Respuesta :
Answer:
the correct result is r = 3.71 10⁸ m
Explanation:
For this exercise we will use the law of universal gravitation
F = [tex]- \frac{m_{1} m_{2} }{r^2}[/tex]
We call the masses of the Earth M, the masses of the moon m and the masses of the rocket m ', let's set a reference system in the center of the Earth, the distance from the Earth to the moon is d = 3.84 108 m
rocket force -Earth
F₁ = - \frac{m' M }{r^2}
rocket force - Moon
F₂ = - \frac{m' m }{(d-r)^2}
in the problem ask for what point the force has the relation
2 F₁ = F₂
let's substitute
2 [tex]2 \frac{M}{r^2} = \frac{m}{(d-r)^2}[/tex]
(d-r) ² = [tex]\frac{m}{2M}[/tex] r²
d² - 2rd + r² = \frac{m}{2M} r²
r² (1 -\frac{m}{2M}) - 2rd + d² = 0
Let's solve this quadratic equation to find the distance r, let's call
a = 1 - \frac{m}{2M}
a = 1 - [tex]\frac{7.36 10^{22} }{2 \ 5398 10^{24}}[/tex] = 1 - 6.15 10⁻³
a = 0.99385
a r² - 2d r + d² = 0
r = [tex]\frac {2d \frac{+}{-} \sqrt{4d^2 - 4 a d^2}} {2a}[/tex]
r = [2d ± 2d [tex]\sqrt{1-a}[/tex]] / 2a
r = [tex]\frac{d}{a}[/tex] (1 ± √ (1.65 10⁻³)) = [tex]\frac{d}{a}[/tex] (1 ± 0.04)
r₁ = \frac{d}{a} 1.04
r₂ = \frac{d}{a} 0.96
let's calculate
r₁ = [tex]\frac{3.84 10^8}{0.99385}[/tex] 1.04
r₁ = 401.8 10⁸ m
r₂ = \frac{3.84 10^8}{0.99385} 0.96
r₂ = 3.71 10⁸ m
therefore the correct result is r = 3.71 10⁸ m
