qz8sol_6839sp10 - m due to the lead sphere M before...

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TA: Tomoyuki Nakayama Tuesday, March 30 th , 2010 PHY 2048: Physic 1, Discussion Section 6839 Quiz 8 (Homework Set # 11) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ The figure below right shows a spherical hollow inside a lead sphere of radius R = 5.00 cm; 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 = 4.00 kg. A small sphere of mass m = 0.500 kg lies at a distance d = 10.0 cm from the center of the lead sphere, on the straight line connecting the center of the spheres and of the hollow? a) What is the magnitude of the gravitational force on the small sphere
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Unformatted text preview: m due to the lead sphere M before hollowing? The gravitational force between spherical objects is the same as that between two particles with all masses concentrated at their centers. Therefore, F M = GMm/d 2 = 1.33 × 10-8 N b) What is the mass of the hollowed out lead sphere? The removed part has radius R/2. Thus its mass is M’ = ρ(4π/3)(R/2) 3 = (1/8) × ρ(4π/3)R 3 = M/8 = 0.500 kg The mass of the hollowed out sphere is M – M’ = 3.50 kg c) What is the magnitude of the gravitational force on the small sphere m by the hollowed out lead sphere? According to the superposition principle, the gravitational force due to the sphere before hollowing is the sum of the gravitational force due to the removed part and the hollowed out sphere. F M-M’ + F M’ = F M F ⇒ M-M’ = F M - F M’ = GMm/d 2 – GM’m/(d-R/2) 2 = 1.04 × 10-8 N...
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This note was uploaded on 01/15/2012 for the course PHY 2048 taught by Professor Field during the Fall '08 term at University of Florida.

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