1.)A lead sphere of radius R = 4.00 cm has a spherical hollow inside it whose su
ID: 2242825 • Letter: 1
Question
1.)A lead sphere of radius R = 4.00 cm has a spherical hollow inside it whose surface passes through the center and the right edge of the sphere as depicted in the figure. A small mass m= 0.462 kg is placed at a distance d = 9.00 cm from the center of the lead sphere, on the straight line connecting the centers of the spheres and of the hollow (see the figure). What is the gravitational force the hollowed-out lead sphere applies on the small mass if the mass of the full sphere before hollowing was M = 3.20 kg?
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Hint: Calculate the gravitational force on the small mass due to the full lead sphere, and then calculate the gravitational force on the small mass due to a lead sphere of the same size of the hollowing and at its location.
2.)At what distance from the Earth center (in units of km) will an object of mass m weigh the same as it does on the surface of the Moon?
A lead sphere of radius R = 4.00 cm has a spherical hollow inside it whose surface passes through the center and the right edge of the sphere as depicted in the figure. A small mass m= 0.462 kg is placed at a distance d = 9.00 cm from the center of the lead sphere, on the straight line connecting the centers of the spheres and of the hollow (see the figure). What is the gravitational force the hollowed-out lead sphere applies on the small mass if the mass of the full sphere before hollowing was M = 3.20 kg? Hint: Calculate the gravitational force on the small mass due to the full lead sphere, and then calculate the gravitational force on the small mass due to a lead sphere of the same size of the hollowing and at its location. At what distance from the Earth center (in units of km) will an object of mass m weigh the same as it does on the surface of the Moon?Explanation / Answer
mass of unaltered sphere
m1 = 3.35
mass of altered sphere is
= 3.35 * (1 - (2/4)^3) = 3.35 * (1 - 1/8) = 3.35 * 7/8
mass removed
mx = 3.35 * 1/8
other mass
m2 = 0.335
9 cm from m1 and 7 cm from mx
F = (G*m1*m2 / 0.09^2) - (G*mx*m2 / 0.07^2)
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