Assuming the radius of the earth to be $6.38\times {10}^{8}cm$, the gravitational constant to be $6.67\times {10}^{-}8c{m}^3/g{s}^2$ , acceleration due to gravity on the surface to be $980cm/s^2$, find the mean density (in $-g/c{m}^3$) of the earth.

Q.

The acceleration due to gravity g and the mean density of earth p are related by which relation.?

{G=gravitational constant, R= radius of earth{

Q.

Acceleration due to gravity 'g' in terms of mean density of the Earth, 'd ' (where R is radius of the Earth and G is universal gravitational constant) is

Q. Radius of earth is equal to $6\times {10}^{6}m$. Acceleration due to gravity is equal to $9.8m/s^2$. Gravitational constant $G$ is equal to $6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$. Then mass of the earth is:-

Q. Solve:
Estimate the mass of the earth, given, radius of the earth $=6.4\times {10}^{6}m$, acceleration due to gravity $=9.8m/s^2$ and gravitational constant $=6.67\times {10}^{-11}S.I.$ units.

Q. At a point above the surface of earth, the gravitational potential is $-5.12\times {10}^{7}J/kg$ and the acceleration due to gravity is $6.4m/s^2$. Assuming the mean radius of the earth to be $6400km$, calculate the height of this point above the earth's surface.