Question

how will the weight of a body of mass 100g change if it is taken from equator to the poles ? give reason for your answer

Answer 1

At eguator due to increase in radius there will be decrease in g.This will result in low weigt in equator as weight=mass*g....At pole is the opposite and the a body of same mass will weigh more there.

Answer 2

Explanation:

At equator due to increase in radius there will be decrease in g. This will result in low weight in equator as

$\text {Weight} = \text {Mass} \times g$

At poles due to force at opposite direction, a body of same mass will weigh more at poles.

g is acceleration due to gravity.

g is indirectly proportional to radius. As radius increases value of g decreases. So, it is least at equator and maximum at poles.

The mass of Earth is roughly $5.972 \times 10^{27} \ g$ and mass of object is 100 g.

The radius of Earth at equator = $6.378 \times 10^{6} \ \mathrm{m}$

The radius of Earth at poles = $6.356 \times 10^{6} \ \mathrm{m}$

$g=\frac{m_{1} m_{2}}{r^{2}}$

$g(a t \text { equator })=\frac{5.972 \times 10^{27} \times 100}{6.378 \times 10^{6}}=93.6 \times 10^{21} \ \mathrm{m} / \mathrm{s}^{2}$

$g(a t \text { poles })=\frac{5.972 \times 10^{27} \times 100}{6.356 \times 10^{6}}=93.9 \times 10^{21} \ \mathrm{m} / \mathrm{s}^{2}$

$\Rightarrow \text { Weight }=\text {Mass} \times g$

$\text {Weight at equator} = 93.6 \times 10^{21} \times 100=93.6 \times 10^{19} \ N$

$\text {Weight at poles} = 93.9 \times 10^{21} \times 100=93.6 \times 10^{19} \ N$