What is that imaginary angular velocity of earth at which a body placed at equator has zero weight?The radius of the earth is $6400$ km and $g = 10 m / s^{2}$ ?

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Solution

$g^{'} = g - r ω^{2} cos λ$
At equator $λ = 0$
$g^{'} = g - r ω^{2}$
Weight of a body at equator, $W = m g^{'} = m (g - r ω^{2}) = 0$
$g = r ω^{2}$
$W = \sqrt{\frac{g}{r}}$
$= \sqrt{\frac{10}{64 \times 10^{5}}}$
$W = 0.00125$ rad/s.

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Q. What is that imaginary angular velocity of the earth at which a body placed at the equator has zero weight? The radius of the earth is 6400 km and

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What is that imaginary angular velocity of earth at which a body placed at equator has zero weight?The radius of the earth is 6400 km and g=10m/s2?

g′=g−rω2cosλAt equator λ=0g′=g−rω2Weight of a body at equator, W=mg′=m(g−rω2)=0g=rω2W=√gr=√1064×105W=0.00125 rad/s.

What is that imaginary angular velocity of earth at which a body placed at equator has zero weight?The radius of the earth is $6400$ km and $g = 10 m / s^{2}$ ?

$g^{'} = g - r ω^{2} cos λ$
At equator $λ = 0$
$g^{'} = g - r ω^{2}$
Weight of a body at equator, $W = m g^{'} = m (g - r ω^{2}) = 0$
$g = r ω^{2}$
$W = \sqrt{\frac{g}{r}}$
$= \sqrt{\frac{10}{64 \times 10^{5}}}$
$W = 0.00125$ rad/s.