A newly discovered planet has a radius twice as large as earth’s and a mass five times as large. What is the free-fall acceleration on its s

Question

A newly discovered planet has a radius twice as large as earth’s and a mass five times as large. What is the free-fall acceleration on its surface?

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bonexptip 6 months 2021-08-04T13:11:59+00:00 1 Answers 18 views 0

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    2021-08-04T13:13:12+00:00

    Answer:

    free-fall acceleration = 12.25 m/s²

    Explanation:

    Formula for free fall acceleration is given by the gravity equation;

    g = Gm/r²

    Where;

    M is mass of the Earth

    r is radius of the Earth

    G is the gravitational constant

    Now, that equation applies to the earth.

    For this new planet, we are told that the mass = 5 × mass of earth

    And radius = 2 × radius of earth.

    Thus, free fall acceleration for this planet is;

    g_p = G(5m)/(2r)²

    g_p = (5/4)(Gm/r²)

    Gravitational constant has a value of 6.674 × 10^(−11) N.m²/kg²

    Mass of the earth = 5.972 × 10^(24) kg

    Radius of the earth = 6378 km = 6378000

    Thus;

    g_p = (5/4)(6.674 × 10^(−11) × 5.972 × 10^(24))/(6378000²) = 12.25 m/s²

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