Halley’s comet takes 76 years to complete an orbit, and its closest approach to the sun is 8.9 × 10¹⁰ meters. Given the mass of the sun is 2 × 10³⁰ kg and the gravitational constant G = 6.67 × 10⁻¹¹ in MKS units, what is the comet’s farthest distance from the sun?

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    <p>1. Halley’s comet has a period of 76 years, and its distance of closest approach to the sun is equal to 8.9×10<sup>10</sup> m. The comet’s farthest distance from the sun, given the mass of the sun is 2×10<sup>30</sup> kg and G = 6.67×10<sup>-11</sup> in MKS units, is:</p>

    Question:

    1. Halley’s comet has a period of 76 years, and its distance of closest approach to the sun is equal to 8.9×1010 m. The comet’s farthest distance from the sun, given the mass of the sun is 2×1030 kg and G = 6.67×10-11 in MKS units, is:

    • a) 2×1012 m
    • b) 2.7×1013 m
    • c) 5.3×1012 m
    • d) 5.3×1013 m

    Answer:

    1. (c) It is self-evident that the orbit of the comet is elliptic with the Sun being at one of the foci. Now, as for elliptic orbits, according to Kepler’s third law:

    T2 = (4π2 a3) / GM

    ⇒ a = ((T2 GM) / (4π2 ))1/3

    a = [((76 × 3.14 × 107) × 6.67 × 10-11 × 2 × 1010) / (4π2)]1/3

    But in the case of an ellipse, 2a = rmin + rmax

    ∴ rmax = 2a – rmin = 2 × 2.7 × 1012 – 8.9 × 1010

    ≅ 5.3 × 1012 m

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