Question

Miranda, a satellite of Uranus, is shown in part a of the figure below. It can be modeled as a sphere of radius 242 km and mass 6.68 1019 kg.

(a) Find the free-fall acceleration on its surface. (m/s2)

(b) A cliff on Miranda is 5.00 km high. It appears on the limb at the 11 o'clock position in part a of the figure above and is magnified in part b of the figure above. A devotee of extreme sports runs horizontally off the top of the cliff at 8.50 m/s. For what time interval is he in flight? (Ignore the difference in g between the lip and base of the cliff.) (s)

(c) How far from the base of the vertical cliff does he strike the icy surface of Miranda? m (d) What is his vector impact velocity? (m/s ° below the horizontal)

Answer 1

(a) Miranda's surface gravity = 0.0761 m/s^2
(b) Sports fan's time of flight = 362 s
(c) Total horizontal distance = 1,160 m
(d) Idiot's impact velocity = 27.6 m/s

How and why these are true...

(a) Surface gravity

a = [GM] / r^2

a = Acceleration
G = Universal Gravitational Constant
M = Mass of Miranda
r = Radial distance from center

Given:

a = ?
G = 6.67428E-11 m^3/kg-s^2
M = 6.68E+19 kg
r = 242,000 m

Solve

a = [ (6.67428E-11 m^3/kg-s^2) * (6.68E+19 kg) ] / (242,000 m)^2
a = [ 4,458,419,040 m^3/s^2 ] / (58,564,000,000 m^2)
a = 0.0761 m/s^2

(b) Time of flight of crazy sports fan... if there's an angle, I don't see it, so horizontal launch

H = 5,000 m
Vo = 8.50 m/s
g = 0.0761 m/s^2 (as above)

t = SQRT { [2H] / g }
t = SQRT { [ 2 * (5,000 m) ] / (0.0761 m/s^2) }
t = SQRT { [ 10,000 m ] / (0.0761 m/s^2) }
t = SQRT { 131356 s^2 }
t = 362 s

(c) Distance from cliff, assuming a perfectly flat landing zone

Vo = 8.50 m/s
t = 362 s

R = Vo * t
R = (8.50 m/s) * (362 s)
R = 3077 m

(d) Impact velocity

Vy = SQRT { 2gH }
Vy = SQRT { 2 * (0.0761 m/s^2) * (5,000 m) }
Vy = SQRT { 761 m^2/s^2 }
Vy = 27.6 m/s