Two objects, with masses m1 and m2 , are originally a distance r apart. The magnitude of the gravitational between is F. The masses are chan

Question

Two objects, with masses m1 and m2 , are originally a distance r apart. The magnitude of the gravitational between is F. The masses are changed to 2m1 and 2m2 , and the distance is changed to 4r. What is the magnitude of the new gravitational force?

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Acacia 5 years 2021-07-29T06:45:34+00:00 2 Answers 523 views 0

Answers ( )

  1. kimcuc
    0
    2021-07-29T06:47:08+00:00
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    According to Newton’s gravitational law, the magnitude of the gravitational force is:

    F=-G\frac{m_1m_2}{r^2}

    In this case, we have: m_1'=2m_1,m_2'=2m_2, r'=4r

    Therefore, replacing this values and solving for F’ in function of F:

    F'=-G\frac{m_1'm_2'}{r'^2}\\F'=-G\frac{2m_1(2m_2)}{(4r)^2}\\F'=-G\frac{4m_1m_2}{16r^2}\\F'=\frac{1}{4}(-G\frac{m_1m_2}{r^2})\\F'=\frac{1}{4}F

  2. kimchi2
    0
    2021-07-29T06:47:09+00:00
    Reply

    Answer:

    \frac{1}{4} F

    Explanation:

    Newton’s law of gravitation states that the force, F, that exists between the two objects of masses m₁ and m₂ is directly proportional to the product of the masses of the objects and inversely proportional to the square of the distance, r, between the bodies. i.e

    F ∝ (m₁m₂ / r²)

    F = Gm₁m₂/r²           ——————(i)

    Where;

    G = proportionality constant called the gravitational constant.

    Now, when the masses are changed to 2m₁ and 2m₂, and the distance is changed to 4r, then according to equation (i) the new force (say F₂) becomes;

    F₂ = G (2m₁)(2m₂) / (4r)²

    F₂ = 4Gm₁m₂ / 16r²

    F₂ = Gm₁m₂ / 4r²

    F₂ = \frac{1}{4} Gm₁m₂/r²     ————————-(ii)

    Comparing equations (i) and (ii), equation (ii) can be re-written as;

    F₂ = \frac{1}{4} F

    Therefore, the magnitude of the new gravitational force is a quarter of the old gravitational force, F. i.e   \frac{1}{4} F

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