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An rocket flies to the Moon (a distance of 300,000 km) at aspeed of 1500 m/s.a) How much time does the one-way trip take in the Earth’s frameof reference?b) ...

Question

An rocket flies to the Moon (a distance of 300,000 km) at aspeed of 1500 m/s.a) How much time does the one-way trip take in the Earth’s frameof reference?b) What β? What is γ? (Yourcalculator may show “1” for γ, so you will have to use the binomialapproximation to find γ.) c) Will the time for the trip measured in the rocket’s frame beslightly more or slightly less than the time measured in the Earthframe?d) What is the difference (in seconds) between the two timemeasurements (Earthâ

An rocket flies to the Moon (a distance of 300,000 km) at a speed of 1500 m/s. a) How much time does the one-way trip take in the Earth’s frame of reference? b) What β? What is γ? (Your calculator may show “1” for γ, so you will have to use the binomial approximation to find γ.) c) Will the time for the trip measured in the rocket’s frame be slightly more or slightly less than the time measured in the Earth frame? d) What is the difference (in seconds) between the two time measurements (Earth’s frame vs rocket’s frame)?



Answers

A spacecraft travels along a straight line from Earth to the Moon, a distance of $3.84 \cdot 10^{8} \mathrm{~m}$. Its speed measured on Earth is $0.50 c$ a) How long does the trip take, according to a clock on Earth? b) How long does the trip take, according to a clock on the spacecraft? c) Determine the distance between Earth and the Moon if it were measured by a person on the spacecraft.

Perform the following conversions for a we're told that the distance from the earth to the moon is 200 and 40,000 miles. What is this distance in meters? You know that one mile, It's equivalent to 1.60 93 kilometers, But that one km problem too. 1000 m. This gives us 3 9 Times 10 to the 8th m for B. If this fall controls at 350 km/h and it could fly to the moon at this speed, how many seconds would it take? So let's convert 350 km/h in two m/s. Mhm. one km to go into 1000 m in one hour, 60 minutes, one minute. Yeah, 60 seconds. Mhm. This gives us 90 97 m/s, Rounded to two significant figures. Yeah. Mhm. And we know that our distance here Is equal to 3.9 times 10- eight. So we know that distance is equal the velocity times the time rearranging here, at the time. Equal distance over velocity. So the distance is 39 times 10 to the eight Years over 97 m/s. This would give us a time of 4.0 times 10 of a six seconds. For part C. How long does it take for light to travel from the Earth to the moon and back again? So. Mhm. Yeah, if we look at the a distance here um from the Earth seven moon and then back again, you're So earth to the moon is 3.9 times 10 to the eight. Yeah, Muse. And back again is 3, 9 times 10-8 m. So combining these two together here, this gives us 7.8 Yeah, Times 10- eight m. And how long would it take here? So, these are formula again, time is equal the distance over velocity. 7.8 times 10 to the eight m. And the speed here is 3.00 Times 10 G eight m/s. Okay, this gives us 2.6 seconds. And the earth travels around the sun At an average speed of 29 700 In 83 km/s, convert this two mph, Starting at 29783 kilometers per second. First convert. Mhm. Uh 1.6 old, 93 km in one mile. There are 60 seconds in one minute And there are 60 minutes in one hour and this is equal to 6 7 times, tend to be four miles her. Our Mhm. Double check my mouth on this one here. Yeah. Mhm. Is this is 29, 7 83. This will be incorrect. 29 783, 1.6093, 60 times 66.7 Terms tend to be fourth. Now, that is correct, 6.7 times 10 to the 4th MPH.

Here we are Fastly calculating a distance in meters, So we have to fall 00000 males multiplied by a conversion factor 1.6093 kilometer divided by one male, Multiplied by 1000 m divided by one km. This is equal to about 3.9 times 10 to the eight m. So the next example we can take a look at, we have the speed in meters per second, so that is equal to 3 50 kilometers divided by one hour, multiplied by 1000 m, divided by one kilometer, Multiplied by one hour, divided by 60 minutes. And then that is multiplied by one minute, Divided by 60 seconds. So what we get approximately out of this is 97 m/s. Here, I've just rounded it two significant figures next week and calculate the time in seconds to travel to the moon. So we have time in seconds is equal to distance divided by speed. So we have 3.9 Times 10 to the 8th. Try that by 97 m/s. What we get is 4.0 times 10 to the six seconds. So in that in past, see we have the time and seconds. Lots of distance divided by the speed. We have 3.9 times 10 to the eight m, divided by 3.0 times 10 to the eighth m per second, that's equal to 1.3 seconds in the last part we have here. So the speed in MPH is 29 783 kilometers, divide that by one second, multiplied by one mile, divide that by 1.6093 kilometer, Multiplied by 60 seconds, divided by one minute, Multiplied by 60 minutes Divided by one hour. So what we get here is about 6662 four mph.

This problem. We're gonna talk about relativistic philosophy transformation. So consider that we have an ass reference ring. And in this reference rain, we have an object that's moving to the left with a speed V. Now, consider that we also have a second reference frame as prime. Yeah, that is moving to the right. So in the opposite direction as the first object with the speed capital. And our goal is to find what is the speed of the black object, the one that has speed little V relative to the S. Brian. Reference from in special activity tells us that if the prime that is the velocity relative to the s prime reference tree is equal to V plus capital V divided by one plus V Capital V Oversee square. And in our problem, we have two rockets. They're both traveling with the speed of 0.75 times the speed of light, and they're traveling in opposite directions. So they are approaching each other. And this velocity that I just wrote is relative to the earth. And our goal is to find what is the speed of the rockets relative to one another s 01 of the rockets. I'm gonna call, uh, I'm going to say that it's addressed relative to the s prime reference from such that the prime capital V is equal to 0.75 times the speed of light. What I'm doing here is to consider that the reference frame is the Earth reference ring. And the s prime reference ring is the rockets referenced ring. Okay, so visa, which is your 0.75 seats on and little V is also 0.75 seat so v prime. That's what we want to calculate is equal to V now a couple of V plus V divided by one plus V Capital view over C squared. So this is 1.5 c, divided by one plus 0.75 squared, and this is equal to 0.96 seat. So this is the speed off one rocket relative to the other

Everyone. What we have here is a rocket that is trying to learn land on the surface of the planet. Well, we're asked us all for us to determine the flight speed. It must have a prime so that landing occurs at B, and we're also has to determine the time it takes the rock. It'll land from a prime to be. We have to assume that the mass of the planet is 0.6 times of the earth. And the way that we're gonna be going about solving this problem is using our knowledge of orbital dynamics and using the relationships that we know for values at perigee and apogee and also using the concepts of conservation of specific momentum and using that quantity to determine the period of our elliptical orbit. So let's go ahead and get started. What we can first do is use the relationship between apogee and perigee on, say, that are a is equal to r p, divided by two g m divided by R. P. V. P. Square minus one. We can solve this equation for V, P and C. The B P is equal two times the square root or square root of two g m r a. All over our p times of quantity, our people. So what we can do now is determine this velocity at perigee, which is gonna be our velocity be that which the rocket has to have landed beat. So we can say that AARP is going to be equal to six times 10 to the sixth and are a is equal to 100 times 10 to the fixed. And these values just closed my figure and that v P is equal to be You're talking thes values of r, p and R A into our expression for V p. We see the V P, which is equal to B B, is equal to 8674.17 meters per second. Next, what we can do is we can use the concept of conservation of specific momentum each and say that R P B B is equal to R A B A crime and we can solve for V a prime and we see that va prime is equal 521 m per second. So here is the velocity of the crime that must be attained to ensure landing at point B next were also asked to solve for the time required to travel from time to be. And we can first do that by saw them for the period of that. Corbett. We know that this is equal to T, which is pi over h time. The quantity R p plus r a times the square root of AARP times are this gives us a total period of 156.7 times 10 to the third seconds. Now, the time travel time required to travel from a time to be is simply going to be half this value so key a prime to be will be equal toe period over two, which is equal the 21.8 hours. And that is our transit time. That's a final answer.


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