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
Channel: Bartleby
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

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter11: Gravity, Planetary Orbits, And The Hydrogen Atom

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Concept explainers

Topic Video

Question

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.

Expert Solution

Step 1

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

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions