has 5 times the mass of the earth and also 5 times the radius. How does the gravitational acceleration on the surface of Driff compare to th

Question

has 5 times the mass of the earth and also 5 times the radius. How does the gravitational acceleration on the surface of Driff compare to the gravitational acceleration on the surface of the earth?

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Nguyệt Ánh 3 years 2021-08-24T02:08:10+00:00 2 Answers 12 views 0

Answers ( )

  1. haidang
    0
    2021-08-24T02:09:20+00:00
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    Answer:

    The gravitational acceleration on earth is 5 times that on Driff.

    Explanation:

    The gravitational acceleration on earth is given as:

    ge = Gm/r²

    Where m = mass of earth, r = radius of earth.

    The gravitational acceleration on Driff is given as:

    gd = GM/R²

    Where M = Mass of Driff, R = radius of Driff.

    Since we’re told that M = 5m and R = 5r:

    gd = G*5m/(5r)²

    gd = Gm/5r²

    Comparing this to the gravitational acceleration on earth:

    ge : gd = Gm/r² : Gm/5r²

    ge:gd = 5:1

    The gravitational acceleration on earth is 5 times that on Driff.

  2. mit
    0
    2021-08-24T02:09:42+00:00
    Reply

    Answer:

    The planet has one-fifth of the gravitational acceleration of the Earth.

    Explanation:

    Newton’s law of gravitational attraction between two bodies of masses m_1 and m_2 separated by a distance d gives

    F = G\dfrac{m_1m_2}{d^2}                       (G is a universal constant)

    For any body of mass m on the Earth surface,

    F = G\dfrac{Mm}{R^2}

    Here, M and R are the mass and the radius of the Earth, respectively.

    But this force is the gravitational force on the body.

    F = mg = G\dfrac{Mm}{R^2}

    g = G\dfrac{M}{R^2}

    For a planet with 5 times the mass of Earth and 5 times its radius,

    g_p = G\dfrac{5M}{(5R)^2} = G\dfrac{M}{5R^2} = \dfrac{1}{5}G\dfrac{M}{R^2} = \dfrac{g}{5}

    Hence the planet has one-fifth of the gravitational acceleration of the Earth.

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