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Chapter 26: Problem 22

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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Most popular questions from this chapter

Radon decays as follows: $^{222} \mathrm{Rn} \rightarrow^{218} \mathrm{Po}+\alpha .$ The mass of the radon-222 nucleus is 221.97039 u, the mass of the polonium- 218 nucleus is \(217.96289 \mathrm{u},\) and the mass of the alpha particle is 4.00151 u. How much energy is released in the decay? \(\left(1 \mathrm{u}=931.494 \mathrm{MeV} / \mathrm{c}^{2} .\right)\)
An engineer in a train moving toward the station with a velocity \(v=0.60 c\) lights a signal flare as he reaches a marker \(1.0 \mathrm{km}\) from the station (according to a scale laid out on the ground). By how much time, on the stationmaster's clock, does the arrival of the optical signal precede the arrival of the train?
A clock moves at a constant velocity of \(8.0 \mathrm{km} / \mathrm{s}\) with respect to Earth. If the clock ticks at intervals of one second in its rest frame, how much more than a second elapses between ticks of the clock as measured by an observer at rest on Earth? [Hint: Use the binomial approximation \(.(1-v^{2} / c^{2})^{-1 / 2} \approx 1+v^{2} /(2 c^{2}) .\)]
A spaceship is traveling away from Earth at \(0.87 c .\) The astronauts report home by radio every 12 h (by their own clocks). (a) At what interval are the reports sent to Earth, according to Earth clocks? (b) At what interval are the reports received by Earth observers, according to their own clocks?
As observed from Earth, rocket Alpha moves with speed \(0.90 c\) and rocket Bravo travels with a speed of \(0.60 c\) They are moving along the same line toward a head-on collision. What is the speed of rocket Alpha as measured from rocket Bravo? (tutorial: adding velocities)
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