Answered: Two 10 kg particles are placed 500 mm…

Name: Two 10 kg particles are placed 500 mm apart. Express the gravitational
Uploaded: 2024-03-20T06:56:46.000Z
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Description: Two 10 kg particles are placed 500 mm apart. Express the gravitational attraction acting on one of the particles as a percentage (%) of its weight on Earth. This means calculate the gravitational attraction, divide it by the weight, and multiply by 1

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Two 10 kg particles are placed 500 mm apart. Express the gravitational attraction acting on one of the particles as a percentage (%) of its weight on Earth. This means calculate the gravitational attraction, divide it by the weight, and multiply by 100%. Show all necessary math and include all units.

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$G i v e n : M a s s, m = 10 k g R a d i u s, r = 500 m m = 0.5 m G r a v i t a t i o n a l c o n s \tan t, G = 6.673 \times 10^{- 11} N m^{2} / k g^{2} F i n d t h e g r a v i t a t i o n a l a t t r a c t i o n . N o w, A c c o r d i n g t o t h e l a w o f u n i v e r s a l g r a v i t a t i o n, a t t r a c t i o n f o r c e, a c t i n g o n t h e s p h e r e, i s e q u a l t o . F_{12} = G \frac{m_{1} m_{2}}{r^{2}} A n d, O n E a r t h w e i g h t o f t h e s p h e r e i s, F_{1} = m_{1} g N o w, \frac{F_{12}}{F_{1}} = \frac{G m_{1} m_{2}}{r^{2} m_{1} g} = \frac{G m_{2}}{r^{2} g} = \frac{6.673 \times 10^{- 11} N m^{2} / k g^{2} \times 100 k g}{{(0.5 m)}^{2} \times 9.81 m / s^{2}} = 2.720 \times 10^{- 9} N A t t r a c t i o n a c t i n g o n t h e s p h e r e i s p e r c e n t a g e (%) o f i t s w e i g h t o n E a r t h % \frac{F_{12}}{F_{1}} = 2.720 \times 10^{- 9} \times 100 N = 2.7208 \times 10^{- 7} %$

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