At what distance above the surface of the earth is the acceleration due to the earth’s gravity 0.995 m/s2 if the acceleration due to gravity

Question

At what distance above the surface of the earth is the acceleration due to the earth’s gravity 0.995 m/s2 if the acceleration due to gravity at the surface has magnitude 9.80 m/s2?

in progress 0
2022-04-24T05:00:11+00:00 2 Answers 0

Answers ( 2 )

  1. Myna
    0
    2022-04-24T05:01:25+00:00

    If we assume the Earth to be of uniform density and spherical, then Newton’s inverse square law of g-field can be applied.

    i.e. g = GM/ru00b2, which means g is inversely proportional to ru00b2
    Or, gru00b2 = constant.

    The Earth’s radius is approximately 6.4×10^6 m

    So, 0.975 x (R’)u00b2 = 9.80 x (6.4×10^6)u00b2, where R’ is the distance from Earth’s centre.
    R’ = 2.03×10^7 m

    Distance above Earth’s surface = 2.03×10^7 – 6.4×10^6 = 1.39×10^7

  2. Allison
    0
    2022-04-24T05:01:32+00:00

    Answer:

    The distance above the surface of the earth is [tex]1.36times 10^7[/tex].

    Explanation:

    Given that,

    Acceleration due to the gravity g’= 0.995 m/su00b2 (at height )

    Acceleration due to the gravity g= 9.80 m/su00b2 (at surface )

    We know that,

    The formula of gravity at height

    [tex]g’=dfrac{g}{(1+dfrac{h}{R})^2}[/tex]

    Where, h = height

    r= radius of the earth

    We substitute the value into the formula

    [tex]0.995=dfrac{9.8}{(1+dfrac{h}{6.378times10^{6}})^2}[/tex]

    [tex]h=6.378times 10^6(sqrt{dfrac{9.8}{0.995}}-1)[/tex]

    [tex]h=1.36times 10^7 m[/tex]

    Hence, The distance above the surface of the earth is [tex]1.36times 10^7[/tex].

Leave an answer

Browse
Browse

45:5+15*4 = ? ( )