Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As ch
ID: 1500485 • Letter: Y
Question
Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements: a 2.50-kgstone thrown upward from the ground at 14.0 m/sreturns to the ground in 4.50 s ; the circumference of Mongo at the equator is 2.60×105 km ; and there is no appreciable atmosphere on Mongo.
Part A
The starship commander, Captain Confusion, asks for the following information: what is the mass of Mongo?
Express your answer with the appropriate units.
Part B
If the Aimless Wanderer goes into a circular orbit 30,000 km above the surface of Mongo, how many hours will it take the ship to complete one orbit?
Explanation / Answer
PART A
The acceleration of gravity at the surface of Mongo is:
a = GM/R^2
where M = mass of Mongo
R = radius of Mongo
When tossed up,
v(t) = v(0) - at, so v(T) = 0 when:
0 = v(0) - aT
=> a = v(0)/T
Since it takes time T to arrive at the high point, it takes another time T to return to the ground.
2*T = 4.5 s
a = 14/(4.5/2)) =6.22 (m/s^2)
6.22 = a = GM/R^2 , so:
M = (6.22)*R^2/G
2.6e5 (km) = 2.6e8 (m) = 2R , so:
R = (2.6)e8/2 (m)
M = (6.22)(2.6e8/2)^2/G = 1.59 e26 (kg)
PART B
R2 = 2.6e8/2 + 3e4 * e3 = 7.27 e7
m^2*R2 = mv^2/R2 = GMm/R2^2
^2 = GM/R2^3
2/T = sqrt(GM/R2^3)
T = 2 * sqrt(R2^3/(GM))
= 2 * sqrt( (7.27 e7^3 * e21)/(6.67e-11 * 1.59 e26) )
= 3.78e4
Since 1 hour = 60*60 = 3600 (s),
T = 3.78e4/3.6e3 = 10.51 hours
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