Time intervals in the reference frame moving relative to each other are connected by the Lorentz transformation
(1) "\\Delta t'= \\Delta t\\cdot \\sqrt{1-\\beta^2}" , where "\\beta=\\frac{V}{c}" . In first case the time interval on Earth is
(2) "\\Delta t_E=\\frac{\\Delta t_1}{\\sqrt{1-\\beta_1^2}}" , where "\\Delta t_1=30hrs" , and "\\beta_1=0.8" . Since the measurement of the time interval was made by the clock in the rocket, relative to which the Earth moved at a speed 0.8c. This interval measured in a rocket moving at a speed of 0.9 s will be
(3) "\\Delta t_2=\\Delta t_E\\cdot \\sqrt{1-\\beta_2^2}=\\Delta t_1\\frac{\\sqrt{1-\\beta_2^2}}{\\sqrt{1-\\beta_1^2}}=30\\frac{\\sqrt{1-0.9^2}}{\\sqrt{1-0.8^2}}=21.8hrs"
Answer: The time interval in the spacecraft travelling at a speed of 0.9c is 21.8hrs.
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