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Deduce Newton's law of universal gravitation from Kepler's laws of planetary motion.

Physics

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Answer

2.22×10−7 Nkg−1\text{2.22} \times 10^{-7} \text{ Nkg}^{-1}

Step-by-step solution

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Ili kukokotoa ukubwa wa nguvu ya uvutano (gravitational field intensity) kwenye sehemu iliyotajwa, tutatumia kanuni ya nguvu ya uvutano kwa kila masi na kisha kuongeza vekta hizo.

  • Masi ya kwanza, M1=800 kgM_1 = 800 \text{ kg}
  • Masi ya pili, M2=600 kgM_2 = 600 \text{ kg}
  • Umbali kutoka M1M_1 hadi sehemu hiyo, r1=0.2 mr_1 = 0.2 \text{ m}
  • Umbali kutoka M2M_2 hadi sehemu hiyo, r2=0.15 mr_2 = 0.15 \text{ m}
  • Kipimo cha uvutano, G=6.67×10−11 Nm2kg−2G = 6.67 \times 10^{-11} \text{ Nm}^2\text{kg}^{-2}
  • Umbali kati ya masi hizo mbili, d=0.25 md = 0.25 \text{ m}

  1. Kokotoa nguvu ya uvutano (E1E_1) kutokana na masi ya 800 kg.

    Tunatumia fomula E=GMr2E = \frac{GM}{r^2}.

    E1=GM1r12E_1 = \frac{GM_1}{r_1^2} E1=(6.67×10−11 Nm2kg−2)(800 kg)(0.2 m)2E_1 = \frac{(6.67 \times 10^{-11} \text{ Nm}^2\text{kg}^{-2})(800 \text{ kg})}{(0.2 \text{ m})^2} E1=533.6×10−11 Nm2kg−10.04 m2E_1 = \frac{533.6 \times 10^{-11} \text{ Nm}^2\text{kg}^{-1}}{0.04 \text{ m}^2} E1=1.334×10−7 Nkg−1E_1 = 1.334 \times 10^{-7} \text{ Nkg}^{-1}

  2. Kokotoa nguvu ya uvutano (E2E_2) kutokana na masi ya 600 kg.

    E2=GM2r22E_2 = \frac{GM_2}{r_2^2} E2=(6.67×10−11 Nm2kg−2)(600 kg)(0.15 m)2E_2 = \frac{(6.67 \times 10^{-11} \text{ Nm}^2\text{kg}^{-2})(600 \text{ kg})}{(0.15 \text{ m})^2} E2=400.2×10−11 Nm2kg−10.0225 m2E_2 = \frac{400.2 \times 10^{-11} \text{ Nm}^2\text{kg}^{-1}}{0.0225 \text{ m}^2} E2≈1.7787×10−7 Nkg−1E_2 \approx 1.7787 \times 10^{-7} \text{ Nkg}^{-1}

  3. Bainisha mwelekeo wa vekta za nguvu ya uvutano.

    Nguvu ya uvutano huelekea kwenye masi inayoizalisha. Ili kujua kama vekta hizi ziko perpendicular, tunachunguza umbali uliotolewa.
    r12+r22=(0.2 m)2+(0.15 m)2=0.04 m2+0.0225 m2=0.0625 m2r_1^2 + r_2^2 = (0.2 \text{ m})^2 + (0.15 \text{ m})^2 = 0.04 \text{ m}^2 + 0.0225 \text{ m}^2 = 0.0625 \text{ m}^2.
    Umbali kati ya masi hizo mbili ni d=0.25 md = 0.25 \text{ m}, hivyo d2=(0.25 m)2=0.0625 m2d^2 = (0.25 \text{ m})^2 = 0.0625 \text{ m}^2.
    Kwa kuwa r12+r22=d2r_1^2 + r_2^2 = d^2, hii inamaanisha kuwa sehemu tunayokokotoa nguvu ya uvutano inaunda pembe ya 90∘90^\circ na masi hizo mbili. Hivyo, vekta za E1E_1 na E2E_2 ziko perpendicular.

  4. Kokotoa ukubwa wa nguvu ya uvutano jumla (EnetE_{net}).

    Kwa kuwa vekta za E1E_1 na E2E_2 ziko perpendicular, ukubwa wa nguvu ya uvutano jumla utapatikana kwa kutumia theorem ya Pythagoras:

    Enet=E12+E22E_{net} = \sqrt{E_1^2 + E_2^2} Enet=(1.334×10−7 Nkg−1)2+(1.7787×10−7 Nkg−1)2E_{net} = \sqrt{(1.334 \times 10^{-7} \text{ Nkg}^{-1})^2 + (1.7787 \times 10^{-7} \text{ Nkg}^{-1})^2} Enet=(1.779556×10−14 N2kg−2)+(3.16375×10−14 N2kg−2)E_{net} = \sqrt{(1.779556 \times 10^{-14} \text{ N}^2\text{kg}^{-2}) + (3.16375 \times 10^{-14} \text{ N}^2\text{kg}^{-2})} Enet=4.943306×10−14 N2kg−2E_{net} = \sqrt{4.943306 \times 10^{-14} \text{ N}^2\text{kg}^{-2}} Enet≈2.223×10−7 Nkg−1E_{net} \approx 2.223 \times 10^{-7} \text{ Nkg}^{-1}

    Ukubwa wa nguvu ya uvutano kwenye sehemu hiyo ni:

    2.22×10−7 Nkg−1\text{2.22} \times 10^{-7} \text{ Nkg}^{-1}
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