A question about sepcial relativity

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SUMMARY

The discussion centers on calculating time dilation effects experienced by an astronaut on the Helios satellite traveling to Jupiter and back in 75 days, as observed from Earth. The satellite's speed is noted to be a quarter of a million km/h, which is approximately 0.23 times the speed of light (c). Participants suggest using the formula for time dilation to determine the time difference in seconds between the Earth clock and the satellite clock during the trip.

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  • Understanding of special relativity concepts, particularly time dilation.
  • Familiarity with the formula for time dilation: Δt' = Δt / √(1 - v²/c²).
  • Basic knowledge of converting days into seconds for time calculations.
  • Ability to calculate speeds as multiples of the speed of light (c).
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  • Study the implications of time dilation in special relativity.
  • Learn how to apply the time dilation formula in various scenarios.
  • Explore the concept of relativistic speeds and their effects on time measurement.
  • Investigate real-world applications of special relativity in space travel.
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An astronaut traveling on the Helios satellite (the fastest human made craft) goes on a trip to Jupiter and back in 75 days as measured by an observer on the Earth. If the average speed of the satellite is a quarter of a million km/h with respect to the Earth , what is the time difference ( in seconds) between the time recorded by the clock on the Earth and the one on board the satellite?



honestly, i am really confused on this part , can somebody help me?

thx so much
 
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Hi wowolala! :smile:

Find the speed, v (as a multiple of c).

Then the clock on the satellite will go slow by … ?

Then multiply by the time (75 days converted into seconds). :smile:
 

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