Question

A body is weighted by a spring balance to be 1.000 kg at the north pole. How much will it weight at the equator. Account for the earth\'s rotation only.

Answer 1

Answer:

hloo mate..

Let g’ be the acceleration due to gravity at equation & that of pole = g

Let g’ be the acceleration due to gravity at equation & that of pole = gg’ = g ω^2 R

Let g’ be the acceleration due to gravity at equation & that of pole = gg’ = g ω^2 R= 9.81 – (7.3 * 10^-5)^2 * 6400 * 10^3

Let g’ be the acceleration due to gravity at equation & that of pole = gg’ = g ω^2 R= 9.81 – (7.3 * 10^-5)^2 * 6400 * 10^3= 9.81 – 0.034

Let g’ be the acceleration due to gravity at equation & that of pole = gg’ = g ω^2 R= 9.81 – (7.3 * 10^-5)^2 * 6400 * 10^3= 9.81 – 0.034= 9.776 m/s^2

Let g’ be the acceleration due to gravity at equation & that of pole = gg’ = g ω^2 R= 9.81 – (7.3 * 10^-5)^2 * 6400 * 10^3= 9.81 – 0.034= 9.776 m/s^2Mg’ = 1 kg * 9.776 m/s^2

Let g’ be the acceleration due to gravity at equation & that of pole = gg’ = g ω^2 R= 9.81 – (7.3 * 10^-5)^2 * 6400 * 10^3= 9.81 – 0.034= 9.776 m/s^2Mg’ = 1 kg * 9.776 m/s^2= 9.776 N or 0.997 kg

Let g’ be the acceleration due to gravity at equation & that of pole = gg’ = g ω^2 R= 9.81 – (7.3 * 10^-5)^2 * 6400 * 10^3= 9.81 – 0.034= 9.776 m/s^2Mg’ = 1 kg * 9.776 m/s^2= 9.776 N or 0.997 kgThe body will weight 0.997 kg at equator.

hope it helps..

Answer 2

GIVEN :

A body is weighted by a spring balance to be 1.000 kg at the north pole.

TO FIND :

weight of the body at the equator.

SOLUTION :

◆The acceleration due to gravity at equator - g'

◆The acceleration due to gravity at pole - g

◆By equation,

g’ = g - w^2 R. ; (w - angular velocity)

= 9.81 – (7.3 × 10 ^ -5)^ 2 × 6.4× 10^6

= 9.81 – 0.034

= 9.77 m /s ^2

◆mg’ = 1 kg × 9.77 m/s^2

= 9.77 N = 0.99 kg

ANSWER :

The weight at equator is 0.99 kg .