Special Relativity and spacecraft

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SUMMARY

The discussion centers on calculating time intervals in special relativity for a spacecraft traveling at different velocities. The spacecraft initially travels at 1.8x108 m/s and observes two events on Earth as 32 hours apart. When the spacecraft's speed increases to 2.82x108 m/s, the time interval observed by the spacecraft is calculated using the formula delta Tm = delta Ts / [1 - (v2 / c2)]. The final answer for the time interval at the higher speed is determined to be 75 hours.

PREREQUISITES
  • Understanding of special relativity concepts
  • Familiarity with time dilation equations
  • Knowledge of the speed of light (c = 3x108 m/s)
  • Ability to perform algebraic manipulations with equations
NEXT STEPS
  • Study the Lorentz transformation equations in special relativity
  • Learn about time dilation effects at relativistic speeds
  • Explore practical applications of special relativity in modern physics
  • Investigate the implications of escape velocity in space travel
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Students of physics, aerospace engineers, and anyone interested in the implications of special relativity on time perception during high-speed travel.

sean-820
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Homework Statement



A spacecraft is traveling around Earth at 1.8x10^8m/s relative to the earth. If the spacecraft determines two events on Earth to be 32H, what time interval would they find if the spacecraft is traveling at 2.82x10^8m/s?

Homework Equations



delta Tm=delta Ts/[1-(v^2)/(c^2)]

Where Tm= time viewed by a moving object
Ts= time viewed y a stationary object
v= velocity
c= speed of light=3x10^8m/s

The Attempt at a Solution



This is the equation i know, but i don't know how to incorporate the 32h from just the spacecraft s frame of reference
 
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sean-820 said:

Homework Statement



A spacecraft is traveling around Earth at 1.8x10^8m/s relative to the earth. If the spacecraft determines two events on Earth to be 32H, what time interval would they find if the spacecraft is traveling at 2.82x10^8m/s?

Homework Equations



delta Tm=delta Ts/[1-(v^2)/(c^2)]

Where Tm= time viewed by a moving object
Ts= time viewed y a stationary object
v= velocity
c= speed of light=3x10^8m/s

The Attempt at a Solution



This is the equation i know, but i don't know how to incorporate the 32h from just the spacecraft s frame of reference

The spacecraft can't be traveling "around" the Earth at this speed if you mean that it is circling the earth. Does 32H mean 32 hours? Are the events at the same location on the earth? Can you give us the exact wording of the question?

AM
 
The spacecraft can't be traveling "around" the Earth at this speed if you mean that it is circling the earth.Yes its path is around Earth at that speed. Does 32H mean 32 hours? yesAre the events at the same location on the earth?yes Can you give us the exact wording of the question?

A spacecraft has a speed of 1.8x10^8m/s with respect to earth. The spacecraft determines two events on Earth to be 32 hours apart. What time interval would they find if the spacecraft is traveling at 2.82x10^8m/s (instead of 1.8x10^8)?
 
sean-820 said:
The spacecraft can't be traveling "around" the Earth at this speed if you mean that it is circling the earth.Yes its path is around Earth at that speed. Does 32H mean 32 hours? yesAre the events at the same location on the earth?yes Can you give us the exact wording of the question?

A spacecraft has a speed of 1.8x10^8m/s with respect to earth. The spacecraft determines two events on Earth to be 32 hours apart. What time interval would they find if the spacecraft is traveling at 2.82x10^8m/s (instead of 1.8x10^8)?
Ok. The question simply states that it is moving at a constant speed relative to the Earth ie. in a straight line. It has more than enough escape velocity so it cannot be going around the earth.

How do you transform time from one inertial frame to another?

AM
 
Last edited:
I solved it. The answer is 75 hours. I think it was more the wording of the question that messed me up. Escape velocity wouldn't matter as it isn't necessarily orbiting earth.To solve, i just used the formula given twice. First time to sub in the initial velocity and time to get the stationary observers time, then did the equation again using this stationary time to find time observed by the spacecraft at its new speed.

Thanks for trying to help
 

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