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An astronaut in a rocket moving at \(0.50 c\) toward the Sun finds himself halfway between Earth and the Sun. According to the astronaut, how far is he from Earth? In the frame of the Sun, the distance from Earth to the Sun is \(1.50 \times 10^{11} \mathrm{m}\).

Short Answer

Expert verified
Answer: Approximately \(0.65 \times 10^{11} \mathrm{m}\).

Step by step solution

01

Find the distance to the midpoint in the Sun's frame

We know the distance between the Earth and the Sun is \(1.50 \times 10^{11} \mathrm{m}\). Since the astronaut is halfway between the Earth and the Sun, we can find the distance to the midpoint by dividing by 2: \(Midpoint\_distance = (1.50 \times 10^{11} \mathrm{m}) / 2 = 0.75 \times 10^{11} \mathrm{m}\)
02

Calculate the length contraction factor

The length contraction factor is given by the formula: \(Length \ contraction \ factor = \sqrt{1 - v^2/c^2}\) where \(v = 0.50 c\) is the speed of the astronaut and \(c\) is the speed of light. In our case, \(Length \ contraction \ factor = \sqrt{1 - (0.50c)^2/c^2} = \sqrt{1 - 0.25} = \sqrt{0.75}\)
03

Calculate the distance in the astronaut's frame

Now we can find the distance from the Earth to the midpoint in the moving frame by multiplying the midpoint distance in the Sun's frame by the length contraction factor: \(Distance\_in\_moving\_frame = Midpoint\_distance \times Length \ contraction \ factor\) \(Distance\_in\_moving\_frame = (0.75 \times 10^{11} \mathrm{m}) \times \sqrt{0.75} \approx 0.65 \times 10^{11} \mathrm{m}\) So according to the astronaut, the distance from Earth is approximately \(0.65 \times 10^{11} \mathrm{m}\).

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