Answer:
The acceleration of the satellite is [tex]0.87 m/s^{2}[/tex]
Explanation:
The acceleration in a circular motion is defined as:
[tex]a = \frac{v^{2}}{r}[/tex] (1)
Where a is the centripetal acceleration, v the velocity and r is the radius.
The equation of the orbital velocity is defined as
[tex]v = \frac{2 \pi r}{T}[/tex] (2)
Where r is the radius and T is the period
For this particular case, the radius will be the sum of the high of the satellite ([tex]1.50x10^{7} m[/tex]) and the Earth radius ([tex]6.38x10^{6} m[/tex]) :
[tex]r = 1.50x10^{7} m+6.38x10^{6}m [/tex]
[tex]r = 21.38x10^{6}m [/tex]
Then, equation 2 can be used:
[tex]T = 8.65 hrs \cdot \frac{3600 s}{1hrs}[/tex] ⇒ [tex]31140 s[/tex]
[tex]v = \frac{2 \pi (21.38x10^{6}m)}{31140s}[/tex]
[tex]v = 4313 m/s[/tex]
Finally equation 1 can be used:
[tex]a = \frac{(4313m/s)^{2}}{21.38x10^{6}m}[/tex]
[tex]a = 0.87 m/s^{2}[/tex]
Hence, the acceleration of the satellite is [tex]0.87 m/s^{2}[/tex]
The acceleration of the satellite is 0.87 m/s²
The given parameters;
distance of the satellite, r₁ = 1.5 x 10⁷ m
radius of the earth, r₂ = 6.38 x 10⁶ m
time of motion, t = 8.65 hours
The radius of the satellite is calculated as;
R = r₁ + r₂
R = (1.5 x 10⁷) m + (6.38 x 10⁶) m
R = 2.138 x 10⁷ m
The time of motion of the satellite in seconds;
T = 8.65 hours x 3600 s
T = 31,140 s
The velocity of the satellite is calculated as;
[tex]V = \frac{2\pi R}{T} \\\\V = \frac{2\pi \times (2.138\times 10^7)}{31, 140} \\\\V = 4314.45 \ m/s[/tex]
The acceleration of the satellite is calculated as;
[tex]a_c = \frac{V^2}{R} \\\\a_c = \frac{4314.45^2}{2.138\times 10^{7}} \\\\a_c = 0.87 \ m/s^2[/tex]
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