Assume the earth to be a nonrotating sphere with mass MEME and radius RERE. If an astronaut weights WW on the ground, what is his weight whe

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Assume the earth to be a nonrotating sphere with mass MEME and radius RERE. If an astronaut weights WW on the ground, what is his weight when he is 2RE2RE above the surface of the earth

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Diễm Thu 4 years 2021-09-02T17:32:54+00:00 1 Answers 77 views 0

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  1. thienthanh
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    2021-09-02T17:34:44+00:00
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    Answer:

    The weight at a distance 2 RE from surface of earth is W/9

    Explanation:

    For the value of acceleration due to gravity (g), we have a formula, that is:

    g = (G)(ME)/(RE)²    —– equation (1)

    where,

    G = Gravitational Constant

    ME = Mass of Earth

    RE = Radius of Earth

    g = Acceleration due to gravity on surface of earth = 9.8 ms²

    When the person goes 2RE, distance above earth’s surface. Then the total distance from center of earth becomes: 2RE + RE = 3RE.

    Therefore, equation (1) becomes:

    gh = (G)(ME)/(3RE)²

    where,

    gh = acceleration due to gravity at height

    gh = (G)(ME)/(RE)²9

    using equation (1), we get:

    gh = g/9

    Now, he weight is given by formula:

    W = mg   ——- equation (2)

    At height 2RE

    Wh = (m)(gh)

    where,

    Wh = Weight at height = ?

    m = mass of astronaut

    Therefore, using vale of gh, we get:

    Wh = mg/9

    Using equation (2), we get:

    Wh = W/9

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