic energy is equal to gamma minus one times m c square and we have gone Our bitches one divided by square hood off one minus. We squared, divided by C squared minus one to M C square. No, let's plug in the values. We have one divided by square root off one minus zero point mine for six. We'll scare, um minus one multiplied by. Well, I am in this case is for multiply by 10 to the power five and speed off. Mind his three year old to buy me 10 to the power it holds square and therefore kinetic energy is equal to 7.5 run multiply by 10 to the power 22 chills.

So here in this problem, we are given that speed of the alien ship. With respect to the Earth, we have given us 0.6 c and the speed of the Landau. With respect to the mothership, we are given us you x dash us 0.8 c. Now we'll be taking the earth as R s free and the alien ship as R s dash frame. And there are four parts in the question will be solving the apart first. Now, in a part, we have to calculate the speed of the lender with respect to the earth. So let it be you, x, And by using the Lorentz transformation equation for velocity the speed of Landau with respect to Earth, you x will be your ex prime or you X dash plus V divide by one plus you X dash V over C squared. Now we have the village. You can plug it here, you x dash. We have 0.8 c plus. We we have 0.6 c divided by one plus you x dash. We have 0.8 c. Times V is 0.6 c. Divide by C square. No goodness all this and we'll get the answer. You excess zero point 94 60. So this dancer for the a part next will be solving the B part. Now, in the B part, we have to calculate additions to the oath. At the moment the landing craft was launched as measured by the aliens. So and we are given that the distance between the lender and the Earth, as observed by the earth, uh, is LP were given us 0.2 little So l is equal to l P times underwrote one minus the square over C square, so it would be equal to L. P. We have 0.2 light years. Times underwrote one minus. The square means 0.6 square, 0.6 c square. Sorry. Divide by C Square as it is. No goodness, All this and we'll get our answer US 0.16 light here. So there's the answer for the B part now. Next will be solving, though. See part now in the C part. We have to calculate the time travel or the time required for the landing craft to reach the Earth as measured by the aliens so basically we have to calculate so t would be called to distance divided by the speed of the lender. Not just the distance is 0.16 light here, 0.16 light here that was major by the aliens divided by the speed and speed would be 0.8 C plus 0.6 c Not going to solve this, and we'll get our answer us 0.11 for you. So this answer for the C part next it will be solving the deep part now in the report were to calculate the kinetic energy as measured in the Earth's reference frame, and the expression for it is given by Canada. Calls to one over on the Route one minus U X square over C square minus one times M C squared. Now we have all the valleys. We can plug it here, and this is the mosque of the Lindo, which we are given us MM recalls to four times 10 to the power five kg. This is given the question. Now I can put the values so one over underwrote one minus U x squared that we have calculate earlier as 0.946 See whole square divided by C square minus one times. And we have four times 10 to the power five and the value of C. We have three times 10 to the power eight media per second. This is the speed of light. I'm going to simplify this and we'll get our answer as 7.5 times 10 to the power 22 June. So this is required Canada Energy. That was asking the question. So I hope you have understood the problem. Thank you.

Okay, so we are moving at a velocity of 0.95 C. We're going a distance of 4.5 light years away. We need to find how long it will take for earth clock first. So one light year is equal to 2.998, 10 to the eight m/s to the speed of light Times How many seconds in a year? Which is 3.16 10 to the 7th second. So this is 9.47 10 to the 15 m. That makes 4.5 light years, B 4.3 10 to the 16th meters. So for part A the change in time on Earth is the distance over the velocity. We just found the distance. We know the velocity. So this is 1.510 to the eighth seconds. So part B for the spacecraft, the time is dilated. So the change in time on the craft is equal to the change in time on Earth. I'm so scared of one minus V squared over C squared Plugging in values, and we give 4.7 10 to the 7th 2nd. So part C. For the distance you have length contraction. So the distance of the craft sees is the distance of the earth times the square root of one minus the square Overseas square Plugging in values. And we get 1.310-16 meters and part B. Their relative speed is the distance of the craft. Over the change in time, Just 2.810 to the 8th meters per second, Which is .95 c. So they will measure the same relative speed.


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