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Calculate the gravitational force between two objects with given masses and distance.

Physics

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Answer

8.3425×10−14 N8.3425 \times 10^{-14} \text{ N}

Step-by-step solution

5 steps

  1. Identify the given values.

    Mass of object A, m1=2,000 kgm_1 = 2,000 \text{ kg}
    Mass of object B, m2=4,000 kgm_2 = 4,000 \text{ kg}
    Distance between the centers of objects A and B, r=80 Kmr = 80 \text{ Km}

  2. Convert the distance to SI units (meters).

    1 Km=1000 m1 \text{ Km} = 1000 \text{ m}
    r=80 Km×1000 m/Km=80,000 mr = 80 \text{ Km} \times 1000 \text{ m/Km} = 80,000 \text{ m}

  3. State the gravitational constant.

    The universal gravitational constant, G=6.674×10−11 N m2/kg2G = 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2

  4. Apply Newton's Law of Universal Gravitation.

    The formula for gravitational force (FF) is:

    F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}

  5. Substitute the values into the formula and calculate the force.

    F=(6.674×10−11 N m2/kg2)(2,000 kg)×(4,000 kg)(80,000 m)2F = (6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \frac{(2,000 \text{ kg}) \times (4,000 \text{ kg})}{(80,000 \text{ m})^2} F=(6.674×10−11)8,000,000 kg26,400,000,000 m2F = (6.674 \times 10^{-11}) \frac{8,000,000 \text{ kg}^2}{6,400,000,000 \text{ m}^2} F=(6.674×10−11)8×1066.4×109 NF = (6.674 \times 10^{-11}) \frac{8 \times 10^6}{6.4 \times 10^9} \text{ N} F=(6.674×10−11)×(1.25×10−3) NF = (6.674 \times 10^{-11}) \times (1.25 \times 10^{-3}) \text{ N} F=8.3425×10−14 NF = 8.3425 \times 10^{-14} \text{ N}

    The gravitational force of attraction between objects A and B is 8.3425×10−14 N\boxed{8.3425 \times 10^{-14} \text{ N}}.

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