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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The value of acceleration due to gravity of the earth is

`98 m//s^2`

`9.8 m//s^2`

`0.98 m//s^2`

`0.098 m//s^2`

The value of acceleration due to gravity of earth :

is the same on equator and poles

is the least on poles

is the least on equator

increase from pole to equator

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

is maximum at the poles

is maximum at the equator

remains constant everywhere on the surface of the earth

is maximum at the international time line