special relativity homework?

Question

First question:
The starship Enterprise leaves Starbase Q at 9:00 in the morning, according to clocks both on the ship and on the station. It then moves away from the station at a constant velocity of 0.59 c.

At 10:00 AM, the communications office on the Starbase sends a radio message to the Enterprise: "What time do your clocks read right now?" As soon as the Enterprise receives the message, they send a reply back.


What is the time in the message?

Second question
Jane gets in her spaceship on her twenty-fifth birthday and flies to the star Vega at a constant velocity v. She is just turning age 32 when she reaches the Vega system.

What is the speed of her ship, as a fraction of the speed of light c?

Third question
Cosmic rays are particles like protons and helium nuclei which fly through space at very, very high speeds. Occasionally, one of these cosmic particles will smash into an atom in the upper atmosphere of the Earth. When that happens, the collision creates many new particles which fly downwards towards the ground at very high speeds. Physicists use particle detectors on the ground to detect "showers" of these secondary particles.

One day, Professor Smith talks to his colleague, Professor Jones, about his research. "Our ground-based detectors find a lot of Bluto particles," he says, "and that is a puzzle."

"Why is that?" asks Professor Jones.

"The Bluto particle is unstable. It decays into other particles with a half-life of only t = 0.7 microseconds. That means that, even if it were travelling at the speed of light, it would only be able to travel

___________________meters

before disappearing. However," continues Professor Smith, "we find plenty of these Bluto particles at the ground level. We believe that they are created in the upper atmosphere at an altitude of H = 39.2 km above the ground. How in the world can they reach the ground before decaying?"

"Ah", answers Professor Jones, "have you taken into account the time dilation factor? If the Bluto particles are created with a high enough initial speed, they could reach the ground during their brief lifetime."

How fast must the particles be moving to reach the ground? Express your answer as a fraction of the speed of light. You must be correct to the sixth digit in order to be marked correct.

_______________times the speed of light

Answer 1

T = t/√[1-(v^2/c^2)]

T= time "difference"

t= time

v= velocity

c= speed of light



Q1) T=1/√[1-(59^2/100^2) = 1/√(1-.3481) = 1/√(.6519) = 1/.8074032 = 1.238

1.238*T(time traveled, 1 hr) = 74.28(that means there is a 14 min 16 second difference in their times)

message sent at 10:00am, message received at 9:45 and 44 seconds if you do not include the time it takes for the message to be sent.



i have to go to work sorry i can onl answer one

Answer 2

Too much, I'll do one.



"E leaves at 9:00 in the morning, It then moves away from the station at a constant velocity of 0.59 c. At 10:00 AM, the base sends a radio message to the Enterprise: "What time do your clocks read right now?" As soon as the Enterprise receives the message, they send a reply back. what is the reply?"



After one hour, E has traveled 0.59 light hour. Now we need to calculate the time it takes the message to overtake them. Overtake speed is 1–0.59C = 0.41C. Time to overtake is 0.59 / 0.41 = 1.439 hours. (note below)



At 0.59C, γ = 1/√(1–V²/c²) = 1.239. So while the earth clocks advanced 1.439 hours, the E's clocks advanced by a smaller amount, 1.439 / 1.239 = 1.1619 hours or 1 hour 9 min 43 sec



so the message they sent back was 10:09 AM and 43 sec ⬅



More detail on overtake speed.

distance taken by message = distance taken by E

distances in light hours, time in hours



(speed of message)(overtake time) = 0.59 light hours + (speed of ship)(overtake time)



(1 LH/H)(t) = 0.59 + (0.59 LH/H)(t)

t = 0.59 + 0.59t

0.41t = 0.59

t = 1.439



________________________



edit, seeing the number of wrong answers, I'll do another one



Jane gets in her spaceship on her twenty-fifth birthday and flies to the star Vega at a constant velocity v. She is just turning age 32 when she reaches the Vega system. What is the speed of her ship, as a fraction of the speed of light c?



So she has traveled for 7 years, her time.

Vega is at a distance of 25.3 LY



so γ = 1/√(1–V²/c²) = 25.3/7 = 3.61

(1–V²/c²) = 0.07655

V/c = 0.961





.

Answer 3

It is too many questions!

Anyway, I'll answer them.

1)

The velocity is 0.59c. (we are assuming the ship has accelerated to 0.59c in no time and so the velocity is constant.



In 1 hr, Enterprise covered a distance of 3600*0.59*c (here, c is the distance covered by light in 1 s)

Distance = 2124c

For estimating the time taken by the radio signal sent by Starbase(sent at 10:00) hrs to reach Enterprise, we need to use the speed of radio wave relative to Enterprise.

Speed of radio wave relative to Enterprise = c -0.59c = .41c

The time taken for the signal to reach Enterprise = 2124c/.41c = 5180.487805 s

OR 1 hr, 26 min 20.49s

The time on the message is 11:26:20.49 (11hrs, 26 min 20.49s in the morning)



2)

Star Vega (OR α Lyrae) is 25 light years from Earth.

As Jane turned from 25 to 32 (7 years) to cover a distance of 25 light years, the velocity as a fraction of speed of light = 25/7 OR 3.571c (ANSWER)



3)



Let us consider the speed of light as 299792.5 km/s

Bluto particles disintegrate in 0.7 micro seconds

If they were traveling at speed of light, they would travel 299792500*0.7e-6 meters before disintegrating to other particles

OR 209.85475 meters before they disintegrate (here, half life business is ignored. The life of the Bluto particle is considered as 0.7 microseconds. Period)

Distance from ground where the particles are created = 39.2 km = 39200 m

If v is the velocity of Bluto particles, so that they can reach the ground,

v = 39200/209.85475 = 186.795867 c , where 'c' is the speed of light. (ANSWER)



Cheers

Answer 4

An observer A who can bear in mind himself at leisure (that means that he's at leisure or in uniform movement, which Einstein confirmed to be the identical factor), and who's watching a a relocating approach which comprises observer B, will bear in mind time to move extra slowly within the relocating approach than it does for observer A. If the relocating approach is relocating uniformly, then observer B is entitled to bear in mind himself to be at leisure, and A to be relocating, and can detect time to be passing extra slowly for observer A than for B. Each considers time to move extra slowly for the opposite one. As lengthy as they hold relocating uniformly there is not any paradox, given that they on no account once more meet. If they're introduced again in combination, it is given that one or the opposite, or probably each, has been subjected to acceleration to difference his path of journey. The standard idea of relativity, which in contrast to the exact idea offers with acceleration and gravity, predicts that if A and B are introduced in combination once more, whichever underwent extra acceleration (I'm oversimplifying, however that is the standard notion could have skilled much less time to move. The so much popular illustration is the "dual paradox." Take 2 same twins (or 2 same stopwatches), go away one on Earth and ship the opposite to Alpha Centauri and again at close to the velocity of sunshine, and evaluate them. The paradox is if each and every is entitled to bear in mind time to run extra slowly for the opposite, then each and every watch must be at the back of the opposite one and each and every dual must be more youthful than the opposite dual, which isn't feasible. The selection is that the famous person-hopping dual underwent enormous acceleration leaving Earth, in turning round at Alpha Centauri, and in slowing down whilst it got here again, so it was once subjected to forces that the Earthbound dual wasn't, and the Alpha Centauri dual might be more youthful than the Earthbound dual.

Answer 5

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