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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

Show that each of these statements implies that \(v<c\) which means that \(v\) can be considered a nonrelativistic speed: (a) $\gamma-1<<1 \text { [Eq. }(26-14)] ;\( (b) \)K<<m c^{2}\( [Eq. \)(26-15)] ;(\text { c }) p<m c[\text { Eq. }(26-16)] ;\( (d) \)K \approx p^{2} /(2 m)$.
Derivation of the Doppler formula for light. A source and receiver of EM waves move relative to one another at velocity \(v .\) Let \(v\) be positive if the receiver and source are moving apart from one another. The source emits an EM wave at frequency \(f_{\mathrm{s}}\) (in the source frame). The time between wavefronts as measured by the source is \(T_{\mathrm{s}}=1 / f_{\mathrm{s}}\) (a) In the receiver's frame, how much time elapses between the emission of wavefronts by the source? Call this \(T_{\mathrm{T}}^{\prime} .\) (b) \(T_{\mathrm{T}}^{\prime}\) is not the time that the receiver measures between the arrival of successive wavefronts because the wavefronts travel different distances. Say that, according to the receiver, one wavefront is emitted at \(t=0\) and the next at \(t=T_{\mathrm{o}}^{\prime} .\) When the first wavefront is emitted, the distance between source and receiver is \(d_{\mathrm{r}}\) When the second wavefront is emitted, the distance between source and receiver is \(d_{\mathrm{r}}+v T_{\mathrm{r}}^{\prime} .\) Each wavefront travels at speed \(c .\) Calculate the time \(T_{\mathrm{r}}\) between the arrival of these two wavefronts as measured by the receiver. (c) The frequency detected by the receiver is \(f_{\mathrm{r}}=1 / T_{\mathrm{r}} .\) Show that \(f_{\mathrm{r}}\) is given by $$f_{\mathrm{r}}=f_{\mathrm{s}} \sqrt{\frac{1-v / \mathrm{c}}{1+v / \mathrm{c}}}$$
An unstable particle called the pion has a mean lifetime of 25 ns in its own rest frame. A beam of pions travels through the laboratory at a speed of $0.60 c .$ (a) What is the mean lifetime of the pions as measured in the laboratory frame? (b) How far does a pion travel (as measured by laboratory observers) during this time?
Suppose your handheld calculator will show six places beyond the decimal point. How fast (in meters per second) would an object have to be traveling so that the decimal places in the value of \(\gamma\) can actually be seen on your calculator display? That is, how fast should an object travel so that \(\gamma=1.000001 ?\) [Hint: Use the binomial approximation.]
Refer to Example \(26.2 .\) One million muons are moving toward the ground at speed \(0.9950 c\) from an altitude of \(4500 \mathrm{m} .\) In the frame of reference of an observer on the ground, what are (a) the distance traveled by the muons; (b) the time of flight of the muons; (c) the time interval during which half of the muons decay; and (d) the number of muons that survive to reach sea level. [Hint: The answers to (a) to (c) are not the same as the corresponding quantities in the muons' reference frame. Is the answer to (d) the same?]
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