Question

$\sf\underline{ ♡Question♡}$

Find the percentage change in the weight of a body when it is taken from the equator to the poles. The radius of the earth at the poles is 6357 km, the radius at the equator is 6378 km.

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Answer 1

Answer:

65.74%

Explanation:

g=GM/R^2

g at pole=6.67 ×10^-11×M/(6357×10^3)^2

g at equator =6.67×10^-11 ×M/(6378×10^3)^2

6378^2×10^6/6357^2 ×10^6

(6378^2-6357^2 )/6378^2 ×100%= 65.74%

Answer 2

Answer:

The value of acceleration due to gravity at equator (g) and that at poles (gp) are given by

GM and (9₂) = GM R²p P (98) = ² RE

Weight of a person at equator. WE = m(g) Weight of the man at pole. Wp = m(gp)

P

The percentage change in the weight of the person in moving from equator to pole = Wp-WE WE

* 100

or percentage change: m(gp)-m(9E); m(9E) 100

= (9p)-(9E) (9E) 100

GM GM R² R²E * .* 100 GM RE

R² E-R² p * R² p 100

(6378)²-(6357) ²* (6357) 2 = 100

12735 * 21 * 6357 * 6357 = 100 =

0.66%