Show the `g=(GM)/(R^(2))`

The escape velocity of a body from the surface of the earth depends upon (i) the mass of the earth M, (ii) The radius of the eath R, and (iii) the gravitational constnat G. Show that v=ksqrt((GM)/R) , using the dimensional analysis.

Assertion : If the radius of earth is decreased kepping its constant, effective value of g may increase or decrease at pole. Reason : Value of g on the surface of earth is given by g= (Gm)/(R^(2)) .

If the gravitational field intensity at a point is given by g = (GM)/(r^(2.5)) . Then, the potential at a distance r is

If the gravitational field intensity at a point is given by g = (GM)/(r^(2.5)) . Then, the potential at a distance r is

If G is the G.M. between two positive numbers a and b, show that (1)/(G^(2) - a^(2)) + (1)/(G^(2) - b^(2)) = (1)/(G^(2)) .

Prove that g = GM/R^2 , where M is mass and R is radius of earth.

If p, q, r, s are in G.P. then show that (q -r )^(2) + ( r -p) ^(2) + ( s-q) ^(2) = ( p -s) ^(2)