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Calculate the gravitational potential energy between the Sun and Earth.

Physics

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Answer

−4.96×1033 J-4.96 \times 10^{33} \text{ J}

Step-by-step solution

5 steps

  1. Identify the given values and constants.

    Mass of Earth (mEm_E) = 5.98×10245.98 \times 10^{24} kg
    Mass of Sun (MSM_S) = 1.99×10301.99 \times 10^{30} kg
    Distance between Earth and Sun (rr) = 160 million km
    Gravitational constant (GG) = 6.674×10−11 N⋅m2/kg26.674 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2

  2. Convert the distance from kilometers to meters.

    1 million km=106 km1 \text{ million km} = 10^6 \text{ km}
    160 million km=160×106 km=1.6×108 km160 \text{ million km} = 160 \times 10^6 \text{ km} = 1.6 \times 10^8 \text{ km}
    Since 1 km=1000 m=103 m1 \text{ km} = 1000 \text{ m} = 10^3 \text{ m}:

    r=1.6×108 km×103 m1 km=1.6×1011 mr = 1.6 \times 10^8 \text{ km} \times \frac{10^3 \text{ m}}{1 \text{ km}} = 1.6 \times 10^{11} \text{ m}

  3. Use the formula for gravitational potential energy.

    The gravitational potential energy UU between two masses MM and mm separated by a distance rr is given by:

    U=−GMmrU = -\frac{GMm}{r}

    Substitute the values for the Sun and Earth:

    U=−(6.674×10−11 N⋅m2/kg2)(1.99×1030 kg)(5.98×1024 kg)1.6×1011 mU = -\frac{(6.674 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2)(1.99 \times 10^{30} \text{ kg})(5.98 \times 10^{24} \text{ kg})}{1.6 \times 10^{11} \text{ m}}

  4. Calculate the numerator.

    GMSmE=(6.674×10−11)×(1.99×1030)×(5.98×1024)G M_S m_E = (6.674 \times 10^{-11}) \times (1.99 \times 10^{30}) \times (5.98 \times 10^{24}) GMSmE=(6.674×1.99×5.98)×(10−11×1030×1024)G M_S m_E = (6.674 \times 1.99 \times 5.98) \times (10^{-11} \times 10^{30} \times 10^{24}) GMSmE≈79.40×10(−11+30+24)G M_S m_E \approx 79.40 \times 10^{(-11+30+24)} GMSmE≈79.40×1043 N⋅m2G M_S m_E \approx 79.40 \times 10^{43} \text{ N} \cdot \text{m}^2

  5. Calculate the gravitational potential energy.

    U=−79.40×1043 N⋅m21.6×1011 mU = -\frac{79.40 \times 10^{43} \text{ N} \cdot \text{m}^2}{1.6 \times 10^{11} \text{ m}} U=−(79.401.6)×(10431011) JU = -\left(\frac{79.40}{1.6}\right) \times \left(\frac{10^{43}}{10^{11}}\right) \text{ J} U=−49.625×10(43−11) JU = -49.625 \times 10^{(43-11)} \text{ J} U=−49.625×1032 JU = -49.625 \times 10^{32} \text{ J} U=−4.9625×1033 JU = -4.9625 \times 10^{33} \text{ J}

    The gravitational potential energy of the Earth is:

    −4.96×1033 J-4.96 \times 10^{33} \text{ J}
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