A question about sepcial relativity

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An astronaut on the Helios satellite travels to Jupiter and back in 75 days, as observed from Earth, at an average speed of 250,000 km/h. To determine the time difference between the Earth clock and the satellite clock, the speed must be expressed as a fraction of the speed of light (c). The time dilation effect of special relativity indicates that the satellite's clock will run slower due to its high speed. The time difference can be calculated by applying the time dilation formula to the 75 days, converting it into seconds. Understanding these calculations is essential for resolving the confusion surrounding the scenario.
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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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