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Given masses: m1 = 1 kg, m2 = 2 kg.

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Part 12b(i): Initial accelerations of the particles

  1. Identify given values and convert units.

    Given masses: m1=1 kgm_1 = 1 \text{ kg}, m2=2 kgm_2 = 2 \text{ kg}.
    Separation distance: r=50 cm=0.50 mr = 50 \text{ cm} = 0.50 \text{ m}.
    Gravitational constant: G=6.67×10−11 N m2/kg2G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2.

  2. Calculate the gravitational force between the particles using Newton's Law of Universal Gravitation.

    The formula is F=Gm1m2r2F = \frac{G m_1 m_2}{r^2}.

    F=(6.67×10−11 N m2/kg2)×(1 kg)×(2 kg)(0.50 m)2F = \frac{(6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \times (1 \text{ kg}) \times (2 \text{ kg})}{(0.50 \text{ m})^2} F=13.34×10−11 N m20.25 m2F = \frac{13.34 \times 10^{-11} \text{ N m}^2}{0.25 \text{ m}^2} F=53.36×10−11 NF = 53.36 \times 10^{-11} \text{ N} F=5.34×10−10 NF = 5.34 \times 10^{-10} \text{ N}

  3. Calculate the initial acceleration of each particle using Newton's Second Law (F=maF = ma).

    For the first particle (m1m_1):

    a1=Fm1=5.34×10−10 N1 kga_1 = \frac{F}{m_1} = \frac{5.34 \times 10^{-10} \text{ N}}{1 \text{ kg}} a1=5.34×10−10 m/s2a_1 = 5.34 \times 10^{-10} \text{ m/s}^2

    For the second particle (m2m_2):

    a2=Fm2=5.34×10−10 N2 kga_2 = \frac{F}{m_2} = \frac{5.34 \times 10^{-10} \text{ N}}{2 \text{ kg}} a2=2.67×10−10 m/s2a_2 = 2.67 \times 10^{-10} \text{ m/s}^2

    The initial accelerations are:

    a1=5.34×10−10 m/s2\boxed{a_1 = 5.34 \times 10^{-10} \text{ m/s}^2} a2=2.67×10−10 m/s2\boxed{a_2 = 2.67 \times 10^{-10} \text{ m/s}^2}

    Part 12b(ii): Escape velocity from the moon

  4. Identify given values and convert units.

    Mass of the moon: M=7.4×1022 kgM = 7.4 \times 10^{22} \text{ kg}.
    Radius of the moon: R=1740 km=1740×103 m=1.74×106 mR = 1740 \text{ km} = 1740 \times 10^3 \text{ m} = 1.74 \times 10^6 \text{ m}.
    Gravitational constant: G=6.67×10−11 N m2/kg2G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2.

  5. Calculate the escape velocity using the formula ve=2GMRv_e = \sqrt{\frac{2GM}{R}}.

    ve=2×(6.67×10−11 N m2/kg2)×(7.4×1022 kg)1.74×106 mv_e = \sqrt{\frac{2 \times (6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \times (7.4 \times 10^{22} \text{ kg})}{1.74 \times 10^6 \text{ m}}} ve=98.616×10111.74×106 m/sv_e = \sqrt{\frac{98.616 \times 10^{11}}{1.74 \times 10^6}} \text{ m/s} ve=56.675×105 m/sv_e = \sqrt{56.675 \times 10^5} \text{ m/s} ve=5.6675×106 m/sv_e = \sqrt{5.6675 \times 10^6} \text{ m/s} ve≈2380.65 m/sv_e \approx 2380.65 \text{ m/s}

  6. Round the result to an appropriate number of significant figures.

    ve≈2380 m/sv_e \approx 2380 \text{ m/s}

    The escape velocity from the moon is:

    2380 m/s\boxed{2380 \text{ m/s}}
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