Question

While describing a circular orbit 300 mi above the earth a space vehicle launches a 6000-lb communications satellite. Determine the additional energy required to place the satellite in a geosynchronous orbit at an altitude of 22,000 mi

Answers

  1. Answer:

    \Delta U = 2.2126039 x 10^{12} J

    Explanation:

    While the satellite is in the space vehicle, it has the next potential energy

    U = -\frac{GmMe}{r}

                 where G is the gravitational constant

                             m is the satellite’s mass in kilograms

                            Me is the earth’s mass

                            r is the orbit’s radius from to the earth’s center in meters

    U = - \frac{6.67x10^{-11}*2721.554*5.972x10^{24} }{482803}

    U = -2.2423x10^{12} J

    The additional energy required is the difference between this energy and the energy that the satellite would have in an orbit with an altitude of 22000 mi

    U = -\frac{6.67x10^{-11}*2721.554*5.792x10^{24} }{35405568}

    U = -29696124610.3 J

    Then

    \Delta U = 2.2126039 x 10^{12} J

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