Question

Re 1 kg of sugar rate more at poles or at the centre of the earth derive an expression for it

Answer 1

Hey mate!!!

$<b >Here is your answer ;</b>$

We know by the newton's $2nd$ law of motion that;
$F = mg$_eq1
(where $<b>g = Acceleration due to gravity, F = weight of the mass m</b>$ )
And also,
By newton's universal law of gravitation,
$F = G( \frac{Mm} {r^2})$ _eq2
Where $<b> M is the mass of earth with radius r</b>$

From eq 1 and 2,
We get;
$mg = G( \frac{Mm} {r^2})$
=>$g = G( \frac{M} {r^2})$
=>$g ∝ \frac{1}{r^2}$

You get the expression for the Acceleration due to gravity.

The earth is geoid in shape, ie. Flattened at the poles.

Weight of an object = $mg$
As mass is constant
Weight of an object ∝ $g$

And radius of equator > radius of the poles.
The Acceleration due to gravity is maximum at the poles.

If the cost of sugar is measured with weight,
Then,
The weight or cost of sugar becomes $<b>maximum at the poles</b>$

$<h1><marquee >Hope it helps you </h1>$