Assuming earth to be spherical, at what approximate height (in km) above the north pole, value of $g$ is same as that on the earth's surface at equator?

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Solution

Assuming the required height to be much smaller than the radius of earth,

$g^{'} = g (1 - \frac{2 h}{R}) = g - ω^{2} R$

$\frac{2 g h}{R} = ω^{2} R$

$h = \frac{ω^{2} R^{2}}{2 g} = \frac{(3600) 4 π^{2} 6400000^{2}}{2 (9.8) (24)} \approx 10 k m$

Answer is $10 k m$ .

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Assuming earth to be spherical, at what approximate height (in km) above the north pole, value of g is same as that on the earth's surface at equator?

Assuming the required height to be much smaller than the radius of earth,g′=g(1−2hR)=g−ω2R2ghR=ω2Rh=ω2R22g=(3600)4π2640000022(9.8)(24)≈10kmAnswer is 10km.

Assuming earth to be spherical, at what approximate height (in km) above the north pole, value of $g$ is same as that on the earth's surface at equator?

Assuming the required height to be much smaller than the radius of earth,

$g^{'} = g (1 - \frac{2 h}{R}) = g - ω^{2} R$

$\frac{2 g h}{R} = ω^{2} R$

$h = \frac{ω^{2} R^{2}}{2 g} = \frac{(3600) 4 π^{2} 6400000^{2}}{2 (9.8) (24)} \approx 10 k m$

Answer is $10 k m$ .