Respuesta :
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:
The final speed of the stone is 18 m/s
Explanation:
From energy principle;
[tex]K_f +U_f=K_i+U_i[/tex]
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
[tex]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[/tex]
when the initial speed of the rock is 30 m/s, then we apply energy principle to determine the final speed;
[tex]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[/tex]
Therefore, the final speed of the stone is 18 m/s
