The magnitude of the free-fall acceleration on the surface of planet Roton is approximately 24.4 m/s².
Step-by-step explanation:
To find the magnitude of the free-fall acceleration on the surface of the planet Roton, we can use the equation for centripetal acceleration. The centripetal acceleration is equal to the gravitational acceleration:
ac = g
Since the satellite is orbiting the planet, its centripetal acceleration is given by:
ac = v2/r
where v is the linear speed of the satellite and r is the radius of its orbit. Rearranging the equation, we can solve for the linear speed:
v = √(ac × r)
Substituting the values provided in the question, we have:
v = √[(9.8 m/s²)(5.0 x 10⁶ m)]
v = 7.0 x 10³ m/s
Therefore, the linear speed of the satellite is 7.0 x 10³ m/s. Since the satellite completes one orbit every 2.8 hours, we can find the angular velocity:
ω = 2π/T
ω = 2π/(2.8 hours)
Converting the time to seconds:
ω = 2π/(2.8 × 3600 seconds)
ω ≈ 0.00349 radians/s
The magnitude of the free-fall acceleration can be obtained by multiplying the linear speed and the angular velocity:
a = v × ω
a = (7.0 x 10³ m/s)(0.00349 radians/s)
a ≈ 24.4 m/s²
Therefore, the magnitude of the free-fall acceleration on the surface of Roton is approximately 24.4 m/s².