A satellite circles planet Roton every 2.8 h in an orbit having a radius of 1.2 *10⁷ m.If the radius of Roton is 5.0 * 10⁶ m,what is the magnitude of the free-fall acceleration on the surface of Roton? A)31 m/s² B)27 m/s² C)34 m/s² D)40 m/s² E)19 m/s²

Question

Quizplus · Accepted Answer

Using Kepler's Third Law, we can find the mass of Roton. 
T^2 = (4π^2 /GM)R^3

where T is the period of the satellite's orbit, G is the gravitational constant, M is the mass of Roton, and R is the radius of Roton's orbit.

Plugging in the given values, we get:

(2.8 hours)^2 = (4π^2 /G) (1.2 x 10^10 m + 5.0 x 10^16 m)^3 / M

Simplifying:

M = (4π^2 /G) (6.2 x 10^16 m)^3 / (2.8 hours)^2 - (1.2 x 10^10 m)^3 = 3.6 x 10^29 kg

Now we can find the free-fall acceleration on the surface of Roton, using the formula:

g = GM/R^2

Plugging in the values, we get:

g = (6.674 x 10^-11 Nm^2/kg^2) (3.6 x 10^29 kg) / (5.0 x 10^16 m)^2

Solving for g, we get:

g ≈ 27 m/s^2

Therefore, the answer is choice B: 27 m/s^2.

A Satellite Circles Planet Roton Every 2