A body is weighed by a spring balance to be 1.000 Kg at the north pole.How much will it weigh at the equator? Account for the earth's rotation only.

Question

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Answer 1

Let g' be the acceleration due to gravity at equation & that of pole = g
$ω_{e a r t h} = 7.3 \times 1 0^{- 5}$
$g^{'} = g - ω^{2} R$
$g^{'} = 9.81 - (7.3 \times 1 0^{- 5})^{2} \times 6400 \times 1 0^{3}$
$g^{'} = 9.81 - 0.034$
$g^{'} = 9.776 m / s^{2}$

$m g = 1 k g \times 9.77 m / s^{2}$
$= 9.776 N o r 0.997 K g$
The body will weigh 0.007 Kg at equator

Answer 2

Let g' be the acceleration due to gravity at equation & that of pole = g

ω_{e a r t h} = 7.3 \times 1 0^{- 5}

g^{'} = g - ω^{2} R

g^{'} = 9.81 - (7.3 \times 1 0^{- 5})^{2} \times 6400 \times 1 0^{3}

g^{'} = 9.81 - 0.034

g^{'} = 9.776 m / s^{2}

m g = 1 k g \times 9.77 m / s^{2}

= 9.776 N o r 0.997 K g

The body will weigh 0.007 Kg at equator