Answer:
The period (T) of a satellite in a circular orbit can be determined using the formula:
T = 2πr/v
where r is the radius of the orbit and v is the speed of the satellite.
We can rearrange this formula to solve for v:
v = 2πr/T
Substituting the given values, we get:
v = 2π(3.5 × 10^7 m)/(6.3 h × 3600 s/h) ≈ 5,099 m/s
The centripetal force (F) that keeps the satellite in orbit is given by:
F = mv^2/r
where m is the mass of the satellite.
The gravitational force (Fg) between the satellite and Roton is given by:
Fg = GmM/R^2
where G is the gravitational constant, M is the mass of Roton, and R is the radius of Roton.
Since the satellite is in a circular orbit, the centripetal force is equal to the gravitational force:
mv^2/r = GmM/R^2
Simplifying and solving for M, we get:
M = v^2r/GR^2
Substituting the given values, we get:
M = (5,099 m/s)^2 × 3.5 × 10^7 m/(6.674 × 10^-11 m^3/kg s^2) × (1.75 × 10^7 m)^2 ≈ 8.35 × 10^23 kg
The free-fall acceleration (g) on the surface of Roton is given by:
g = GM/R^2
Substituting the calculated value of M and the given value of R, we get:
g = (6.674 × 10^-11 m^3/kg s^2) × (8.35 × 10^23 kg)/(1.75 × 10^7 m)^2 ≈ 8.73 m/s^2
Therefore, the magnitude of the free-fall acceleration on the surface of Roton is approximately 8.73 m/s^2