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12 December, 17:18

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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  1. 12 December, 17:44
    0
    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.
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