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?

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.