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IOAA 2022, Теория — Задача 4. Journey Between Galaxies

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International Olympiad on Astronomy and Astrophysics 2022 Georgia — Theoretical Competition · 16 август 2022 г. · 10 т.

Условие

We start a long journey to a planet located at a distance of $d_0 = 10\,\mathrm{Mpc}$ at the start of our journey. During our travel, the universe keeps expanding according to Hubble-Lemaître’s law. (Assume that Hubble’s constant ( $H$ ) does not change.)

(a) Write an expression for the distance between Earth and the planet at a time $t$ from the start of our journey. [2 т.]

(b) What minimum constant speed $v_0$ should the rocket have to reach the destination? Assuming that the speed of our rocket is $v_0 = 1000\,\mathrm{km/s}$ , can we reach the planet? Write YES or NO. If yes, how long will our journey take?

Hint: A simplified way to account for the expansion of our universe is to introduce a “scale factor” $a(t)$ , which relates distance between two objects $l(t)$ at a time $t$ to a distance $l_0$ at a time $t = 0$ as $l(t) = a(t)l_0$ [8 т.]

Отговори

Решение

(a) By the Hubble’s Law, the rate of recession is

v(t) = \frac{\Delta d}{\Delta t} = Hd(t) \qquad \mathbf{1.0}

Thus, by referring to the given relation,

d(t) = Ce^{Ht}

at time $t = 0$ , $d(0) = d_0 = C$ . Thus, the distance at a time $t$ :

d(t) = d_0 e^{Ht} \qquad \mathbf{1.0}

(b) It is easier to solve this problem, if we consider coordinate system that changes its scale in such a way that distance between earth and our destination does not change (called co-moving frame). Let $v(t)$ be the velocity in this frame at time $t$ and let $l(t)$ be the total distance traveled until the moment $t$ .

\begin{aligned} v(t)d(t) &= v_0 d_0 \\ v(t) &= v_0 \frac{d_0}{d(t)} = v_0 \frac{d_0}{d_0 e^{Ht}} \\ v(t) &= \frac{\Delta l}{\Delta t} = -v_0 e^{-Ht} && \mathbf{3.0} \end{aligned}

l(t) = -\frac{v_0}{-H} e^{-Ht} + C

for specific cases:

l(t_0) = \frac{v_0}{H} e^{-Ht_0} + C

l(0) = \frac{v_0}{H} + C

subtracting yields:

l(0) = \frac{v_0}{H} - \frac{v_0}{H} e^{-Ht_0}

thus:

\therefore l_{\mathrm{Total}} = d_0 = -\frac{v_0}{H}(e^{-Ht_0} - 1) \qquad \mathbf{2.0}

Rearranging,

t_0 = -\frac{1}{H} ln \left(1 - \frac{Hd_0}{v_0}\right) \qquad \mathbf{1.0}

Thus, condition for reaching the planet at all:

\begin{aligned} 1 - \frac{Hd_0}{v_0} &> 0 \\ v_0 &> Hd_0 = 70\,\mathrm{km/s} && \mathbf{1.0} \end{aligned}

(However at this speed it will take eternity to reach it)

For $v_0 = 1000\,\mathrm{km/s}$ we can reach the planet, in:

t_0 \approx 5.3 \times 10^{16}\,\mathrm{s} = 17\,\mathrm{Gyr} \qquad \mathbf{1.0}

Оригинал в Архива: Theory Problems.pdf · официални решения: Theory Solutions.pdf