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If the radius of the Earth shrinks by 2%...

If the radius of the Earth shrinks by `2%`, mass remaing same, then how would the have of acceleration due to gravity change?

A

decrease by 2%

B

increased by 2%

C

incresed by 4%

D

decrease by 4%

Text Solution

Verified by Experts

The correct Answer is:
C
As `Gg=(GM)/(R^(2))` if R decreases then g increases taking logarithm of both the sides we get
`log g = log G+log M -2 log R`
Differentaiting it we get
`(dg)/(g)=0 +0-(2dR)/(r )=-2(-2)/(100)=(4)/(100)`
` therefore` % increasing in `g=(dg)/(g)xx100=4/100xx100=4%`
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Knowledge Check

  • If the radius of the earth shrinks by 1.5% ( mass remaining same) , then the value of acceleration due to gravity changes by

    A
    0.01
    B
    0.03
    C
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    D
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  • If the radius of the earth shrinks by 1.5% (mass remaining same), then the value of acceleration due to gravity changes by

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    `1%`
    B
    `2%`
    C
    `3%`
    D
    `4%`

  • 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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