Akg object and a kg object are separated by m. find the magnitude of the net gravitational force exerted by these objects on a kg object placed midway between them. 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?

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

the answer is a. at the equator