The planet Earth travels in a circular orbit at constant speed around the Sun. What is the net work done on the Earth by the gravitational a

Question

The planet Earth travels in a circular orbit at constant speed around the Sun. What is the net work done on the Earth by the gravitational attraction between it and the Sun in one complete orbit

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Thanh Hà 3 months 2021-09-02T14:48:39+00:00 1 Answers 0 views 0

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    2021-09-02T14:50:30+00:00

    Answer:

    WT =  3.32*10^34 J

    Explanation:

    The work done by the gravitational attraction between the Sun and the Earth in one complete orbit of the Earth can be calculated by using the following formula:

    W_T=\int F_g dr  (1)

    Fg: gravitational force between Sun and Earth

    The gravitational force is given by:

    F_g=G\frac{m_sm_e}{r^2}  (2)

    G: Cavendish’s constant = 6.674*10^-11 m^3 kg^-1 s^-2

    ms: mass of the sun = 1.989*10^30 kg

    me: mass of the Earth = 5.972 × 10^24 kg

    r: distance between Earth and Sun, this value is a constant r = R = 149,597,870 km

    You replace the formula (2) in (1):

    W_T=\int G\frac{m_sm_e}{R^2}dr=G\frac{m_sm_e}{R^2}\int dr\\\\W_T=G\frac{m_sm_e}{R^2}(2\pi R)=2\pi G\frac{m_sm_e}{R}

    Next, you replace the values of all variables and solve obtain WT:

    W_T=2\pi (6.674*10^{-11}m^3kg^{-1}s^{-2})\frac{(1.989*10^{30}kg)(5.972*10^{24}kg)}{(149597870*10^3 m)}\\\\W_T=3.32*10^{34}J

    hence, the work done on the Earth, in one orbit, is 3.32*10^34 J

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