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A body is taken from the equator to the poles. Which of the following will be the percentage change in the weight of the body? The equatorial radius of the earth is $21 km$ more than the polar radius of the earth. Polar radius, $R_{p} = 6357 km$ (Gravitational constant $G = 6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}$ )

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0.6 %

1.6 %

6 %

0.06 %

answer is A.

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Detailed Solution

The percentage change in the weight of the body in polar region compared to equatorial region of the earth is

0.6 %

under the given condition.
As per Newton’s law of gravitation we know that

F = \frac{G m_{1} m_{2}}{d^{2}} . . . . . . . . . . . . . . (a)

where

G

is the gravitational constant (

6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}

)

m_{1}

is the mass of the first object.

m_{2}

is the mass of the second object .

d

is the distance between them.
Here

d

can be written as

R

which is the radius of the earth.

m_{1} = M

as the mass of earth.

m_{2} = m

as the mass of the body.
So from (a) we can write

F = \frac{GMm}{R^{2}} . . . . . . . . . . . . . (i)

At the same time

W = mg

F = mg . . . . . . . . . . . . . (ii)

where

F

is the force of attraction.

m

is the mass of the body.

W

is the weight of the body.

g

is the acceleration due to gravity.
We can equate the equations (i) and (ii), We get

\frac{GMm}{R^{2}} = mg

\Rightarrow g = \frac{GMm}{R^{2} m}

\Rightarrow g = \frac{GM}{R^{2}} . . . . . . . . . . . . . (iii)

Using the equation (iii) we can write the gravity of the earth at the equator and at the pole.

g_{e} = \frac{GM}{{R_{e}}^{2}} . . . . . . . . . . . (b)

where

g_{e}

is the gravity at the equator and

R_{e}

is the radius of earth at the equator.

g_{p} = \frac{GM}{{R_{p}}^{2}} . . . . . . . . . . . . . . (c)

where

g_{p}

is the gravity at the equator and

R_{p}

is the radius of earth at the pole.
We are asked to find the percentage difference of the weight of the body at the pole from the equator. Hence, dividing the weight of the body at the pole with the weight of the body at the equator, We get

\frac{W_{P}}{W_{e}} = \frac{m g_{p}}{m g_{e}}

\Rightarrow \frac{W_{P}}{W_{e}} = \frac{g_{p}}{g_{e}}

Substituting value of

g_{e}

and

g_{p}

from (b) and (c)

\Rightarrow \frac{\frac{GM}{{R_{p}}^{2}}}{\frac{GM}{{R_{e}}^{2}}} = \frac{{R_{e}}^{2}}{{R_{p}}^{2}}

\Rightarrow \frac{W_{P}}{W_{e}} = {(\frac{R_{e}}{R_{p}})}^{2}

\Rightarrow \frac{W_{P}}{W_{e}} = 1.006

\Rightarrow W_{P} = 1.006 W_{e}

Therefore, change in weight is

{= W}_{P} - W_{e}

= 1.006 W_{e} - W_{e}

= 0.006 W_{e}

∴ %

change in weight

= \frac{0.006 W_{e}}{W_{e}} \times 100

\Rightarrow %

change in weight

= 0.6 %

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A body is taken from the equator to the poles. Which of the following will be the percentage change in the weight of the body? The equatorial radius of the earth is 21 km more than the polar radius of the earth. Polar radius, Rp=6357 km (Gravitational constant G =6.67×10-11 N m 2kg-2)

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Detailed Solution

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