The figure shows a spherical hollow inside a lead sphere of radius R 4.2 m; the

Question

The figure shows a spherical hollow inside a lead sphere of radius R 4.2 m; the surface of the hollow passes through the center of the sphere and "touches" the right side of the sphere. The mass of the sphere before hollowing was M = 426 kg, with what gravitational force does the hollowed- out lead sphere attract a small sphere of mass m = 47 kg that lies at a distance d 19 m from the center of the lead sphere, on the straight line connecting the centers of the spheres and of the hollow? in Number Units the tolerance is +/-290 Click if you would like to Show Work for this question: Open Show Work SHOW HINT SHOW SOLUTION

Explanation / Answer

mass of hollow part before getting removed = 426*(1/2)^3 = 53.25 kg

Now by law of gravitation,

Force on m = summation Gm1m2/r^2

= 6.67e-11*426*47/19^2 + 6.67e-11*-53.25*47/(19-4.2/2)^2

= 3.115*10^-9 N answer

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