Answer:
F₂ = 0 N
x = 0.5*d = 0.5 m
Explanation:
Given
m₁ = m₂ = m₃ = m = 1 Kg
d = distance between m₁ and m₃ = d₁₃ = 1 m
d₁₂ = 0.5*d = 0.5*(1 m) = 0.5 m = d₃₂
G = 6.673*10⁻¹¹ N*m²/ Kg²
a) Find the magnitude of the net gravitational force exerted by these objects on a kg object (F₃) placed midway between them.
we can apply
F₂ = F₁₂ + F₃₂
then we use the formula
F₁₂ = G*m₁*m₂ / d₁₂² = G*m² / (0.5*d)² = 4*G*m² / d²
⇒ F₁₂ = 4*G*m² / d² = 4*(6.673*10⁻¹¹ N*m²/ Kg²)*(1 Kg)² / (1 m)²
⇒ F₁₂ = 2.6692*10⁻¹⁰ N (←)
and
F₃₂ = G*m₃*m₂ / d₃₂² = G*m² / (0.5*d)² = 4*G*m² / d²
⇒ F₃₂ = 4*G*m² / d² = 4*(6.673*10⁻¹¹ N*m²/ Kg²)*(1 Kg)² / (1 m)²
⇒ F₃₂ = 2.6692*10⁻¹⁰ N (→)
we get
F₂ = F₁₂ + F₃₂ = (- 2.6692*10⁻¹⁰ N) + (2.6692*10⁻¹⁰ N) = 0 N
b) At what position other than an infinitely remote one can the kg object be placed so as to experience a net force of zero from the other two objects?
From a) It is known that x = 0.5*d = 0.5*(1 m) = 0.5 m
Nevertheless, we can find the distance as follows
If
F₂ = 0 ⇒ F₁₂ + F₃₂ = 0 ⇒ F₁₂ = - F₃₂
If
x = distance between m₁ and m₂
d - x = distance between m₂ and m₃
we get
F₁₂ = G*m² / x²
F₃₂ = G*m² / (d - x)²
If F₁₂ = - F₃₂
⇒ G*m² / x² = - G*m² / (d - x)²
⇒ 1 / x² = - 1 / (d - x)²
⇒ (d - x)² = - x²
⇒ d²- 2*d*x + x² = - x²
⇒ 2*x² - 2*d*x + d² = 0
⇒ 2*x² - 2*1*x + 1² = 0
⇒ 2*x² - 2*x + 1 = 0
Solving the equation we obtain
x = 0.5 m = 0.5*d
