Share

## 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?

in progress
0

Physics
3 years
2021-07-16T00:09:04+00:00
2021-07-16T00:09:04+00:00 1 Answers
312 views
0
## 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.