Gravitation

Since gravitational force is inversely proportional to the square of distance between the particles/bodies and according to given conditions of question, the distance between two body is now three times more than before, hence it is F/(3*3).

m, a m, a
Here, F = G mm/(2a)2 = Gm2/4a4 = G/4a2 × (D × 4/3 × πa3)2 = (4/9)Gπ2D2a4
 

Effective acceleration due to gravity is given by, g’ = g-Rω² cos2Ө.
Hence, change in acceleration due to gravity is given as, ∆g = g’-g = - Rω² cos2Ө
At equator Ө = 0.
Hence, g’-g = -R ω2 = -6.4×10⁶× (7.27×10^-5)²
= -0.0338 m/s²
Or, g’-g = -3.38 cm/s²
The (-) ve sign signifies that the decrease in gravitational acceleration is due to the rotation of earth.

Fcentripetal = Fgravitation
mω2R = mg or, ω = √ (g/R) or, T = 2π√(R/g)
Since g = 9.8 m/s2 and r = R = 6.4×106 (radius of the earth)
Now, T = 2π√(R/g) = 2π√ (6.4×106/9.8) = 84 min (approx.)
 

It is known that gravitational potential is a scalar quantity.