A satellite wishes to orbit the earth at a height of 100 km (approximately 60 miles) above the surface of the earth. Determine the speed, acceleration and orbital period of the satellite. (Given: Mearth = 5.98 x 1024 kg, Rearth = 6.37 x 106 m)
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Answer:
Practice Problem #1
A satellite wishes to orbit the earth at a height of 100 km (approximately 60 miles) above the surface of the earth. Determine the speed, acceleration and orbital period of the satellite. (Given: Mearth = 5.98 x 1024 kg, Rearth = 6.37 x 106 m)
Like most problems in physics, this problem begins by identifying known and unknown information and selecting the appropriate equation capable of solving for the unknown. For this problem, the knowns and unknowns are listed below.
Given/Known:
R = Rearth + height = 6.47 x 106 m
Mearth = 5.98x1024 kg
G = 6.673 x 10-11 N m2/kg2
Unknown:
v = ???
a = ???
T = ???
Note that the radius of a satellite's orbit can be found from the knowledge of the earth's radius and the height of the satellite above the earth. As shown in the diagram at the right, the radius of orbit for a satellite is equal to the sum of the earth's radius and the height above the earth. These two quantities can be added to yield the orbital radius. In this problem, the 100 km must first be converted to 100 000 m before being added to the radius of the earth. The equations needed to determine the unknown are listed above. We will begin by determining the orbital speed of the satellite using the following equation:
v = SQRT [ (G•MCentral ) / R ]
The substitution and solution are as follows:
v = SQRT [ (6.673 x 10-11 N m2/kg2) • (5.98 x 1024 kg) / (6.47 x 106 m) ]
v = 7.85 x 103 m/s
The acceleration can be found from either one of the following equations:
(1) a = (G • Mcentral)/R2
(2) a = v2/R
Equation (1) was derived above. Equation (2) is a general equation for circular motion. Either equation can be used to calculate the acceleration. The use of equation (1) will be demonstrated here.
a = (G •Mcentral)/R2
a = (6.673 x 10-11 N m2/kg2) • (5.98 x 1024 kg) / (6.47 x 106 m)2
a = 9.53 m/s2
Observe that this acceleration is slightly less than the 9.8 m/s2 value expected on earth's surface. As discussed in Lesson 3, the increased distance from the center of the earth lowers the value of g.
Finally, the period can be calculated using the following equation:
The equation can be rearranged to the following form
T = SQRT [(4 • pi2 • R3) / (G*Mcentral)]
The substitution and solution are as follows:
T = SQRT [(4 • (3.1415)2 • (6.47 x 106 m)3) / (6.673 x 10-11 N m2/kg2) • (5.98x1024 kg) ]
T = 5176 s = 1.44 hrs
Explanation:
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