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F1=m1g                                      ............................................ accelaration due to gravity 
F2=GM1 M2/R²
F= gravitational force
g= gravitational accelaration
R= distance means height 
both shows force so equal them 
F1= F2

now part 2nd 
we have mg=Gm1m2/r²
Density = mass / volume 
volume of  sphere = 4/3 \pi
density =D
Mass=4/3 \pi r³×D
 we have already proved that g=GM2/R²
put the value of M2 we drived
g=G4/3 \pi R³/R²
g=G4/3 \pi R
When the object is at a depth d below the surface of earth, it is attracted by the sphere of radius (R-d) only as the attraction due to the rest of earth cancel out. (At the centre of earth, the object would be attracted equally in all directions and the net force experienced by the object and hence the value of g would be zero)
M1=4/3 \pi (R-d)³
g1=4/3 \pi (R-d)³/(R-d)²
g1=4/3 \pi (R-d)

hence -------------g1=4/3π(R-d)
g1=g{(R-d)R}.............................hence prove 

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