A 7910-kg satellite has an elliptical orbit, as in Figure 6.9b. The point on the
ID: 1362000 • Letter: A
Question
A 7910-kg satellite has an elliptical orbit, as in Figure 6.9b. The point on the orbit that is farthest from the earth is called the apogee and is at the far right of the drawing. The point on the orbit that is closest to the earth is called the perigee and is at the left side of the drawing. Suppose that the speed of the satellite is 3460 m/s at the apogee and 7170 m/s at the perigee. Find the work done by the gravitational force when the satellite moves from (a) the apogee to the perigee and (b) the perigee to the apogee.
Explanation / Answer
Here ,
speed of satellite apogee , va = 3460 m/s
speed of satellite perigee , vp = 7170 m/s
mass of satellite , m = 7910 Kg
work done by gravitational force = change in kinetic energy
work done by gravitational force = 0.5 * 7910 * (7170^2 - 3460^2)
work done by gravitational force = 3.12 *10^11 J
the work done by gravitational force is 3.12 *10^11 J
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for the perigee to apogee
work done by gravitational force = 2 * change in kinetic energy
work done by gravitational force = 2 * 0.5 * 7910 * (- 7170^2 + 3460^2)
work done by gravitational force = - 3.12 *10^11 J
the work done by gravitational force is - 3.12 *10^11 J
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