Question

A seen from the Earth, the distance from Earth toQuick answer please. THANKS.

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Answer #1

Part A :

given

d = 1.5 X 1011 m

t = 9.62 minutes

t = 9.62 X 60

t = 577.2 sec

speed of the particle is v = d / t

v = 1.5 X 1011 / 577.2

v = 2.59 X 108 m/s

v = 2.59 X 108 / 3 X 108

v = 0.8633 c

Part B :

\Delta tp = t / ( \sqrt{}1 - v2 / c2 )

v = 0.8633 c

\Delta tp = 577.2 / ( \sqrt{}1 - 0.86332 c2 / c2 )

\Delta tp = 1143.6696 sec

\Delta tp = 1143.6696 / 60

\Delta tp = 19.06 min

Part C :

\Delta Lp = L X ( \sqrt{}1 - v2 / c2 )

\Delta Lp = 1.5 X 1011 X ( \sqrt{}1 - 0.86332 c2 / c2 )

\Delta Lp = 1.5 X 1011 X 0.504

\Delta Lp = 7.57 X 1010 m

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