Question

The body weighs 98N on a spring balance on the north pole. The same body is shifted to the equator, the weight of the body at the equator on the same scale is (assume g=9.8m/sec2 and radius of the earth=6400km)

Answer 1

acceleration due to gravity at pole (g') = 10 m/s^2

Weight = 98N

mass = W/g' = 98/10 = 9.8 kg

At equator, acceleration due to gravity(g) = 9.8 m/s^2

Weight = mg = 9.8 × 9.8 = 96.04 N

Weight of body at equator is 96.04 N

Answer 2

Answer:

Explanation:At poles, the apparent weight is same as the true weight.

Thus,

98N=mg=m(9.8m/s^2)

m=10kg

At the equator, the apparent weight is mg'=mg-mw^2R

The radius of the earth is 6400km and the angular speed is

w= (2*22/7rad)/24*60*60s=7.27*10^-5rad/s

Thus,

mg'=98N - (10kg)(7.27*10^-5 s^-1)^2 (6400km)

=97.66N