The figure shows a spherical hollow inside a lead sphere of radius R = 2.6 m; th
ID: 2195577 • Letter: T
Question
The figure shows a spherical hollow inside a lead sphere of radiusR= 2.6 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 wasM= 439 kg. With what gravitational force does the hollowed-out lead sphere attract a small sphere of massm= 29 kg that lies at a distanced= 18 m from the center of the lead sphere, on the straight line connecting the centers of the spheres and of the hollow?
Explanation / Answer
change the numbers and follow the process You can do this by subtracting the force of the hollow (as if it were filled) from the force from the sphere with no hollow. I think. Otherwise this is lots of integrals. density Lead 11340 kg/m³ Volume of total sphere = 4/3pr³ = 4/3p4³ = 268 cm³ = 268e-6 m³ Volume of void = 4/3p2³ = 33.5 cm³ = 33.5e-6 m³ mass of total sphere = 268e-6 m³ x 11340 kg/m³ = 3.039 kg mass of void, if it were filled = 33.5e-6 m³ x 11340 kg/m³ = 0.380 kg Gravitational attraction in newtons F = G m1m2/r² G = 6.674e-11 m³/kgs² m1 and m2 are the masses of the two objects in kg r is the distance in meters between their centers F = G(0.424)(3.039) / (0.09)² – G(0.424)(0.380) / (0.07)² F = (6.674e-11)(0.424) [ (3.039/0.0081) – (0.380/0.0049) ]
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