Answer to Question #99452 in Physics for ROHIT SHARMA

Question

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

Answer 1

$\frac{}{}$

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

$$\frac{∂}{∂x}\frac{a}{r}=-a\frac{x}{r}r^{-2}=\frac{-ax}{r^3}$$

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

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

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

Thus,

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