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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 a
F/2 b. F/8 c. 7F/8 d. 17F/8

Preeti Kharel , 5 Years ago
Grade 12th pass
anser 1 Answers
Ashutosh Kumar Singh

Last Activity: 5 Years 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.
 

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