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If the radius of the earth were to shrin...

If the radius of the earth were to shrink by 1% its mass remaining the same, the acceleration due to gravity on the earth's surface would

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`g=(Gm)/(R^2)` if R decreases by 1% it becomes `(99)/(100)R`
`g'= (GM)/((99R)^2)=1.02(Gm)/(R^2)= (1+0.02)(Gm)/(R^2)`
g' increases by `0.02 (Gm)/(R^2)`, therefore increases by 2%.
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Knowledge Check

  • If the radius of the earth where to shrink by 1% with its mass remaining the same, the acceleration due to gravity on the earth's surface would :

    A
    decrease by 1%
    B
    decrease by 2%
    C
    increase by 1%
    D
    increase by 2%

  • If the radius of the earth where to shrink by 1% with its mass remaining the same, the acceleration due to gravity on the earth's surface would :

    A
    decrease by 1%
    B
    decrease by 2%
    C
    increase by 1%
    D
    increase by 2%

  • The Earth is not a perfect sphere. Due to the flattening of Earth at the poles, the radius of Earth is minimum at the poles and hence the value of g is maximum at the poles. One the other hand, the radius of the Earth is maximum at the equator of the Earth. As the mass and radius of moon are smaller than that of the Earth, so the value of g on moon is 1 . 63 m//s ^(2) . As we go up from the surface of the Earth, the distance from the centre of the Earth increases and hence the value of g decreases. The value of g decreases as we go down inside the Earth and it become zero at the centre of the Earth . If the radius of the Earth were to shrink by 1% and its mass remaining the same, the acceleration due to gravity on the Earth's surface would

    A
    decreases
    B
    increases
    C
    remain unchanged
    D
    will decrease by ` 9 . 8 % `

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