Question

The value of acceleration due to gravity

• A. is least on equator.
• B. is least on poles.
• C. increases from pole to equator.
• D. is same on equator and poles.

Answer 1

Answer: (A)

Because of centripetal acceleration you will weigh a tiny amount less at the equator than at the poles. Try not to think of centripetal acceleration as a force though; what's really going on is that objects which are in motion like to go in a straight line and so it takes some force to make them go round in a circle. So some of the force of gravity is being used to make you go round in a circle at the equator (instead of flying off into space) while at the pole this is not needed.

The centripetal acceleration at the equator is given by $\frac{4{\pi }^{2}r}{T^2}$, where, $r$ is radius of the Earth and $T$ is period of rotation of the Earth.

The period of rotation is 24 hours (or 86400 seconds) and the radius of the Earth is about 6400 km. This means that the centripetal acceleration at the equator is about $0.03m/s^2$. Compare this to the acceleration due to gravity which is about $10m/s^2$ and you can see how tiny an effect this is - you would weigh about 0.3% less at the equator than at the poles!

There is an additional effect due to the oblateness of the Earth. The Earth is not exactly spherical but rather is a little bit like a "squashed" sphere, with the radius at the equator slightly larger than the radius at the poles (this shape can be explained by the effect of centripetal acceleration on the material that makes up the Earth, exactly as described above). This has the effect of slightly increasing your weight at the poles (since you are close to the center of the Earth and the gravitational force depends on distance) and slightly decreasing it at the equator.

Taking into account both of the above effects, the gravitational acceleration is $9.78m/s^2$ at the equator and $9.83m/s^2$ at the poles, so you weigh about 0.5% more at the poles than at the equator.