Answer to Question #277650 in Physics for Kristine

Question

Three uniform spheres are fixed at the positions shown in Figure 90. (a) What are the magnitude and direction of the force on a 0.0150-𝑘𝑔 particle placed at 𝑃? (b) If the spheres are in deep outer space and a 0.0150-𝑘𝑔 particle is released from rest 300 𝑚 from the origin along a line 45 below the – 𝑥-axis, what will the particle’s speed be when it reaches the origin?

Answer 1

We have three spheres of mass M located in a line with a distance of 1 m between each two successive spheres. So, the force is

$F=GMm\bigg(\frac{1}{r^2}+\frac{1}{(r+1)^2}+\frac{1}{(r+2)^2}\bigg).$

The acceleration and velocity are

$a=F/m=GM\bigg(\frac{1}{r^2}+\frac{1}{(r+1)^2}+\frac{1}{(r+2)^2}\bigg),\\\space\\ v=\sqrt{2ad}=\sqrt{2GM\bigg(\frac{1}{r^2}+\frac{1}{(r+1)^2}+\frac{1}{(r+2)^2}\bigg)L\cos\theta.}$