Question

spaceship of mass m travels from the Earth to the Moon along a line that passes through the center of the Earth and the center of the Moon. (a) At what distance from the center of the Earth is the force due to the Earth twice the magnitude of the force due to the Moon

Answer 1

Answer:

the correct result is r = 3.71 10⁸ m

Explanation:

For this exercise we will use the law of universal gravitation

          F = $- \frac{m_{1} m_{2} }{r^2}$

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 $2 \frac{M}{r^2} = \frac{m}{(d-r)^2}$

          (d-r) ² = $\frac{m}{2M}$ 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 - $\frac{7.36 10^{22} }{2 \ 5398 10^{24}}$ = 1 - 6.15 10⁻³

           a = 0.99385

         

            a r² - 2d r + d² = 0

           r =  $\frac {2d \frac{+}{-} \sqrt{4d^2 - 4 a d^2}} {2a}$

           r = [2d ± 2d $\sqrt{1-a}$] / 2a

           r = $\frac{d}{a}$   (1 ± √ (1.65 10⁻³)) =  $\frac{d}{a}$ (1 ± 0.04)

           r₁ = \frac{d}{a} 1.04

           r₂ = \frac{d}{a} 0.96

let's calculate

           r₁ = $\frac{3.84 10^8}{0.99385}$ 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