Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As ch

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 14.0 m/s returns to the ground in 5.70 s ; the circumference of Mongo at the equator is 2.60×105 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? 1.26×1026 kg

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?

I can figure out part A but not B. I keep getting 8 and 18 but both are wrong.

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 = 5.7 (s), so:
a = 14/(5.7/2) = 22/5 (m/s^2)
28/5.7 = a = GM/R^2 , so:
M = (28/5.7)*R^2/G
2.6e5 (km) = 2.6e8 (m) = 2R , so:
R = 1.3e8/ (m)
Therefore:
M = (28/5.7)(1.3e8/)^2/G
= 1.26e26 (kg)
In a circular orbit of radius R2, where
R2 = 1.3e8/ + 4e4 * e3 = 1.3e8/ + 4e7 = 4.14e7 + 4e7 = 8.14e7
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( (8.14^3 * e21)/(6.67e-11 * 1.26e26) )
= 5.0e4 (s)
Since 1 hour = 60*60 = 3600 (s),
T = 5.0e4/3.6e3 = 13.89 hours
a) M = 1.26e26 (kg)
b) T = 13.89 hours
(Earth hours, of course!)

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