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

Answer 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

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