Which sections of the heating curve illustrate this process?
1 answer:
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Refer to the diagram shown below.
Given:
R = 6.37 x 10⁶ m, the radius of the earth
h = 3.58 x 10⁷ m, the height of the satellite above the earth's surface.
Therefore
R + h = 4.217 x 10⁷ m
In geosynchronous orbit, the period of rotation is 1 day.
Therefore the period is
T = (24 h)*(60 min/h)*(60 s/min) = 86400 s
The angular velocity is
ω = (2π rad)/(86400 s) = 7.2722 x 10⁻⁵ rad/s
Part (a)
The tangential speed is
v = (R+h)*ω
= (4.217 x 10⁷ m)*(7.2722 x 10⁻⁵ rad/s)
= 3066.7 m/s
= 3.067 km/s
Part (b)
The centripetal acceleration is
a = v²/(R+h)
= (3066.7 m/s)²/(4.217 x 10⁷ m)
= 0.223 m/s²
Answers:
(a) The speed is 3.067 km/s
(b) The acceleration is 0.223 m/s²
Given:
R = 6.37 x 10⁶ m, the radius of the earth
h = 3.58 x 10⁷ m, the height of the satellite above the earth's surface.
Therefore
R + h = 4.217 x 10⁷ m
In geosynchronous orbit, the period of rotation is 1 day.
Therefore the period is
T = (24 h)*(60 min/h)*(60 s/min) = 86400 s
The angular velocity is
ω = (2π rad)/(86400 s) = 7.2722 x 10⁻⁵ rad/s
Part (a)
The tangential speed is
v = (R+h)*ω
= (4.217 x 10⁷ m)*(7.2722 x 10⁻⁵ rad/s)
= 3066.7 m/s
= 3.067 km/s
Part (b)
The centripetal acceleration is
a = v²/(R+h)
= (3066.7 m/s)²/(4.217 x 10⁷ m)
= 0.223 m/s²
Answers:
(a) The speed is 3.067 km/s
(b) The acceleration is 0.223 m/s²