Spherical Ball - Variable Mass Density

Classical Mechanics Level 5

A solid spherical ball of radius 1 1 has its center at the origin of the x y z xyz coordinate system. The ball's variable volume mass density is expressed as ( x + y + z ) 2 (x+y+z)^2 .

A test mass is positioned at ( x , y , z ) = ( 0 , 0 , 2 ) (x,y,z) = (0,0,2) . For simplicity, assume that the universal gravitational constant and test mass are both numerically equal to 1 1 .

What is the magnitude of the gravitational force exerted by the ball on the test mass?


The answer is 0.63.

Steven Chase by Steven Chase , 37, USA

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1 solution

Mark Hennings
May 8, 2019

The force is ∭ x 2 + y 2 + z 2 ≤ 1 ( x + y + z ) 2 ( x 2 + y 2 + ( z − 2 ) 2 ) 3 2 ( x y z − 2 ) d x d y d z = ( 0.0448799 0.0448799 − 0.628318 ) \iiint_{x^2+y^2+z^2\le 1} \frac{(x+y+z)^2}{\big(x^2+y^2+(z-2)^2\big)^{\frac{3}{2}}}\left(\begin{array}{c} x \\ y \\ z-2 \end{array}\right)\,dx\,dy\,dz \; = \; \left( \begin{array}{c} 0.0448799 \\0.0448799 \\-0.628318\end{array}\right) which has magnitude 0.631516 \boxed{0.631516} . Since the mass distribution is not spherically symmetric, we cannot make a point particle approximation for the ball.

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