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    Truong Gia Binh IAAC Pre-Final Solution 2024

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    Khàn Tả Trần

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    Truong Gia Binh IAAC Pre-Final Solution 2024

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    Khàn Tả Trần
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    Truong Gia Binh IAAC pre-final solution 2024

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    Truong Gia Binh IAAC Pre-Final Solution 2024

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    of 5

    IAAC Pre-final round solution

    Truong Gia Binh

    A1

    a)

    As the rocket is launch from the equator,it can take full advantage of the Earth's self-rotational
    speed which is about 1670 km/h In the coordinate system, the center of the earth is the origin

    b)

    Consider that the motion of an object located on the equator is a circular motion whose diameter is
    the diameter of the earth. We have the equation of relation between linear velocity and angular
    velocity 𝑣 = 𝑤𝑟

    With 𝑣 is the speed of the object, 𝑟 is the diameter of the circular motion and 𝑤 is angular velocity
    2𝜋 𝜋
    A full rotation of the Earth take 24 hours and a full cirle is 2𝜋 radiant → 𝑤 = 24
    = 12 (𝑟𝑎𝑑𝑖𝑎𝑛𝑡/ℎ𝑜𝑢𝑟)
    𝜋
    → 𝑣𝑒𝑞𝑢𝑎𝑡𝑜𝑟 = 𝑤𝑟 = 6371 ≈ 1668 (𝑘𝑚/ℎ)
    12
    The earth rotates around its principal axis and surfaces built on latitude lines are perpendicular to
    that axis, which make the angular velocity w of any object that lies on any altitude the same.

    So the velocity of the Eath’s rotation at the


    latitude 𝜙° is 𝑣(𝜙) = 𝑤𝑟ϕ

    As the figure below shows, 𝑟ϕ = cos( ϕ) 𝑟𝑒𝑎𝑟𝑡ℎ

    → 𝑣(𝜙) = 𝑤𝑟ϕ = 𝑤𝑟𝑒𝑎𝑟𝑡ℎ cos( ϕ)


    = 𝑤𝑟0° cos( ϕ)
    = 𝑣𝑒𝑞𝑢𝑎𝑡𝑜𝑟 cos( ϕ)
    ≈ 1668 cos(ϕ) (𝑘𝑚/ℎ)
    → 𝒗(𝝓) = 𝒘𝒓𝛟 ≈ 𝟏𝟔𝟔𝟖 𝐜𝐨𝐬(𝛟) (𝒌𝒎/𝒉)

    c)

    Apply the equation above 𝑣(𝜙) ≈ 1668 cos(ϕ) (𝑘𝑚/ℎ)

    To calculate the rotational speed at 5oS: 𝒗(𝟓°) ≈ 𝟏𝟔𝟔𝟖 𝐜𝐨𝐬(𝟓°) ≈ 𝟏𝟔𝟔𝟏 (𝒌𝒎/𝒉)
    To calculate the rotational speed at 80oN: 𝒗(𝟖𝟎°) ≈ 𝟏𝟔𝟔𝟖 𝐜𝐨𝐬(𝟖𝟎°) ≈ 𝟐𝟗𝟎 (𝒌𝒎/𝒉)

    A2

    a)

    The amount did ISS descend between December 2023 and April 2024:
    (416.6 − 414.4) + (417.8 − 415.7) + (418.5 − 416.7) + (418.8 − 415.7) + (419 − 416.9) + (419.9 − 417)
    = 14.2 (𝑘𝑚)

    So ISS descended about 14.2 km between December 2023 and April 2024

    The amount did ISS ascend between December 2023 and April 2024:
    (417.8 − 414.4) + (418.5 − 415.7) + (418.8 − 416.7) + (419 − 415.7) + (419.9 − 416.9) = 14.6 (𝑘𝑚)

    So ISS ascended about 14.6 km between December 2023 and April 2024

    b)

    During 4 months from December 2023 to April 2024, the ISS descended by 14.2 km. So the average descent
    rate of the ISS is 14.2/4 or 3.55 km/month.

    c)

    At April 2024, the ISS currently at the height of 417 km above ground, so it has to descend 317 km to descend
    to the edge of space.

    Time for the ISS to descend to the edge of space: 317 ÷ 3. .55 ≈ 𝟖𝟗. 𝟑 (𝒎𝒐𝒏𝒕𝒉𝒔)

    B1

    a)

    Let x represents rate of descent 3.55 km/month = 1.369 x 10-3 m/s

    R is the distance from the ISS to the center of the Earth.

    mearth is the mass of the earth.

    𝐺𝑚𝑒𝑎𝑟𝑡ℎ 𝑑𝑣 −𝐺𝑚𝑒𝑎𝑟𝑡ℎ 𝑑𝑣 +𝐺𝑚𝑒𝑎𝑟𝑡ℎ


    𝑣2 = ⇒ 2𝑣 = 2
    ⇒ 2𝑣 =
    𝑅 − 𝑥𝑡 𝑑(𝑅 − 𝑥𝑡) (𝑅 − 𝑥𝑡) 𝑡𝑥𝑑𝑥 (𝑅 − 𝑥𝑡)2

    𝐺𝑚𝑒𝑎𝑟𝑡ℎ 𝑚𝐼𝑆𝑆 𝑥 1 𝐺𝑚𝑒𝑎𝑟𝑡ℎ 𝑚𝐼𝑆𝑆 𝑥 1 𝐺𝑚𝑒𝑎𝑟𝑡ℎ


    ⇒ 𝐹𝑑𝑟𝑎𝑔 = 2
    = 𝜌𝐶𝑑 𝐴𝑣 2 ⇔ 2
    = 𝜌𝐶𝑑 𝐴
    2𝑣(𝑅 − 𝑥𝑡) 2 2𝑣(𝑅 − 𝑥𝑡) 2 𝑅 − 𝑥𝑡
    𝑥𝑚𝐼𝑆𝑆
    ⇔ = 𝜌𝐶𝑑 𝐴
    𝑣(𝑅 − 𝑥𝑡)
    𝑥𝑚𝐼𝑆𝑆 𝑥𝑚𝐼𝑆𝑆
    ⇔𝜌= =
    𝑣(𝑅 − 𝑥𝑡)𝐶𝑑 𝐴
    √𝐺𝑚𝑒𝑎𝑟𝑡ℎ (𝑅 − 𝑥𝑡)𝐶𝑑 𝐴
    𝑅 − 𝑥𝑡
    1.369 × 10−3 × 4.5 × 105
    ≈ ≈ 𝟏. 𝟗 × 𝟏𝟎−𝟏𝟐 (𝒌𝒈𝒎−𝟑 )
    −11 × 5.97 × 1024
    √ 6.67 × 10 (6788000 − 1.369 × 10−3 × 𝑡) × 1.3 × 4800
    6788000 − 1.369 × 10−3 × 𝑡
    𝜌 ≈ 𝟏. 𝟗 × 𝟏𝟎−𝟏𝟐 (𝒌𝒈𝒎−𝟑 )
    b)

    𝑚𝑠𝑒𝑐𝑜𝑛𝑑 = 𝜌𝐴𝑣 ≈ 1.9 × 10−12 × 4800 × 7659 ≈ 7 × 10−5 (𝑘𝑔/𝑠)

    ⇒ 𝑚𝑑𝑎𝑦 ≈ 7 × 10−5 × 3600 × 24 ≈ 𝟔. 𝟎𝟒𝟖 (𝒌𝒈)

    B2

    a)

    3 regions are bulge (A), disk (B) and halo (C).

    b)

    Write down the equation into the form that I feel easier to read:
    𝑟 𝑟 𝑟
    (Ω𝐴 − ) (Ω𝐵 − ) (Ω𝐶 − )
    𝜌(𝑟) = Ψ (𝑒 𝑅𝐴 +𝑒 𝑅𝐵 +𝑒 𝑅𝐶 )

    Using the given model parameters:


    𝑟 𝑟 𝑟
    (21− ) (−3− ) (−8− )
    𝜌(𝑟) = 10−4 (𝑒 20 +𝑒 12000 +𝑒 50000 )

    c)

    Through the double logarithmic plot of the density distribution decrease continuously as it gets
    further from the center of milky way.

    If we take a is a constant a which 0<α<130000 (ly) the p(α) represents the density of every location
    that stay within the milkyway and have the distance 130000 light-years away from the center, and
    represents everypoint on it we can get a circular wall (consider it as an perfect wall) that have a
    height of 1000 light-years and a radius of α light-years.
    So the number of stars in this volume of wall is approximately p(α) x volume of the wall

    So the number of stars in Milky Way would be ≈ ∑(𝑝(𝑎) × 𝑉𝑤𝑎𝑙𝑙 ) ( as long as the sum of the wall is
    equal to volume of the Milky Way)

    When we consider thickness of the wall is nearly 0 (volume is not exit at this point anymore) , nearly
    infinite walls will be need to use to fulfill volume of milky way. When it comes to calculate this, it is
    also why intergrate was born:
    130000 130000
    ∫ (𝑝(𝛼) × 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑎𝑙𝑙)𝑑𝛼 = ∫ (𝑝(𝛼) × 2𝜋𝛼 × 1000)𝑑𝛼
    0 0
    ≈ 𝟑. 𝟑𝟏𝟒𝟔 × 𝟏𝟎𝟏𝟏 (𝒔𝒕𝒂𝒓𝒔)

    C1

    a) The object in the article is a massive, compact, quiescent galaxy at a photometric redshift
    of zphot ≈2 with a complete Einstein ring (JWST-ER1). It was discovered in James Webb Space
    Telescope NIRCam observations in the context of the COSMOS-Web project.
    b) The figure 1.a show the JWST-ER1, which consists of a compact early type galaxy (JWST-
    ER1g) that is the small shine red and yellow dot or circle in the middle of the picture. The
    JWST-ER1r is the Einstein ring that surround the JWST-ER1g and it has two red concentration
    on the side. There also a smaller dot outside the ring.
    c) our Milky Way age: about 13.6 Gyr ; mass: 1.15×1012 M☉ ; radius: 13.4 kpc
    JWST-ER1 age: maybe about 10 GYr mass:1.02×1012 radius: 6.6 kpc
    We can see that with the radius about half of our Milky Way, its mass is nearly the same as
    our mass
    d) sà
    e) IMF is a method to measure mass of stars of stars cluster . In the article, authors use 3
    types of IMFs: Charbier, Salpeter, steeper-Salpeter. As in Fig3a, IMF measurements have
    something base on the light of the stars and it results might be precisely to what is expected
    base on the type of IMF used.
    f) .

    C2

    a) They use new observations of Io’s polar regions by the Juno spacecraft Jovian Infrared
    Auroral Mapper to complete near-infrared coverage, revealing the global distribution and
    magnitude of thermal emission from Io’s currently erupting volcanoes because full mapping
    of the moon has not previously been possible.
    b) volcanic thermal emission from the poles of Io as compared to those at lower latitudes are
    a sign of lithospheric dichotomies that prevent the volcanism from advancing towards the
    poles, especially in the south polar area.
    c) Loki Patera and Tvashtar Paterae are two powerful volcanic features on Io; both of them are
    known for its intense volcanic activity.
    d) 20 in the North polar cap and 12 in the South polar cap relatively.
    e) Mechanism that lead to volcanic on Io is the tidal heating, which is created by the
    interaction between the Io and other planets or asteroids.
    f)

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