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

Answer 1

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

Answer 2

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