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# The gravitational force on a small particle lying on earths surface is F. If a concentric spherical cavity of radius R/2 is made inside earth, then force on the particle is aF/2  b. F/8  c. 7F/8  d. 17F/8

Ashutosh Kumar Singh
28 Points
one year ago
$F=mg=GM_{e}m/R^{2}$
$g=\frac{GM_{e}}{R^2}$
Mass = volume * density
after removal of the cavity, the effective mass will be:
$M'=M_{e}-\frac{4}{3}\Pi R^3\rho _{e}$
$M'=\frac{7}{8}M_{e}$
New gravitational acceleration will be:
$g'=\frac{GM'}{R^2}=\frac{7}{8}g$
New force will be:
$F'=mg'=m*\frac{7}{8}g =\frac{7}{8}F$
Note - while calculating the gravitational acceleration the effective distance is still R because the position of the center of mass is not changing only effective mass is changing.