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Calculate the value of acceleration due ...

Calculate the value of acceleration due to gravity at a place of latitude `30^(@)`. Radius of the Earth `6.4xx 10^(6)m`.The value of acceleration due to gravity on Earth is `9.8 m//s^(2)`.

Text Solution

Verified by Experts

Here, `lambda = 60^(@), R = 6.38 xx 10^(3) km = 6.38 xx 10^(6) m`
Acceleration due to gravity at the lititude `lambda` is
`g' = g - R omega^(2) cos^(2) lambda = g - R(4pi^(2))/(T^(2)) cos^(2) lambda`
`= 9.8 - (6.38 xx 10^(6)) xx (4 xx (3.14)^(2))/((24 xx 60 xx 60)^(2)) xx cos^(2) 60^(@)`
`= 9.8 - (6.38 xx 10^(6) xx (3.14)^(2))/((24 xx 60 xx 60)^(2))`
`= 9.8 - 0.0084 = 9.79 ms^(-2)`
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Knowledge Check

  • The value of acceleration due to gravity of the earth is

    A
    `98 m//s^2`
    B
    `9.8 m//s^2`
    C
    `0.98 m//s^2`
    D
    `0.098 m//s^2`

  • The value of acceleration due to gravity of earth :

    A
    is the same on equator and poles
    B
    is the least on poles
    C
    is the least on equator
    D
    increase from pole to equator

  • The value of acceleration due to gravity at the surface of earth

    A
    is maximum at the poles
    B
    is maximum at the equator
    C
    remains constant everywhere on the surface of the earth
    D
    is maximum at the international time line