Problem 6.9 Part A Four 85-kg spheres are located at the corners of a square of

Question

Problem 6.9 Part A Four 85-kg spheres are located at the corners of a square of side 0 70 m Calculate the magnitude of the gravitational force exerted on one sphere which is in left and down corner of the square by the other three Express your answer using two significant figures 2.5.10 Submit My Answers Give Up Incorrect: Try Again; 29 attempts remaining Part B What is it's direction? Assume that z axis is directed to the right Express your answer using two significant figures. from the z-axs Submit My Answers Give Up

Explanation / Answer

Let the sphere on which we have to calculate the force be sphere 1

the sphere right to it will be sphere 2

sphere directly above it let it be sphere 3 and sphere diagonally opposite be sphere 4.

So magnitide of force due to sphere 2 and 3 on sphere 1 will be equal so

the direction of resultant force will be in the direction of the force exerted by sphere 4.

Force on sphere 1 due to 2 and 3 will be

(G*M*m)/R^2

So put, G= 6.67*10^(-11)

M=m=8.5Kg

R=0.7

So force will be

983.484*10^(-11)N

so their resultant will be

983.484*10(^-11)*1.414

1390.646*10^(-11)

Now, force due to sphere 4 will be

F=

[6.67*10^(-11)*8.5*8.5]/.7*.7*1.414*1.414

491.89*10^(-11)N

So thier resultant will be

[1390.646+491.89]*10^(-11)

1882.536*10^(-11)N

And it's direction will be 450 with respect to the line joinig sphere 1 and 2 in the counter clockwise sense, or you can say along the line joining sphere 1 and 4.(diagoal line)

mass of earth is 5.9742*10^(24)kg

radius of earth is 6400km or 6400000m

So force on aircraft will be

{[6.67*10^(-11)]*[5.9742*10^(24)]*[1400]}/19200000*19200000

0.00015133*10^(19)N

or 1.5133*10^(15) N

Value of acceleration due to gravity is

GM/R^2

Here R will be

radius of earth+height

so

R=6400000+6500

R=6406500m

So

g= {[6.67*10^(-11)]*[5.9742*10^(24)]}/6406500*6406500

g=9.708

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