If the Earth were a perfect sphere of radius `6.37 xx 10^(6) m`, rotating about its polar exis with a period of `1` day `(= 8.64 xx 10^(4) s)` how much would the acceleration due to gravity differ from the poles to equator?

If the earth were a perfect sphere of radius 6.37 xx 10^(6) m , rotating about its axis with a period of 1 day ( = 8.64 xx 10^(6) s) , how much would the acceleration due to gravity differ from the poles to the equator.

Considering the earth to be a perfect sphere of radius 6.4 xx 10^(6)m , rotating about its axis with a period of 1 day (= 8.64 xx 10^(4)s) , calculate the difference in acceleration due to gravity at the poles and the equator.

If the earth were a perfect sphere of radius 6.4 xx 10^6 m rotating its axis with the perod of one day (8.64 xx 10^4 s) . What is the difference in accelaration due to gravity from poles to equator?

Suppose Earth is perfect sphere of radius 6.4 xx 10^(6) m . It is rotating about its polar axis with a period of 1 day. What is the difference in the value of acceleration due to gravity on pole and at a place of latitude 60^(@) ?

The mass of the Earth is 6 xx 10^(24) kg and its radius is 6400 km. Find the acceleration due to gravity on the surface of the Earth.

If the earth considered as a solid sphere of iron of radius 6.37xx10^6 m and of density 7.86 g*cm^(-3) , what will be the magnitude of the acceleration due to gravity on the earth's surface ? Gravitational constant =6.58xx10^(-8) CGS unit.

Find the mass of earth (Given the radius of earth R= 6.4 xx 10^6 m) .

If the Radius of earth were shrink by 10% its mass remaining same, the acceleration due to gravity on earth.s surface would