Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As ch
ID: 1499345 • 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-kg stone thrown upward from the ground at 11.0 m/s returns to the ground in 5.40 s ; the circumference of Mongo at the equator is 2.30×105 km ; and there is no appreciable atmosphere on Mongo.
The starship commander, Captain Confusion, asks for the following information: what is the mass of Mongo? Express your answer with the appropriate units.
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
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 = 5.4 (s), so:
a = 11/(5.4 /2) = 22/5.4 (m/s^2)
22/5.4 = a = GM/R^2 , so:
M = (22/5.4)*R^2/G
the circumference of magno = 2.3*10^5 km = 2.3*10^8 m = 2*pi*R , so:
R = 2.3*10^8/2*pi (m)
Therefore:
M = (22/5.4)*(2.3*10^8/2*pi)^2/G
= (22/5.4)*(2.3*10^8/2*pi)^2/(6.67*10^(-11))
= 7.97*10^27 kg
In a circular orbit of radius R2, where
R2 = 2.3*10^8/2*pi + 4*10^4*10^3 = 76590909.1 m
mw^2*R2 = mv^2/R2 = GMm/R2^2
w^2 = GM/R2^3
2*pi/T = sqrt(GM/R2^3)
T = 2*pi * sqrt(R2^3/(GM))
= 2*pi * sqrt( (76590909.09)^3/(6.67*10^(-11) * 7.97*10^27) )
= 5778.67 (s)
Since 1 hour = 60*60 = 3600 (s),
T = 5778.67/3600= 1.605 hours
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