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

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0.7 %

0.3 %

0.8 %

answer is B.

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Detailed Solution

The radius of the earth at the poles is

6357 km

, and the radius at the equator is

6378 km

. The percentage change in the weight of the body when it is taken from the equator to the poles is

0.7 %

.
Given the radius of the earth at the equator,

R_{E} = 6378 km

Radius of the earth at the pole,

R_{P} = 6357 km

Let the mass of earth be

M

.
Gravitationla force at the equator,

g_{E} = \frac{GM}{{R_{E}}^{2}}

Gravitationla force at poles,

g_{P} = \frac{GM}{{R_{P}}^{2}}

Weight of a person having mass

m

at the equator,

W_{E} = m g_{E}

Weight of that person at the pole,

W_{P} = m g_{P}

The percentage change in the weight of the person in moving from the equator to pole

= \frac{W_{P} - W_{E}}{W_{E}} \times 100

= \frac{{mg}_{P} - {mg}_{E}}{{mg}_{E}} \times 100

= \frac{g_{P} - g_{E}}{g_{E}} \times 100

= \frac{{R_{E}}^{2} - {R_{P}}^{2}}{{R_{P}}^{2}} \times 100

= \frac{{(6378)}^{2} - {(6357)}^{2}}{{(6357)}^{2}} \times 100

≃ 0.7 %

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