Suppose that F is the inverse square force field below, where c is a constant Fi

Question

Suppose that F is the inverse square force field below, where c is a constant Find the work done by F in moving an object from a point Pi along a path to a point P2 in terms of the distances d and c/2 from these points to the An example of an inverse square field is the gravitational field F = -(mMG)r/|r|3. Use part (a) to find the work done by the gravitational field when the earth moves from aphelion (at a maximum distance of 1.52 x 108 km from the sun) to perihelion (at a minimum distance of 1.47 x 108 km). Use the values m = 5.97 x 1024 kg, M = 1.99 x 1030 kg, and G = 6.67 X 10-11 N-m2/kg2. (Round your answer to two decimal places.) Another example of an inverse square field is the electric force field F = sgQr/|r|3. Suppose that an electron with a charge of -1.6 X 10"19 C is located at the origin. A positive unit charge (1 C) is positioned a distance 10-12 m from the electron and moves to a position half that distance from the electron. Use part (a) to find the work done by the electric force field. Use the value s = 8.985 x 109. (Round your answer to the nearest hundred joules.)

Explanation / Answer

F = cr / ||r||³

Work = [PP] F•dr = c [PP] r•dr/||r||³


r / ||r||³ = (x,y,z)(x²+y²+z²)¹

= ( {(x²+y²+z² ) }/x, {(x²+y²+z² ) }/y, {(x²+y²+z² ) }/z )

= ((x²+y²+z² )) = where =1/d


So F is conservative and work is change in c from P to P

= c/d (c/d) = c/d c/d

so workdone is   W = c(1/ - 1/) answer for part a

b) F= (mMG)/r^2

= (5.97*1.99*6.67)* (10^(24+30-11)/(1.52*10^11)^2 since 1km =1000m

=34.297*10^(21)

Now we   know that W= F*r

= 34.297*1.52*10^21*10^11 joules

=34.297*10^32

=34.30 * 10^32   joules answer

c)   F = (8.985*1.6*1)(10^(9-19-12+36) joules

=14.376*10^14 joules

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