The escape speed from a very small asteroid is only 24 m/s. If you throw a rock away from the asteroid at a speed of 30 m/s, what will be it

Question

The escape speed from a very small asteroid is only 24 m/s. If you throw a rock away from the asteroid at a speed of 30 m/s, what will be its final speed? Entry field with incorrect answer 25.50 m/s

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Phúc Điền 4 years 2021-08-31T13:03:57+00:00 2 Answers 88 views 0

Answers ( )

  1. Amity
    0
    2021-08-31T13:05:47+00:00
    Reply

    Answer:

    The final speed of the stone is 18 m/s

    Explanation:

    From energy principle;

    K_f +U_f=K_i+U_i

    where;

    Kf and Ki are the final and initial kinetic energy respectively

    Uf  and Ui are the final and initial gravitational energy respectively

    Given escape speed of the asteriod as 24 m/s

    0 = \frac{1}{2}mv^2 - \frac{GMm}{R} \\\\ \frac{1}{2}mv^2 = \frac{GMm}{R}\\\\ \frac{1}{2}v^2= \frac{GM}{R}\\\\\frac{1}{2}(24)^2 = \frac{GM}{R} = 288

    when the initial speed of the rock is 30 m/s, then we apply energy principle to determine the final speed;

    K_f +U_f=K_i+U_i\\\\\frac{1}{2}mv_f^2 +0 = \frac{1}{2}mv_i^2 - \frac{GmM}{R} \\\\\frac{1}{2}mv_f^2 = m(\frac{1}{2}v_i^2 - \frac{GM}{R})\\\\\frac{1}{2}v_f^2 =\frac{1}{2}v_i^2 - \frac{GM}{R}\\\\v_f^2 =v_i^2 - \frac{2*GM}{R}\\\\v_f = \sqrt{v_i^2 - \frac{2*GM}{R}} = \sqrt{(30)^2 - 2(288)}\\\\v_f =\sqrt{324} =18 \ m/s

    Therefore, the final speed of the stone is 18 m/s

  2. thnahdat
    0
    2021-08-31T13:05:52+00:00
    Reply

    Given Information:  

    Initial speed of rock = vi = 30 m/s  

    escape speed of the asteroid = ve = 24 m/s  

    Required Information:  

    final speed of rock = vf = ?

    Answer:  

    vf = 18 m/s

    Explanation:  

    As we know from the conservation of energy

    KEf + Uf = KEi + Ui

    Where KE is the kinetic energy and U is the potential energy

    0 + 0 = ½mve² – GMm/R

    When escape speed is used, KEf is zero due to vf being zero. Uf is zero because the object is very far away from mass M, therefore, the equation becomes

    GMm/R = ½mve²

    m cancels out

    GM/R = ½ve²

    GM/R = ½(24)²

    GM/R = 288

    KEf + Uf = KEi + Ui

    ½mvi² + 0 =  ½vf² – GMm/R

    m cancels out

    ½vi² =  ½vf² – GM/R

    Substitute the values

    ½(30)² =  ½vf² – (288)

    ½vf² = 450 – 288

    vf² = 2(162)

    vf = √324

    vf = 18 m/s

    Therefore, the final speed of the rock is 18 m/s

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