The acceleration relationships in kinematics allow finding the result for the centripetal acceleration of the satellite is:
a = 2.10 10⁻³ rad / s²
Given parameters
- Orbit period T = 23.0 d (24h / 1d) (3600s / 1 h) = 1.987 10⁶ s
- Distance from the center of the moon to the surface of the planet r₁ = 2,060 10⁵ km = 2,060 10⁸ m.
- The radius of the planet of r₂ = 4,000 10³ km = 4,000 10⁶ m
To find
The centripetal or radial acceleration is the acceleration that changes the direction of the velocity, but maintains its constant magnitude
a = [tex]\frac{v^2}{r}[/tex]
Where a is the centripetal acceleration, v the linear velocity and r the distance to the center of the circle
The linear and angular variables are related
v = w r
we substitute
a = w² R
As the satellite has a circular orbit, its speed is constant, we look for the angular velocity
[tex]w = \frac{2\pi }{T} \\w= \frac{2 \pi }{1.987 \ 10^6}[/tex]
w = 3.162 10⁻⁶ rad / s
The distance from the center of the satellite to the center of the planet is
r = r₁ + r₁
r = 2,060 10⁸ + 4,000 10⁶
r = 2.10 10⁸ m
Let's calculate the centripetal acceleration
a = (3.162 10⁻⁶) ² 2.10 10⁸
a = 2.10 10-3 rad / s²
In conclusion, using the acceleration relations in kinematics we can calculate the centripetal acceleration of the satellite is:
a = 2.10 10⁻³ rad / s²
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