Çözüldü Integral Application to Physics

Konusu 'Akademik Soru Çözümleri ve Kaynakları' forumundadır ve Bulnir tarafından 13 Mayıs 2020 başlatılmıştır.

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  1. Bulnir

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    HW-3.jpeg Yardım edebilir misiniz?

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  3. Honore

    Honore Yönetici Yönetici

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    Müh. (Elk./Bilg.)
    Taking Earth's radius as r from its center and a very small thickness of it as dr, the differential volume is dV = 4π(r^2)dr, and the density as a function of radius is ρ(r). Then the differential mass is dm = [ ρ(r) ]·dV = [ ρ(r) ]·4π(r^2)dr .
    So, the actual mass is found by integrating dm from its center to surface by using MKS units;
    m = (lower limit r1 = 0, upper limit r2 = 6.37·10^(6), ∫ [ ρ(r) ]·4π(r^2) dr
    m = (lower limit r1 = 0, upper limit r2 = 6.37·10^(6), 4π· ∫ [ ρ(r) ]·(r^2) dr....(I)
    Now that the density linearly decreases, then its line equation is supposed to have a negative slope. Therefore, the equation of the density function is, with positive real coefficients m and r;
    ρ(r) = -m·r + n....(II)
    Recalling from analytic geometry, the equation of the line passing through two points;
    [ ρ(r) - ρ(0) ] / [ ρ(0) - ρ(6370 km) ] = (r - 0) / (0 - 6370 km)
    [ ρ(r) - 12500 km ] / [ 12500 km - 950 (kg / m^3) ] = (r - 0) / (0 - 6370 km) and organizing in the form of ρ(r) = -m·r + n, converting from km to meter unit and writing in the scientific exponential form;
    ρ(r) = -{ 11550 / [ 6.37·10^(6) ] }·r + 12500
    ρ(r) ≈ -1.81·[ 10^(-3) ]·r + 12500....(III)
    Comparing (III) to (II), the m and n coefficients are m = 1.81·[ 10^(-3) ]....(IV), n = 12500....(V)
    Completing the integration at (I) using (II);
    m ≈ (lower limit r1 = 0, upper limit r2 = 6.37·10^(6), 4π·∫ (-m·r + n)(r^2) dr
    m ≈ (lower limit r1 = 0, upper limit r2 = 6.37·10^(6), 4π·∫ [ -m·(r^3) + n·(r^2) ] dr
    m ≈ (lower limit r1 = 0, upper limit r2 = 6.37·10^(6), 4π·|-m·(r^4) / 4 + n·(r^3) / 3|
    m ≈ 4π·{ -m·[ 6.37·10^(6) ]^4 / 4 + n·[ 6.37·10^(6) ]^3 / 4 }....(VI)
    Plugging the values of m and n at (IV) and (V) respectively;
    m ≈ 4π·{ -1.81·[ 10^(-3) ]·(6.37·10^6)^4 / 4 + 12500·(6.37·10^6)^3 / 3 }
    m ≈ 4.17·[ 10^(24) ] kg

    Note: Since the real mass of the Earth is 5.97·[ 10^(24) ] kg, it would be correct to conclude that the real change of the density is different than the linear decrease assumption supposed in this homework.
    ---
    Sorunun Yedeği.: https://i.ibb.co/r5wX8dq/HW-3.jpg
    Bulnir bunu beğendi.
  4. Bulnir

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