We need to find the initial velocity (impulse speed) a body needs from a galaxy at distance D so that it reaches "the origin" (us, presumably at χ=0) only under its own inertia (no forces, geodesic motion) in the "cosmic dust" cosmological model (pressureless fluid, no pressure). With Hubble constant H₀. Interpretation: In an FLRW universe with pressureless dust, the scale factor evolves as a(t) ∝ t (for a matter-dominated universe with Ω = 1, i.e., Einstein–de Sitter: a(t) = (t/t₀), H = 1/t). Actually if "cosmic dust (fluid without pressure)" and we assume a flat matter-dominated universe, a(t) = a₀ (t/t₀) with H(t) = 1/t, H₀ = 1/t₀. A body released from a comoving galaxy at distance D (proper distance at t₀ = 1/H₀) will follow a geodesic: comoving coordinate χ constant if no initial velocity (Hubble flow). To reach the origin, it must be given an initial peculiar velocity. Under its own inertia (geodesic), the peculiar velocity decays as 1/a. We need the minimal (or exact?) speed such that it arrives at origin. Actually with geodesic motion in expanding universe, a particle given a peculiar velocity v at t₀ will have physical distance r(t) = a(t) χ(t) where dχ/dt = v/a(t), with v decaying as v(t) = v₀ a₀/a(t) (for nonrelativistic) or for relativistic, momentum decays as 1/a. Total comoving distance traveled from t₀ to ∞: χ_∞ = ∫_{t₀}^{∞} v(t)/a(t) dt = ∫ v₀ a₀ / a² dt. With a(t) = t/t₀: χ_∞ = ∫_{t₀}^{∞} v₀ a₀ t₀² / t² dt = v₀ a₀ t₀² [1/t₀] = v₀ a₀ t₀ = v₀ a₀ / H₀. Setting proper distance at t₀: D = a₀ χ_∞ ⇒ v₀ = D H₀. So the required speed is v₀ = H₀ D — exactly the Hubble velocity! Interesting: the speed needed equals the Hubble flow speed H₀D. That's a nice result: In Einstein–de Sitter (dust, flat), if you launch a particle toward us at speed v = H₀D, it will just reach the origin asymptotically (arriving at t→∞). Any slower and it never arrives; faster arrives in finite time. But wait — perhaps the question wants the particle to reach the origin in finite time? "logre llegar al origen" - achieve reaching the origin. If launched with speed exactly H₀D, it reaches origin only as t→∞. Perhaps we should present general solution: launched with speed v₀, total comoving distance covered χ_∞ = v₀ a₀/H₀ (in EdS). For arrival at origin, need χ_∞ = D ⇒ v₀ = H₀ D. If you want arrival at finite time t_f, then integrate: Δχ = ∫_{t₀}^{t_f} v₀ t₀²/(a₀ t²) dt = v₀ a₀ t₀² /a₀ (1/t₀ - 1/t_f) = v₀ (H₀⁻¹ - t_f⁻¹) ... let's compute: a(t) = a₀ t/t₀, v(t) = v₀ a₀/a = v₀ t₀/t. Then dχ/dt = v/a = v₀ t₀/(t · a₀ t/t₀) = v₀ t₀²/(a₀ t²). Integrate from t₀ to t_f: Δχ = (v₀ t₀²/a₀)(1/t₀ - 1/t_f) = (v₀/a₀)(t₀ - t₀²/t_f). Set a₀ Δχ = D: D = v₀ (t₀ - t₀²/t_f) = v₀ (1/H₀ - 1/H_f) since t_f = 1/H_f. So v₀ = D / (1/H₀ - 1/H_f) = H₀ D / (1 - H_f/H₀). Interesting: to reach in finite time t_f, need v₀ = D H₀ H_f/(H_f - H₀)... Let me redo: v₀ = D / (t₀ - t₀²/t_f). With t₀ = 1/H₀. t₀²/t_f = H₀²/H_f²? No: t₀²/t_f = (1/H₀²)·(1/t_f) = H_f/H₀² where H_f = 1/t_f. So v₀ = D/(1/H₀ - H_f/H₀²) = D H₀²/(H₀ - H_f) = D H₀/(1 - H_f/H₀). As t_f → ∞, v₀ → H₀ D. So minimum speed is H₀D, arriving asymptotically at t→∞. Hmm, but the phrase "solo bajo su inercia" (only under its own inertia) suggests geodesic motion with no forces—i.e., the particle is "impulsado" (given a kick) and then coasts. In an expanding universe, "coasting" means peculiar velocity decays. So the answer: v₀ = H₀ D is the minimum; this equals the Hubble velocity. There's a cute coincidence: the minimum speed equals the recession speed of the galaxy at launch (Hubble flow speed). This is the "anti-Hubble" launch. Alternatively, maybe the question is about the particle moving with the Hubble flow? If the body is at distance D and is at rest with respect to the Hubble flow, it recedes. To come to us, must be launched toward us. Let me carefully derive with general a(t) for dust: For dust (p=0), a(t) ∝ t^{2/3(1+w)} = t^{2/3}. But if the universe includes dark energy (Λ), a(t) behaves differently at late times (accelerating), and then χ_∞ converges and even the minimum speed changes because the integral ∫dt/a² converges. Actually the question says "modelo cosmológico de polvo cósmico (fluido sin presión)" — cosmic dust model = Einstein–de Sitter? Or maybe they intend the "coasting" solution? Hmm, "polvo cósmico" is just matter-dominated. The standard "dust" universe: a(t) = a₀ (t/t₀)^{2/3} for Ω_m=1 (EdS). Wait, I earlier used a ∝ t (coasting). Let me redo for dust: For a flat matter-dominated universe, Friedmann: (ȧ/a)² = (8πG/3)ρ, ρ ∝ a⁻³ ⇒ ȧ ∝ a^{-1/2} ⇒ a ∝ t^{2/3}. So a(t) = a₀ (t/t₀)^{2/3}, H(t) = 2/(3t), t₀ = 2/(3H₀). Then peculiar velocity decays v = v₀ a₀/a. dχ/dt = v/a = v₀ a₀/a² = v₀ a₀/(a₀² (t/t₀)^{4/3}) = (v₀/a₀)(t₀/t)^{4/3}. χ_∞ = ∫_{t₀}^{∞} (v₀/a₀) t₀^{4/3} t^{-4/3} dt = (v₀/a₀) t₀^{4/3} · 3 t₀^{-1/3} = 3 (v₀/a₀) t₀. Proper distance today D = a₀ Δχ ⇒ D = 3 v₀ t₀ = 3 v₀ · (2/(3H₀)) = 2 v₀/H₀ ⇒ v₀ = D H₀/2. So in EdS, the minimum speed is v₀ = H₀D/2! And arrival happens at t→∞. Hmm, so which is intended? "modelo cosmologico de polvo cosmico (fluido sin presion)" — the pressureless fluid model. The standard result for Ω_m = 1 flat dust universe: a ∝ t^{2/3}. But some textbooks consider "coasting universe" a ∝ t which corresponds to w = -1/3 or Ω_tot = 1 with curvature k=0 empty (Milne). Actually Milne universe is empty (a∝t). Dust with k=-1? For open dust universe, a(t) = ... For Ω_m < 1, a(t) = A(cosh η - 1) etc. Given "polvo cósmico (fluido sin presión)" with Hubble constant H₀, the simplest assumption: flat matter-dominated Einstein–de Sitter. Then answer v₀ = H₀ D / 2, arriving asymptotically. But maybe the question intends something else: perhaps "llegar al origen" means reaching us at the same cosmic time? Or maybe it's like the classic problem: "A stone is thrown from a galaxy at distance D toward the origin; find speed so that it arrives exactly when..." Hmm. Alternatively, maybe the problem is posed in terms of recession: a galaxy at distance D recedes with velocity H₀D. If we launch a body toward us at speed v, its net velocity relative to us initially is v - H₀D (naive). In a linear Hubble law universe (constant H, i.e., a ∝ e^{Ht}? No—de Sitter has H constant). In de Sitter, a ∝ e^{Ht}, peculiar velocity decays, and there's a horizon. In de Sitter, χ_∞ = ∫_{t₀}^{∞} v₀ a₀/a² dt = v₀ a₀/(H a₀²) = v₀/(H a₀). Setting a₀ χ = D gives v₀ = H D. Again H₀D! Interesting: for pure exponential expansion (de Sitter), minimum speed is H₀D, and the particle arrives asymptotically at t→∞. Actually for de Sitter, dχ/dt = v₀ e^{-H(t-t₀)}/a₀... integrate: ∫ v₀/a₀ e^{-H(t-t₀)} dt from t₀ to ∞ = v₀/(a₀ H). Yes. So the answer depends on a(t). For "dust" (EdS, a∝t^{2/3}): v_min = H₀D/2? Let me double check the EdS H(t): H = 2/(3t). t₀ = 2/(3H₀). v₀ = D/(3 t₀) = D H₀/2. Yes. Hmm wait, but maybe the problem intends the particle to travel at constant speed (inertia in flat space)? "solo bajo su inercia" = coasting at constant velocity in comoving coordinates? If space expansion is ignored and only inertia: v = D/T, needs arrival time. Not well posed. Alternatively the problem might be from a course on cosmology discussing "Can we catch a photon/reach a galaxy?" Or the classic problem from Ryden or Liddle: "A particle is thrown from Earth toward a distant galaxy; show that in a matter-dominated universe it can reach..." Hmm. Actually, I recall a classic problem: "In the Einstein–de Sitter universe, show that a projectile launched from Earth with speed v can reach objects only out to distance d_max = 2v/H₀" or something like that. Yes! There's a known result: the maximum proper distance reachable in an EdS universe by a projectile launched with peculiar velocity v is d_max = 2v/H₀. Because total comoving distance χ_∞ = v₀·(3t₀) and proper distance = a₀ χ_∞... wait let me redo: χ_∞ = ∫_{t₀}^∞ (v₀/a₀)(t₀/t)^{4/3} dt = (v₀/a₀)·3 t₀. Proper distance covered (in today's units) = a₀ χ_∞ = 3 v₀ t₀ = 3 v₀ · (2/3H₀) = 2 v₀/H₀. Yes, d_max = 2v₀/H₀. Equivalently, minimum speed to reach distance D is v₀ = H₀D/2. This is a known exercise (e.g., in "Introduction to Cosmology" by Barbara Ryden, problem about "escape velocity" analog). Hmm, I think there's an exercise in Ryden: "Consider a projectile launched from the origin... in EdS, how far can it travel?" The answer: comoving distance ξ_max = 3 v_e t₀? Something like that. So the answer: v = H₀ D / 2 (in EdS dust model), reached asymptotically as t→∞. If we want it to arrive by a finite time t_f, then v = (H₀D/2)/(1 - (t₀/t_f)^{1/3})... let's compute: Δχ(t_f) = (v₀/a₀) t₀^{4/3} ∫_{t₀}^{t_f} t^{-4/3} dt = (v₀/a₀) t₀^{4/3} · 3 (t₀^{-1/3} - t_f^{-1/3}) = (3v₀/a₀)(t₀ - t₀^{4/3} t_f^{-1/3}). Set a₀ Δχ = D: D = 3 v₀ t₀ [1 - (t₀/t_f)^{1/3}] ⇒ v₀ = D/[3t₀ (1 - (t₀/t_f)^{1/3})] = (H₀D/2)/[1 - (t₀/t_f)^{1/3}]. So minimal speed (arrival at t→∞): v_min = H₀ D/2. I think the intended answer is v = H₀D/2 for the dust (matter-dominated, Ω=1) model. But wait—should we consider that the "origin" is a specific comoving location (Earth), and the body is at comoving coordinate χ = D/a₀. Under geodesic motion with initial peculiar velocity toward origin, it moves in comoving coordinate. The expansion doesn't force it away; but its peculiar velocity decays, so it covers finite comoving distance. Need χ_∞ ≥ D/a₀. Alternatively, if the body is launched "with the Hubble flow" i.e., it just coasts in proper coordinates... no. Let me also double-check with relativistic treatment: if the required speed is small (D ~ 100 Mpc, H₀D ~ 7000 km/s ≪ c), nonrelativistic is fine. The problem likely expects nonrelativistic peculiar velocities. General formula: For power-law a ∝ t^n, χ_∞ = ∫_{t₀}^∞ (v₀ a₀/a²) dt = v₀ a₀ t₀^{2n}/(a₀² (2n-1)) t₀^{1-2n} = v₀ t₀/(a₀ (2n-1)). Proper: a₀ χ_∞ = v₀ t₀/(2n-1). With H₀ = n/t₀: proper = v₀/(H₀(2n-1)/n) = v₀ n/(H₀ (2n-1)). For dust n = 2/3: proper = v₀ (2/3)/(H₀(1/3)) = 2v₀/H₀. ✓. For a ∝ t (n=1, Milne/coasting): proper = v₀/H₀ ⇒ v_min = H₀ D. For de Sitter (exponential): v_min = H₀ D. So dust EdS gives H₀D/2. Thus answer: v₀ = H₀ D/2 (EdS). And it arrives only asymptotically (t→∞), i.e., that's the minimal speed; any v > H₀D/2 arrives in finite time t_f satisfying v₀ = (H₀D/2)/(1-(t₀/t_f)^{1/3}). Hmm, but maybe the question is phrased "para que logre llegar al origen" — to be able to reach the origin — so the minimal speed. Provide v_min = H₀D/2, noting equality with Hubble speed: interestingly, the galaxy recedes at H₀D = 2v, twice the launch speed. Cute. But wait: maybe they want the general solution including the condition that the particle arrives exactly (any v ≥ H₀D/2 works; v = H₀D/2 arrives as t→∞). Let me present derivation clearly: Setup: Flat matter-dominated FLRW (Einstein–de Sitter): a(t) = a₀(t/t₀)^{2/3}, with H₀ = 2/(3t₀). Galaxies are at fixed comoving coordinates; the galaxy in question is at χ_g = D/a₀. A body launched from the galaxy toward the origin with peculiar velocity v₀ (measured by a comoving observer at the galaxy) follows a geodesic. Its peculiar velocity decays as v(t) = v₀ a₀/a(t) (momentum redshift). Its comoving position: dχ/dt = v(t)/a(t). Total comoving distance it can ever cover: χ_max = ∫_{t₀}^∞ [v₀ a₀/a(t)²] dt = (v₀/a₀) ∫_{t₀}^∞ (t₀/t)^{4/3} dt = (v₀/a₀)·3t₀. Condition to reach origin: a₀ χ_max ≥ D ⇒ 3 v₀ t₀ ≥ D ⇒ v₀ ≥ D/(3t₀) = H₀D/2. Thus v_min = H₀D/2. Equality: arrives at origin as t → ∞. If arrival within time t_f: v₀ = (H₀D/2)·[1 - (t₀/t_f)^{1/3}]^{-1}. Also note the neat physical interpretation: distance reachable with launch speed v is d_max = 2v/H₀; this is like a "cosmic escape velocity" problem in reverse. Also, we might sanity check numbers: D = 100 Mpc, H₀ = 70 km/s/Mpc ⇒ v_min = 3500 km/s. Alternatively, if the intended model is a(t) ∝ t (constant H, "coasting"), answer is H₀D. The phrase "polvo cósmico (fluido sin presión)" strongly suggests dust = matter-dominated ⇒ EdS with a ∝ t^{2/3} (if flat, Ω_m=1). The mention of H₀ only (no Ω, no k) suggests they want us to assume the flat EdS model. So answer: v = H₀D/2. Hmm, but let me reconsider: could the problem be about a universe of dust where the test particle also gravitates? "solo bajo su inercia" - only under its inertia: no forces act on it (we neglect gravity on the particle? Or the particle follows geodesic which includes cosmic gravity but no non-gravitational forces). In cosmology, free particles follow geodesics; the "inertia" phrase means we give it a kick and then it coasts. The dust background affects geometry (a(t)), which affects the coasting. So yes, my treatment stands. One more nuance: The body is launched from a galaxy that is receding from us. To reach us, it must move toward us in comoving coordinates. Since peculiar velocities decay, there is a max comoving distance; the condition gives v_min. Good. Alternatively, some might solve using energy conservation in Newtonian cosmology: For a shell of dust, specific energy E = v²/2 - GM/r. A particle launched from distance D toward origin with speed v reaches origin if... Actually in Newtonian EdS, consider particle at distance r moving in the gravitational field of enclosed mass M(r) = (4π/3)ρ r³. Its equation: d²r/dt² = -GM/r². Energy: E = ½ ṙ² - GM/r. At launch: r = D, ṙ = v - H₀D (net inward speed relative to origin). At arrival r → 0: potential → -∞, so any inward energy reaches? Hmm, in Newtonian gravity, falling to r=0 happens for any E since potential → -∞. So Newtonian picture gives "any speed suffices" — but that's because the particle keeps falling into the potential well. However, the question says "solo bajo su inercia" — only inertia, i.e., no gravity acting on it? If gravity of the dust acts, then the particle accelerates and always arrives (Newtonian). But in GR, the geodesic in FLRW: the peculiar velocity decays due to expansion; comoving coordinates are "free-falling". Actually both descriptions are consistent: in Newtonian terms, the particle at rest with the Hubble flow at some radius has zero energy (EdS is marginally bound). A particle launched inward with peculiar speed v₀ has energy E = ½v₀² - GM/D relative to the comoving frame... hmm, this Newtonian analysis is subtle because the "enclosed mass" changes as it moves inward (it sweeps through shells). Actually for a uniform density universe, a test particle's motion: comoving coordinates conserved? No—a particle with peculiar velocity has changing comoving coordinate. Let's do Newtonian: equation of motion for a free particle in EdS background: r̈ = -(GM( H₀D/2, there is finite t_f with χ(t_f) = 0, i.e., it arrives in finite time. So the minimal speed for actual arrival is v₀ > H₀D/2, with equality being a limiting case that never truly arrives (it asymptotes to our worldline in comoving space but proper distance diverges... wait, can that be? If χ→0 and the body asymptotically joins our worldline... but proper distance diverging means it doesn't join us. Contradiction? If χ(t) → 0, the worldline approaches our worldline in comoving coordinates. The proper distance at time t between us and the body is a(t)χ(t) → ∞. As t→∞, in the limit, the body's worldline coincides with ours in comoving coordinates, but the limit of the distance is infinite because the limit and the distance function don't commute: lim_{t→∞} a(t)χ(t) ≠ a(∞)·lim χ. It's a singular limit. Physically, the body sweeps toward us in comoving space but the expansion outruns it, so it never reaches us: it just keeps getting farther while its comoving coordinate shrinks. Hmm, wait — that seems paradoxical: if its comoving coordinate is shrinking toward 0, at some point it should cross... no: χ(t) = (D/a₀)(t₀/t)^{1/3} > 0 for all finite t, approaching 0 asymptotically. So it never crosses. And yes its proper distance grows. Both statements are consistent because a(t)χ(t) with χ ~ t^{-1/3}, a ~ t^{2/3} gives r ~ t^{1/3}. Interesting: this shows that with v₀ = H₀D/2, the body never reaches us — it just barely fails to, in the sense that arrival time → ∞. So the true answer: to reach the origin, need v₀ > H₀D/2 (strictly), arriving at finite time t_f determined by v₀ = (H₀D/2)/(1 - (t₀/t_f)^{1/3}). The infimum of required speeds is H₀D/2, approached asymptotically. In practice, we quote v_min = H₀D/2 as the threshold. Hmm wait, but hold on: is it correct that arrival happens at finite time for v₀ slightly greater? χ(t_f) = 0: (D/a₀) - (3v₀t₀/a₀)(1-(t₀/t_f)^{1/3}) = 0 ⇒ 1 - (t₀/t_f)^{1/3} = D/(3v₀t₀) = (H₀D/2)/v₀. If v₀ > H₀D/2, RHS < 1, so (t₀/t_f)^{1/3} = 1 - H₀D/(2v₀) > 0 ⇒ t_f = t₀ [1 - H₀D/(2v₀)]^{-3}. Finite ✓. As v₀ → H₀D/2⁺, t_f → ∞. ✓. So answer: v must exceed H₀D/2; the threshold speed is v* = H₀D/2 (arrival only at t→∞). Nice closed form: t_f = H₀^{-1}(2/3)[1 - H₀D/(2v₀)]^{-3}... with t₀ = 2/(3H₀): t_f = (2/(3H₀)) (1 - H₀D/(2v₀))^{-3}. Let me double check Δχ computation once more. v(t) = v₀ a₀/a. In EdS a = a₀(t/t₀)^{2/3} ⇒ a₀/a = (t₀/t)^{2/3} ⇒ v(t) = v₀ (t₀/t)^{2/3}. dχ/dt = v/a = v₀ (t₀/t)^{2/3} / (a₀ (t/t₀)^{2/3}) = (v₀/a₀) (t₀/t)^{4/3}. ✓. ∫_{t₀}^∞ t^{-4/3} dt = 3 t₀^{-1/3}. So χ_∞ = (v₀/a₀) t₀^{4/3} · 3 t₀^{-1/3} = 3 (v₀ t₀/a₀). ✓. a₀ χ_∞ = 3 v₀ t₀ = 3v₀ · 2/(3H₀) = 2v₀/H₀. ✓. So max reachable proper distance (in present-day units... well, a₀χ) = 2v₀/H₀. Setting = D: v₀ = H₀D/2. ✓. Great. Also the relativistic version: if speeds are relativistic, momentum p ∝ 1/a, and v = pc²/E with E² = p²c² + m²c⁴. Then the integral changes; the max comoving distance is ∫ (p(a)/... ) Actually dχ/dt = v/a, with v as function of a. For ultra-relativistic, v ≈ c: χ_∞ = ∫_{t₀}^∞ c/a dt = (c/a₀)∫ (t₀/t)^{2/3} dt = (c/a₀)·3 t₀^{2/3} t^{1/3}|_{t₀}^∞ → diverges! Photons always reach in EdS. So for photons no limit. For massive particles nonrelativistic at launch, result above. Since problem presumably about a "cuerpo" (body) with some impulse speed, nonrelativistic treatment is standard. I might mention: for the body launched with speed v₀ ≪ c; if v₀ comparable to c, use relativistic formula, and the threshold is still v₀ → c? No—the threshold is still H₀D/2 as long as H₀D/2 < c, i.e., D < 2c/H₀ ≈ 8.6 Gpc (Hubble sphere ×2). If D exceeds 2c/H₀ (~8.6 Gpc for H₀=70), then even light launched toward us... wait, photons launched from galaxies beyond the event horizon never reach us. In EdS there's no event horizon (a(t) grows as power law ⇒ event horizon infinite: ∫_{t₀}^∞ dt/a(t) = ∫ t^{-2/3} dt diverges ⇒ event horizon = ∞). So photons from any comoving distance eventually reach us in EdS. Yes: in EdS, χ_∞ for light = (c/a₀)·3t₀^{2/3}[t^{1/3}]_{t₀}^{∞} = ∞. So any galaxy's light eventually reaches us (particle horizon finite, event horizon infinite). ✓. So for massive body with v₀ < c... the max distance: ∫_{t₀}^∞ (v(a)/a) dt where v(a) = c/√(1+(a m c/p₀)²)... For launch speed v₀ at a₀: p₀ = γ₀ m v₀. p(a) = p₀ a₀/a. v(a) = p c²/E, E = √(p²c²+m²c⁴). As a→∞, v→0 like p c²/(mc²) = p₀a₀ c/(m a)... anyway v(a)/a → const/a² integrable. So finite reach for any finite v₀ < c. And the reach = ? ∫ (v(a)/a) dt with dt = da/ȧ, ȧ = H₀... in EdS ȧ = (2/3)(a/t₀)·... a(t)=a₀(t/t₀)^{2/3} ⇒ ȧ = (2/3) a₀ t₀^{2/3} t^{-1/3} = (2/3) a/t. So dt = (3/2)(t/a) da... simpler: dt/da = 3 t₀^{1/3}... hmm let me param by a: t = t₀ (a/a₀)^{3/2}, dt = (3/2) t₀ (a/a₀)^{1/2} da/a₀. ∫ v(a)/a dt = ∫ v(a)/a · (3/2)(t/a)(...)... Let me just: dt = da/ȧ, ȧ = (2/3)(a/t) ⇒ dt = (3/2)(t/a) da = (3/2)(t₀/a₀^{3/2}) a^{1/2} da. Then χ_∞ = ∫_{a₀}^∞ [v(a)/a] (3/2)(t₀/a₀^{3/2}) a^{1/2} da = (3/2)(t₀/a₀^{3/2}) ∫_{a₀}^∞ v(a) a^{-1/2} da. With v(a) = c/√(1+(mc a/(p₀ a₀))²)... define α = p₀a₀/(mc) (dimensionless), v(a) = c (α a₀... hmm let me set u = a/a₀: p = p₀/u ⇒ v = c β₀/√(β₀² + u²) where β₀ = p₀c/E₀ = v₀/c (since E₀ = γ₀mc², p₀c = γ₀ m v₀ c = E₀ β₀). Indeed v(a) = c β₀/√(β₀² + u²). So χ_∞ = (3/2)(t₀/a₀) ∫_1^∞ [c β₀/√(β₀²+u²)] u^{-1/2} du... wait I need to redo: χ_∞ a₀ = proper reach R = a₀ χ_∞ = (3/2) t₀ ∫_1^∞ v(a) a^{-1/2} da with a in units of a₀: R = (3/2) t₀ c β₀ ∫_1^∞ du/(√(β₀²+u²) √u). For β₀→1 (launch at ~c): integral ∫_1^∞ du/(√(1+u²)√u) converges (integrand ~ u^{-3/2}) to some value ≈ ∫_1^∞ u^{-3/2} du = 2 plus correction; roughly 2.6? Let's not go down this path; nonrelativistic is the intended regime. I'll mention assumption v₀ ≪ c and that H₀D/2 must be ≪ c. Numerical: H₀ = 70 km/s/Mpc, D = 100 Mpc ⇒ v = 3500 km/s ≪ c ✓. Now let me also double check the claim "peculiar velocity decays as 1/a": geodesic in FLRW gives p ∝ 1/a; for nonrelativistic, v = p/m ∝ 1/a. ✓. Also might be worth noting the general formula for a power-law a ∝ t^n: required speed v_min = D(2n-1)/t₀·(1/n)·... let me recompute: a₀χ_∞ = v₀ t₀/(2n-1)? Earlier: χ_∞ = v₀ t₀/(a₀(2n-1))? Let me redo: χ_∞ = ∫_{t₀}^∞ (v₀a₀/a²)dt = (v₀/a₀) t₀^{2n} ∫ t^{-2n} dt = (v₀/a₀) t₀^{2n} · t₀^{1-2n}/(2n-1) = (v₀ t₀)/(a₀ (2n-1)). Proper reach = a₀χ_∞ = v₀ t₀/(2n-1) = v₀ n/(H₀(2n-1)) (using t₀ = n/H₀). Set = D ⇒ v_min = D H₀ (2n-1)/n. For n=2/3 (dust): v_min = D H₀ (1/3)/(2/3) = H₀D/2 ✓. For n=1: H₀D ✓ (Milne/de Sitter-type coasting). Good, could include as a remark. Also worth presenting the intuitive picture: the projectile's peculiar velocity decays like 1/a; the total distance it can ever cover (in comoving terms scaled to today) is finite, R_max = 2v₀/H₀; requiring R_max = D gives the answer. Equivalent known result: "In an EdS universe, a projectile with speed v can reach at most 2v/H₀." Also nice cross-check with the known result about maximum reach of projectile in EdS: In Ryden's book problem: "In the Einstein–de Sitter model, a projectile shot from the origin with speed v₀... maximum comoving distance ξ_max = ...". I recall ξ_max = 2 v₀ t₀ a₀^{1/2}... something; in any case proper today's distance = 2v₀/H₀. Let me double-check with time-reversal symmetry: A projectile launched from origin at t=0 (Big Bang) with... hmm not needed. Alternative derivation via "Hubble sphere" style: Not needed. Let me also present the arrival time formula for completeness: t_f = t₀ [1 - H₀D/(2v₀)]^{-3}, r(t) trajectory: r(t) = a(t)[χ_g - Δχ(t)]. Also sanity: the body's proper distance initially recedes (since H₀D > v₀ needed? With v₀ = H₀D/2, ṙ(0) = a₀χ̇ + ȧχ = -v₀ + H₀D = +H₀D/2 > 0, so it initially recedes, turns around when? ṙ = 0 when v(t) = H(t) r(t) i.e., peculiar inflow equals expansion outflow. Then it comes back in comoving coordinates but r grows... at turnaround ṙ=0, afterwards ṙ<0? ṙ = aχ̇ + ȧχ = v(t) - H(t) r(t). With v(t) = v₀(t₀/t)^{2/3}, r(t) as computed: for the critical case r(t) = D(t/t₀)^{1/3}, H(t) r(t) = (2/(3t)) D (t/t₀)^{1/3} = (2D/(3t₀)) (t/t₀)^{-2/3} = 2 v₀ (t₀/t)^{2/3}·... since v₀ = D/(3t₀): H r = 2 (D/(3t₀)) (t/t₀)^{-2/3}·... let me compute: H(t) = 2/(3t). H r = 2D/(3t) · (t/t₀)^{1/3} = (2D/3) t^{-2/3} t₀^{-1/3}. v(t) = (D/(3t₀)) (t₀/t)^{2/3} = (D/3) t₀^{-1/3} t^{-2/3}. So H r = 2 v(t). So ṙ = v - Hr = -v < 0?! But earlier we computed r(t) = D(t/t₀)^{1/3} increasing! Contradiction. Let me recheck. r(t) = a(t)χ(t), χ(t) = (D/a₀)(t₀/t)^{1/3}, a(t) = a₀(t/t₀)^{2/3}. r(t) = D (t/t₀)^{2/3 - 1/3} = D (t/t₀)^{1/3}. ṙ = (D/(3t₀))(t/t₀)^{-2/3} > 0. So r increases. But v(t) - H(t)r(t): v(t) = peculiar velocity = a χ̇. χ̇ = (D/a₀)(t₀^{1/3})(-1/3) t^{-4/3} = -(D/(3a₀)) t₀^{1/3} t^{-4/3}. a χ̇ = a₀ (t/t₀)^{2/3} · (-(D/(3a₀)) t₀^{1/3} t^{-4/3}) = -(D/3) t₀^{1/3-2/3} t^{2/3-4/3} = -(D/3) t₀^{-1/3} t^{-2/3}. And v(t) from decay law: v₀(t₀/t)^{2/3} = (D/(3t₀)) t₀^{2/3} t^{-2/3} = (D/3) t₀^{-1/3} t^{-2/3}. ✓ consistent: v(t) = +? sign: the body moves toward origin, so its peculiar velocity is negative if positive r direction is away. Magnitude |v| = (D/3) t₀^{-1/3} t^{-2/3} = v₀ (t₀/t)^{2/3}. ✓. H r: H(t) = 2/(3t). r = D(t/t₀)^{1/3}. H r = (2D/(3t))(t/t₀)^{1/3} = (2D/3) t^{-2/3} t₀^{-1/3} = 2 |v(t)|. So ṙ = v_peculiar (inward, negative) + H r (outward) = -|v| + 2|v| = +|v| > 0. ✓ r increases. OK consistent: ṙ = +|v|. Interesting: in the critical case, the body's proper distance always increases at rate |v(t)|. Funny. And it never arrives (χ→0 only at ∞). Good. So the body launched at exactly H₀D/2 toward us always recedes in proper distance yet its comoving coordinate → 0. Physically: never arrives. For v₀ > H₀D/2, arrives at finite t_f. Let me double check arrival: χ(t_f) = 0 with χ(t) = D/a₀ - (3v₀t₀/a₀)(1-(t₀/t)^{1/3}). Set = 0: 1 - (t₀/t_f)^{1/3} = D/(3v₀t₀). Need RHS < 1 ⇒ v₀ > D/(3t₀) = H₀D/2 ✓. So t_f = t₀/(1 - D/(3v₀t₀))³. Check: as v₀ → ∞, t_f → t₀: a very fast launch arrives almost immediately (after traveling proper distance ≈ D... it arrives at time ~t₀; makes sense since it must traverse D at ~v; with v huge, time → 0 extra... t_f - t₀ → 0 ✓.) Let me also express with H₀: t₀ = 2/(3H₀), and D/(3v₀t₀) = D/(3v₀·2/(3H₀)) = H₀D/(2v₀). So t_f = (2/(3H₀)) [1 - H₀D/(2v₀)]^{-3}. And v₀ = (H₀D/2)·[1-(t₀/t_f)^{1/3}]^{-1} = (H₀D/2)[1-(2/(3H₀t_f))^{1/3}]^{-1}. Now write the final answer in Spanish, with derivation. Structure: 1. Modelo: EdS dust: a(t) = a₀(t/t₀)^{2/3}, H₀ = 2/(3t₀). 2. Geodésica: p ∝ 1/a ⇒ v(t) = v₀ a₀/a (v₀ ≪ c). 3. Comoving distance: dχ/dt = v/a; integrate to ∞: reach = 2v₀/H₀. 4. Condition: D = 2v₀/H₀ ⇒ v₀ = H₀D/2. 5. Comments: this is the threshold; with it, arrival is asymptotic (χ→0 as t→∞); for finite arrival time t_f, v₀ = (H₀D/2)/[1-(t₀/t_f)^{1/3}] = (H₀D/2)/[1-(2/(3H₀t_f))^{1/3}]. 6. Interpretation: reach with speed v is d_max = 2v/H₀ (like "distance reachable" result); the galaxy recedes at H₀D = 2v_min. 7. Numeric example. 8. Note relativistic generalization & condition v_min ≪ c ⇒ D ≪ 2c/H₀ (~8.5 Gpc); photons (v→c limit) always arrive in EdS since no event horizon. Also mention alternative: if instead the model were coasting a∝t (constant H), answer would be v₀ = H₀D; but for dust, H₀D/2. Actually wait — should I double check whether the intended answer might be exactly H₀D? Some textbook problems: "Show that in a matter-dominated universe, a galaxy at distance D can be reached by a probe launched at speed v if D < 2v/H₀." Hmm, I recall the problem from Ryden Chapter (problem 5.x?): "The maximum distance a projectile can travel..." Let me recall: Ryden Problem: "Consider a projectile launched from the origin at time t with speed v... In EdS, the maximum comoving coordinate it reaches..." I believe result: x_max = 3 v t^{1/3} t₀^{2/3}...? Let me verify with my formula: launch at t₀ with v₀: χ_∞ = 3v₀t₀/a₀. In comoving units: 3v₀t₀/a₀. Proper at t₀: 3v₀t₀ = 2v₀/H₀. Yes. Alternatively, the problem may come from a Spanish-language course ("modelo cosmológico de polvo cósmico") — likely a course on cosmology where they derived that in the dust model a(t) ∝ t^{2/3}. The phrase "(fluido sin presión)" clarifies dust. So the expected answer: v = H₀D/2. Hmm, but is the expected answer maybe H₀D? If the student is expected to reason "the body must cover distance D while peculiar velocity decays"... The factor depends on n=2/3. I'll present the full derivation and the general result so either way the reader can adapt. But the headline answer: v_min = H₀D/2 (EdS dust, flat). I should clearly state assumptions: flat matter-dominated (Ω_m = 1), nonrelativistic speeds, and that "reach" means comoving arrival at our worldline. Hmm, one more consideration: "para que logre llegar al origen" — maybe they want it to arrive at origin exactly (marginal) — i.e., the minimal speed such that arrival is possible at all: v = H₀D/2, arriving at t = ∞. I'll present that as the answer with the caveat. Let me also double-check the decay law via geodesic equation quickly: u^μ comoving... standard result: for FLRW, p ∝ 1/a for geodesic massive particle (since u_φ etc., energy redshift E ∝ 1/a for u in Hubble frame). Peculiar velocity relative to comoving observers: v = p c²/E_comoving... For nonrelativistic: v ∝ 1/a ✓. Also the momentum conservation argument: physical momentum of free particle p_phys ∝ 1/a (like photon redshift). ✓. Also I could derive via Newtonian: (a²χ̇)˙ = 0 as shown. Nice. Now also compute the trajectory for the general v₀, maybe give r(t) formula: χ(t) = D/a₀ − (3v₀t₀/a₀)[1 − (t₀/t)^{1/3}]. r(t) = a(t)χ(t) = (t/t₀)^{2/3}·D − (t/t₀)^{2/3}·3v₀t₀[1 − (t₀/t)^{1/3}] = D(t/t₀)^{2/3} − 3v₀t₀[(t/t₀)^{2/3} − (t/t₀)^{1/3}]. Arrival when χ=0. For the writeup, I'll do: **Planteamiento**: EdS: ρ = ρ_c, p=0 ⇒ a(t) = a₀(t/t₀)^{2/3}, H(t) = 2/(3t), t₀ = 2/(3H₀). Galaxy at χ_g = D/a₀. **Física del proyectil**: free geodesic; peculiar velocity decays: v(t) = v₀ a₀/a(t) = v₀ (t₀/t)^{2/3}. **Distancia comóvil recorrida**: Δχ(t) = ∫ v/a dt = (3v₀t₀/a₀)[1−(t₀/t)^{1/3}] → 3v₀t₀/a₀ as t→∞. **Condición de llegada**: max reach in proper distance today: d_max = a₀Δχ_∞ = 3v₀t₀ = 2v₀/H₀ ≥ D ⇒ **v₀ ≥ H₀D/2**. So v_min = H₀D/2: half the Hubble recession speed of the source galaxy. **Tiempo de llegada** (if v₀ > H₀D/2): t_f = t₀[1 − H₀D/(2v₀)]^{-3}. At exactly v₀ = H₀D/2, t_f → ∞ (arrives only asymptotically; its proper distance actually keeps growing, funny fact, but its comoving coordinate → 0). **Ejemplo numérico**: D = 100 Mpc, H₀=70: v ≈ 3500 km/s. **Comentarios**: - Equiv: max reach d_max = 2v/H₀ (the "twin" of this problem). - If model were coasting a ∝ t (w=−1/3) the answer would be H₀D; for radiation-dominated n=1/2... check: n=1/2 ⇒ 2n−1 = 0 ⇒ diverges! Interesting: in radiation domination (a∝t^{1/2}), a²χ̇ = const... χ_∞ = ∫ (v₀/a₀)(t₀/t) dt diverges logarithmically ⇒ any infinitesimal inward speed eventually arrives (since deceleration a∝ t^{1/2} is fast). Fun remark. For dust n=2/3 factor 1/2. Could include a small table: v_min = H₀D(2n−1)/n for a∝t^n. Dust: H₀D/2. - Assumptions: v₀ ≪ c (D ≪ 2c/H₀ ≈ 8.6 Gpc h⁻¹... for H₀=70, 2c/H₀ ≈ 8570 Mpc ≈ 8.6 Gpc). If D that large, need relativistic treatment; photons always arrive in EdS. Also mention that "under its inertia" = geodesic, no rocket, and the expansion doesn't exert force but redshifts its momentum; that's why a finite launch speed has finite reach. I should present derivation cleanly. Also define "origen" as our worldline χ=0. Potential subtlety: The problem might intend the projectile launched at the galaxy toward the origin with the galaxy receding; but also one might think the projectile must "catch up" from behind? No—the galaxy is away from us; projectile launched toward us. Double-check the decay v ∝ 1/a applies to peculiar velocity measured by comoving observers: yes. Also, present the general result: χ̇ a² = const ⇒ v a = const. Also alternative derivation via energy in expanding universe? Not needed. Let me also double check t₀ = 2/(3H₀) for EdS: H = 2/(3t) ⇒ t₀ = 2/(3H₀) ✓. Age ~9.3 Gyr for H₀=70. ✓. Compute example: H₀D/2 with D=100 Mpc: 70×100/2 = 3500 km/s ✓. Also express answer in terms of Hubble distance: v_min = cD/(2 D_H) with D_H = c/H₀. Now, write the answer in Spanish (the question is in Spanish). Use clear math. Include: - Setup and assumptions. - Key steps with equations. - Final boxed answer: v₀ = H₀D/2 (mínima), con llegada asintótica; fórmula para llegada en tiempo finito. - Remarks. Also perhaps derive a(t) from Friedmann for dust: ρ ∝ a⁻³, Friedmann flat: ȧ/a = H₀ a^{-3/2}... a(t) = (3H₀t/2)^{2/3} a₀. Provide. Let me make sure about "impulsado... para que logre llegar al origen solo bajo su inercia": the impulse gives initial velocity; afterwards no forces (geodesic). Yes. Also note: since we ask "a qué velocidad debe ser impulsado" — the launch speed relative to the galaxy (peculiar velocity at launch) is the meaningful quantity. Answer: v₀ = H₀D/2 toward the origin. Note that relative to us, it initially recedes at H₀D − H₀D/2 = H₀D/2; yet it eventually reaches us (if v₀ slightly above) because the Hubble flow decelerates (ä<0) while... hmm, actually the Hubble flow deceleration is what allows the comoving approach: since ä<0, the expansion decelerates, so the galaxy's recession slows; the projectile's peculiar velocity decays too, but the net effect must be computed—the integral result stands. Wait, is it right that with v₀ > H₀D/2 it arrives even though initially receding? χ̇ < 0 from the start (it moves inward in comoving coordinates immediately, since dχ/dt = −v/a < 0). Yes, comoving coordinate decreases from t₀. Its proper distance may initially grow (if H r > v) but χ decreases monotonically and hits 0 at t_f. At arrival, r=0 trivially. Before arrival, r could have grown; but arrival is defined in comoving terms (r=0 exactly when χ=0). Actually r = aχ ⇒ r=0 iff χ=0. So arrival at t_f means r(t_f)=0. Fine. Sanity check with the "turnaround": ṙ = v_pec − H r. At launch: −v₀ + H₀D. If v₀ < H₀D it initially recedes. Since v₀_min = H₀D/2 < H₀D, it initially recedes for any near-threshold launch, then comes back. The turnaround in proper distance occurs when v(t) = H(t) r(t). OK. Also nice cross-check: the known result that in EdS a projectile can reach at most d_max = 2v/H₀ is exactly time-reverse of: an object at rest at distance D "falls" ... hmm not needed. Potential alternative interpretation: maybe "llegar al origen" by t→∞? Or maybe they want it to arrive "eventually". Our answer covers. Also maybe they'd like the answer expressed as v = H₀D/2 and note the interesting fact that this is half the Hubble velocity. Or maybe they intend a different model: "polvo cósmico" maybe they have in mind the "dust" solution a ∝ t^{2/3} — yes. Let me also provide the derivation of v(t) ∝ 1/a via geodesic or via "momentum redshift", and the Newtonian shortcut (a²χ̇ = const). Structure of final answer (in Spanish): # Planteamiento - Modelo EdS: a(t)=a₀(t/t₀)^{2/3}, H₀=2/(3t₀); galaxia en χ_g=D/a₀. - "Solo bajo su inercia" ⇒ geodésica; singularidad: velocidad peculiar decae v∝1/a. # Desarrollo 1) Geodésica ⇒ a·v = cte ⇒ v(t)=v₀(a₀/a). 2) dχ/dt = v/a ⇒ χ(t) integrals; total reach χ_∞=3v₀t₀/a₀. 3) Condición a₀χ_∞ ≥ D ⇒ v₀ ≥ H₀D/2. # Resultado v₀,min = H₀D/2 (= mitad de la velocidad de Hubble de la galaxia). - Con esa velocidad exacta llega solo asintóticamente (t→∞; de hecho su distancia propia nunca deja de crecer, pero su coordenada comóvil →0). - Con v₀ > H₀D/2 llega en tiempo finito t_f = t₀[1−H₀D/(2v₀)]⁻³, es decir v₀ = (H₀D/2)[1−(t₀/t_f)^{1/3}]⁻¹. # Ejemplo D=100 Mpc, H₀=70 ⇒ v₀ ≈ 3.5×10³ km/s. # Observaciones - Equivalencia: alcance máximo d_max=2v₀/H₀ (problema inverso). - General a∝tⁿ: v_min=H₀D(2n−1)/n ⇒ polvo (n=2/3): H₀D/2; "coasting" (n=1): H₀D. - Validez: v₀≪c ⇒ D ≪ 2c/H₀; luz siempre llega en EdS (sin horizonte de eventos). I think that's thorough. Also mention assumption of flatness (Ω=1): the "cosmic dust model" typically means EdS. If open dust universe, a(t) = A sinh^{2/3}(...), results change slightly; but stick to flat. Should I double-check the open dust case quickly for interest? Open k=−1 dust: a(η) = (GM... ) parametric: a = A(cosh η − 1), t = A(sinh η − η) with... H(t) varies; at late times a ∝ t (Milne). Then χ_∞ = ∫ v₀a₀/a² dt. Late-time behavior a ∝ t ⇒ integral converges (as n=1 case). Early-time a ∝ t^{2/3} within integral. Result: reach = v₀ ∫_{t₀}^∞ a₀ dt/a(t)²... = v₀ ∫ (a₀/a)² dt. With a₀/a = 1/(... ) For open universe with Ω₀, a(t) = (Ω₀/(2(1−Ω₀))^{1/3}...)... skip; not needed. The problem says only H₀ given ⇒ assume flat EdS. One more check of the integral for χ(t) and the statement "proper distance never stops growing in critical case" — fine as a curiosity; maybe include briefly. Alternatively the expected answer might ignore the subtlety and simply state v = H₀D/2 with "llega en el límite t→∞". Good. Also note: some textbooks (e.g., Ryden problem 5.6? or "problem: How far can a spaceship travel?") give exactly d_max = 2v₀/H₀ in EdS. Actually I recall Ryden Problem: "A projectile is shot with speed v₀ from the origin at t=t₀ in EdS; show its maximum comoving coordinate is ξ_max = 3(v₀/c)(...)"? Let me just trust my derivation; I can double check by computing a specific integral numerically-ish: v₀=1, t₀=1, a₀=1: χ_∞ = 3. Also compute Δχ(2t₀) = 3(1)(1−(1/2)^{1/3}) = 3(1−0.7937)=0.619. And with formula 3v₀t₀[1−(t₀/t)^{1/3}] ✓. Let me also present the derivation of momentum redshift: For FLRW metric ds² = −c²dt² + a²(t)dχ², geodesic: du^χ/dτ + 2H u^χ = 0 ⇒ u^χ ∝ a⁻²; peculiar velocity v = a u^χ/... anyway v∝1/a. Newtonian shortcut: χ̈ + 2Hχ̇=0. Also, define that "impulse" gives speed v₀ measured locally (by comoving observer at galaxy). Time to write the answer. Keep it reasonably concise but complete, with formulas in LaTeX-ish. Spanish language. Also include the neat physical summary: "En el modelo de polvo, un proyectil lanzado con velocidad v puede cubrir (en distancia propia de hoy) como máximo 2v/H₀; por tanto, para alcanzar una galaxia (u origen) a distancia D se necesita v ≥ H₀D/2. Con el mínimo, la llegada es asintótica." Also mention that this equals half the galaxy's recession speed — a curious result. Also the finite-time formula and trajectory: χ(t) = D/a₀ − 3(v₀t₀/a₀)[1 − (t₀/t)^{1/3}]. Set χ(t_f)=0. Write final answer accordingly. Use headings, math. Since user wrote in Spanish, answer in Spanish. Format: medium length; headers or bold; equations displayed. Probably ~400-600 words with equations. Use LaTeX notation with $$ ... $$? The chat renders LaTeX? Likely fine. I'll use $...$ inline and display equations. Let me draft. --- **Modelo (polvo, plano = Einstein–de Sitter)** Con p=0 y k=0: ρ∝a⁻³, Friedmann ⇒ a(t)=a₀(t/t₀)^{2/3}, H(t)=2/(3t), t₀=2/(3H₀). Galaxia en coordenada comóvil χ_g = D/a₀. El origen: χ=0. **Movimiento inercial (geodésica)** Para un cuerpo libre, la velocidad peculiar decae: $$v(t) = v_0\,\frac{a_0}{a(t)} = v_0 \left(\frac{t_0}{t}\right)^{2/3}$$ (v₀ = velocidad de lanzamiento medida localmente en la galaxia). Esto sigue de la geodésica (o de χ̈+2Hχ̇=0 ⇒ (a²χ̇)'=0, con v=aχ̇). **Distancia comóvil que puede cubrir** dχ/dt = v/a = (v₀/a₀)(t₀/t)^{4/3} $$\Delta\chi(t) = \frac{3v_0t_0}{a_0}\left[1-\left(\frac{t_0}{t}\right)^{1/3}\right] \xrightarrow{t\to\infty} \frac{3v_0t_0}{a_0}.$$ **Condición de llegada** En distancia propia (a escala de hoy): alcance máximo d_max = a₀Δχ_∞ = 3v₀t₀ = 2v₀/H₀. ⇒ Llega al origen ⟺ d_max ≥ D: $$\boxed{v_0 \ge \frac{H_0 D}{2}}$$ velocidad de impulso mínima = la mitad de la velocidad de recesión Hubble de la galaxia (H₀D). **Tiempo de llegada** (si v₀ > H₀D/2): χ(t_f)=0 ⇒ $t_f = t_0\left[1-\frac{H_0D}{2v_0}\right]^{-3} = \frac{2}{3H_0}\left[1-\frac{H_0D}{2v_0}\right]^{-3}$. Con exactamente v₀=H₀D/2, t_f→∞: llega solo asintóticamente. **Ejemplo**: D=100 Mpc, H₀=70 km s⁻¹ Mpc⁻¹ ⇒ v₀ ≈ 3500 km/s ≪ c ✓. **Comentarios**: - Resultado dual: con velocidad v se puede llegar como máximo a d = 2v/H₀ (en EdS). - Para una expansión general a∝tⁿ: v_min = H₀D(2n−1)/n: polvo (n=2/3) ⇒ H₀D/2; coasting a∝t ⇒ H₀D. - Válido mientras v₀≪c (D ≪ 2c/H₀ ≈ 8.6 Gpc). La luz siempre llega en EdS (no hay horizonte de eventos). Also clarify assumption: universe flat matter-dominated; if their "dust model" is exactly this, answer stands. I think adding the derivation that peculiar velocity decays: brief justification: geodesic: χ̈ + 2(ȧ/a)χ̇ = 0 ⇒ a²χ̇ = cte ⇒ v = aχ̇ ∝ 1/a. Provide both GR/Newtonian statements. Also maybe note the funny fact: launched with the minimal speed, the body always recedes in proper distance (r(t) = D(t/t₀)^{1/3}) while approaching in comoving coordinates — the arrival is at t=∞. Maybe include as a footnote/curiosity. It emphasizes that "reaching" should be understood in comoving terms. Hmm, this could confuse; but it's honest and interesting. I'll include briefly. Actually, wait. Let me reconsider whether "llegar al origen" with v=H₀D/2 at t→∞ is physically meaningful. The body's worldline: χ(t)→0. As t→∞ both worldlines (ours and the body's) exist; do they intersect? Only at t=∞, never at finite time. In the conformal diagram, EdS: conformal time η ∝ t^{1/3}; χ(t) ∝ t^{-1/3} ∝ 1/η. So in (η, χ) the body's path: χ = k/η → 0 as η→∞. Never reaches χ=0 at finite η. So genuinely never arrives — approaches asymptotically. So the strict statement: arrival requires v₀ > H₀D/2; the threshold value gives asymptotic approach. For the problem's purpose, the answer "v₀ = H₀D/2" is the standard expected threshold ("mínima velocidad"), with the caveat noted. Also maybe the problem intends "llegar" as "eventually arrive" — the infimum speed. I'll present both. Additionally, I might mention that if the launch happens at cosmic time t₀ = today with D the present proper distance. Fine. Also note: relative speed of approach measured at the origin initially is v₀ − H₀D < 0 (receding) for threshold launch. Interesting: you throw it toward us, yet it recedes at first — it must outlast the decelerating expansion. Actually does it always initially recede for v₀ slightly > H₀D/2? ṙ(t₀) = −v₀ + H₀D < 0? If v₀ < H₀D yes recedes initially. Since required v₀ is typically ≪ H₀D? For D=100 Mpc: v₀=3500, H₀D=7000 ⇒ recedes initially, turnaround, comes back. OK. Turnaround time: solve v(t) = H(t) r(t). Not needed. Write the final answer now. Keep math clean. Use Spanish. Use display math. Reasonable length. I'll include a short "resultado" box via bold. One more consideration: Should the "speed" be the peculiar speed measured at the galaxy, or the speed seen from the origin? Peculiar at launch is the natural "impulse" speed. State it. Also express answer also as v₀ = (1/2) H₀ D, and note v_recession = H₀D. Final answer structure: 1. Supuestos/modelo 2. Geodésica y decaimiento de v 3. Alcance comóvil 4. Condición y resultado (box): v_min = H₀D/2 5. Llegada en tiempo finito (fórmula t_f) 6. Ejemplo numérico 7. Observaciones (dual result, general a∝tⁿ, validez relativista, luz siempre llega) Go. ## Planteamiento En el modelo de **polvo cósmico** (fluido sin presión, universo plano = Einstein–de Sitter), la Friedmann da $$a(t)=a_0\left(\frac{t}{t_0}\right)^{2/3},\qquad H(t)=\frac{2}{3t},\qquad t_0=\frac{2}{3H_0}.$$ La galaxia está en la coordenada comóvil $\chi_g = D/a_0$, y el origen en $\chi=0$. "Solo bajo su inercia" significa que tras el impulso el cuerpo sigue una **geodésica**: ninguna fuerza actúa sobre él, pero la expansión le "hace corrimiento al rojo" a su momento. ## Paso 1: la velocidad peculiar decae como $1/a$ La geodésica en FLRW da (equivalentemente, en lenguaje newtoniano, $\ddot\chi + 2H\dot\chi = 0$): $$\frac{d}{dt}\left(a^2\dot\chi\right)=0 \quad\Rightarrow\quad v(t)\,a(t)=\text{cte} \quad\Rightarrow\quad v(t)=v_0\frac{a_0}{a}=v_0\left(\frac{t_0}{t}\right)^{2/3},$$ donde $v_0$ es la velocidad de lanzamiento medida localmente en la galaxia. ## Paso 2: alcance comóvil total $$\frac{d\chi}{dt}=\frac{v}{a}=\frac{v_0}{a_0}\left(\frac{t_0}{t}\right)^{4/3}$$ $$\Delta\chi(t)=\frac{3v_0t_0}{a_0}\left[1-\left(\frac{t_0}{t}\right)^{1/3}\right]\;\xrightarrow{t\to\infty}\;\frac{3v_0t_0}{a_0}.$$ El cuerpo solo puede cubrir una distancia comóvil **finita**: en distancia propia (a escala de hoy), $$d_{\max}=a_0\,\Delta\chi_\infty = 3v_0t_0 = \frac{2v_0}{H_0}.$$ ## Paso 3: condición de llegada Para que alcance el origen se necesita $d_{\max}\ge D$: $$\boxed{\,v_0 \ge \frac{H_0 D}{2}\,}$$ Es decir: hay que lanzarlo hacia nosotros con **la mitad de la velocidad de recesión de Hubble** de la galaxia ($v_{\rm rec}=H_0D$). Con exactamente $v_0=H_0D/2$ la llegada es asintótica ($t\to\infty$); de hecho su distancia propia nunca deja de crecer, pero su coordenada comóvil $\chi\to 0$. **Llegada en tiempo finito:** si $v_0>H_0D/2$, imponiendo $\chi(t_f)=0$: $$t_f = t_0\left[1-\frac{H_0D}{2v_0}\right]^{-3} =\frac{2}{3H_0}\left[1-\frac{H_0D}{2v_0}\right]^{-3}.$$ ## Ejemplo numérico $D=100$ Mpc, $H_0=70\ \mathrm{km\,s^{-1}Mpc^{-1}}$: $$v_0 \gtrsim \tfrac12(70)(100)\approx 3500\ \mathrm{km/s}\ll c \quad\checkmark$$ ## Observaciones - **Resultado dual:** en el universo de polvo, un proyectil lanzado con velocidad $v$ puede llegar como máximo a $d=2v/H_0$. - Para una expansión general $a\propto t^n$ se obtiene $v_{\min}=H_0D\,\dfrac{2n-1}{n}$: polvo ($n=2/3$) $\Rightarrow H_0D/2$; "coasting" $a\propto t$ $\Rightarrow H_0D$. - Válido mientras $v_0\ll c$, es decir $D\ll 2c/H_0\approx 8.6$ Gpc. La luz (límite $v\to c$) siempre llega en este modelo, pues el universo de polvo no tiene horizonte de eventos: $\int_{t_0}^\infty dt/a(t)=\infty$.

A que velocidad debe ser impulsado un cuerpo desde una galac...