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Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements:
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 13.0 m/s returns to the ground in 4.50 s; the circumference of Mongo at the equator is 2.00×10^5km; and there is no appreciable atmosphere on Mongo. The starship commander, Captain Confusion, asks for the following information:
a. What is the mass of Mongo?
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?
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Physics
3 years
2021-07-16T00:09:04+00:00
2021-07-16T00:09:04+00:00 1 Answers
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Answers ( )
Given Information:
Initial speed = v₁ = 13 m/s
time = t = 4.50 sec
Circumference of Mongo = C = 2.0×10⁵ km = 2.0×10⁸ m
Altitude = h = 30,000 km = 3×10⁷ m
Required Information:
a) mass of Mongo = M = ?
b) time in hours = t = ?
Answer:
a) mass of Mongo = M = 8.778×10²⁵ kg
b) time in hours = t = 11.08 h
Explanation:
We know from the equations of kinematics,
v₂ = v₁t – ½gt²
0 = 13*4.50 – ½g(4.50)²
58.5 = 10.125g
g = 58.5/10.125
g = 5.78 m/s²
Newton’s law of gravitation is given by
M = gC²/4π²G
Where C is the circumference of the planet Mongo, G is the gravitational constant, g is the acceleration of planet of Mongo and M is the mass of planet Mongo.
M = 5.78*(2.0×10⁸)²/(4π²*6.672×10⁻¹¹)
M = 8.778×10²⁵ kg
Therefore, the mass of planet Mongo is 8.778×10²⁵ kg
b) From the Kepler’s third law,
T = 2π*(R + h)^3/2/(G*M)^1/2
Where R = C/2π
T = 2π*(C/2π + h)^3/2/(G*M)^1/2
T = 2π*((2.0×10⁸/2π) + 3×10⁷)^3/2/(6.672×10⁻¹¹*8.778×10²⁵)^1/2
T = 39917.5 sec
Convert to hours
T = 39917.5/60*60
T = 11.08 hours
Therefore, it will take 11.08 hours for the ship to complete one orbit.