Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As ch
ID: 1493865 • Letter: Y
Question
Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements: a2.50-kg stone thrown upward from the ground at14.0 m/s returns to the ground in 4.80 s ; the circumference of Mongo at the equator is 2.70×105 km ; and there is no appreciable atmosphere on Mongo.
(a) The starship commander, Captain Confusion, asks for the following information: what is the mass of Mongo?
Express your answer with the appropriate units.
(b) If the Aimless Wanderer goes into a circular orbit 40,000 km above the surface of Mongo, how many hours will it take the ship to complete one orbit?
Explanation / Answer
solution:
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.
Thus 2*T = 4.8 s
so:
a = 14/(4.8/2)) = 28/4.8 (m/s^2)
28/4.8 = a = GM/R^2 , so:
M = (28/4.8)*R^2/G
2.7e5 (km) = 2.7e8 (m) = 2R , so:
R = (2.7)e8/2 (m)
Therefore:
M = (28/4.8)(2.7e8/2)^2/G
= (10.63125)/(^2) * e16/(6.67e-11)
= 1.62e26 (kg)
In a circular orbit of radius R2, where
R2 = 2.7e8/2 + 4e4 * e3 =8.3e7
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( (8.3^3 * e21)/(6.67e-11 * 1.62e26) )
= 2 * sqrt(8.3^3/(6.67*1.62)) * e3 = 4.56e4 (s)
Since 1 hour = 60*60 = 3600 (s),
T = 4.56e4/3.6e3 = 12.69 hours
a) M = 1.62e26 (kg) ans
b) T = 12.69 hours ans
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