Question #99452
THE POTENTIAL OF GRAVITATIONAL FORCE FIELD IS GIVEN AS

f(x,y,z) = - (Gm^2)/r (r>0)

Where r= {(x-x0) ^2 +(y-y0) ^2 + (z-z0) ^2} ^1/2

And r is the distance between two particles.

Show that f(x,y,z) satisfies the equation

(∂^2f)/(∂x^2) + (∂^2f)/(∂y^2) + (∂^2f)/(∂z^2) = 0
1
Expert's answer
2019-11-27T10:16:47-0500

\frac{}{}

xar=arxr1r\frac{∂}{∂x}\frac{a}{r}=a\frac{∂r}{∂x}\frac{∂r^{-1}}{∂r}

xar=axrr2=axr3\frac{∂}{∂x}\frac{a}{r}=-a\frac{x}{r}r^{-2}=\frac{-ax}{r^3}

2x2ar=a3x2r2r5\frac{∂^2}{∂x^2}\frac{a}{r}=a\frac{3x^2-r^2}{r^5}

2y2ar=a3y2r2r5\frac{∂^2}{∂y^2}\frac{a}{r}=a\frac{3y^2-r^2}{r^5}

2z2ar=a3z2r2r5\frac{∂^2}{∂z^2}\frac{a}{r}=a\frac{3z^2-r^2}{r^5}

Thus,

a3x2r2r5+a3y2r2r5+a3z2r2r5=0a\frac{3x^2-r^2}{r^5}+a\frac{3y^2-r^2}{r^5}+a\frac{3z^2-r^2}{r^5}=0


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