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14 April, 06:40

a kg 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?

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  1. 14 April, 08:52
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    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
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