Given/Known:
R = Rearth + height = 6.47 x 106 m
Mearth = 5.98x1024 kg
G = 6.673 x 10-11 N m2/kg2
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
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
The period of the moon is approximately 27.2 days (2.35 x 106 s). Determine the radius of the moon's orbit and the orbital speed of the moon. (Given: Mearth = 5.98 x 1024 kg, Rearth = 6.37 x 106 m)
Like Practice Problem #2, this problem begins by identifying known and unknown values. These are shown below.
Given/Known:
T = 2.35 x 106 s
Mearth = 5.98 x 1024 kg
G = 6.673 x 10-11 N m2/kg2
The radius of orbit can be calculated using the following equation:
The equation can be rearranged to the following form
R3 = [ (T2 • G • Mcentral) / (4 • pi2) ]
The substitution and solution are as follows:
R3 = [ ((2.35x106 s)2 • (6.673 x 10-11 N m2/kg2) • (5.98x1024 kg) ) / (4 • (3.1415)2) ]
R3 = 5.58 x 1025 m3
By taking the cube root of 5.58 x 1025 m3, the radius can be determined as follows:
R = 3.82 x 108 m
The orbital speed of the satellite can be computed from either of the following equations:
(1) v = SQRT [ (G • MCentral ) / R ]
(2) v = (2 • pi • R)/T
Equation (1) was derived above. Equation (2) is a general equation for circular motion. Either equation can be used to calculate the orbital speed; the use of equation (1) will be demonstrated here. The substitution of values into this equation and solution are as follows:
v = SQRT [ (6.673 x 10-11 N m2/kg2)*(5.98x1024 kg) / (3.82 x 108 m) ]
v = 1.02 x 103 m/s
(Given: Mearth = 5.98x1024 kg, Rearth = 6.37 x 106 m)
Just as in the previous problem, the solution begins by the identification of the known and unknown values. This is shown below.
Given/Known:
T = 86400 s
earth = 5.98x1024 kg
earth = 6.37 x 106 m
G = 6.673 x 10-11 N m2/kg2
The equation can be rearranged to the following form
R3 = [ (T2 * G * Mcentral) / (4*pi2) ]
The substitution and solution are as follows:
R3 = [ ((86400 s)2 • (6.673 x 10-11 N m2/kg2) • (5.98x1024 kg) ) / (4 • (3.1415)2) ]
R3 = 7.54 x 1022 m3
By taking the cube root of 7.54 x 1022 m3, the radius can be determined to be
R = 4.23 x 107 m
The radius of orbit indicates the distance that the satellite is from the center of the earth. Now that the radius of orbit has been found, the height above the earth can be calculated. Since the earth's surface is 6.37 x 106 m from its center (that's the radius of the earth), the satellite must be a height of
4.23 x 107 m - 6.37 x 106 m = 3.59 x 107 m
