Question

QuestionDetails:

Answer 1

To solve the two parts of this problem, we will begin by considering the expressions given for gravitational potential energy and finally kinetic energy (to find velocity). From the potential energy we will obtain its derivative that is equivalent to the Force of gravitational attraction. We will start considering that all the points on the ring are same distance:

$r = \sqrt{x^2+R^2}$

Then the potential energy is

$U = \frac{-GMm}{\sqrt{x^2+R^2}}$

PART A) The force is excepted to be along x-axis.

Therefore we take a derivative of U with respect to x.

$F = -\frac{dU}{dx}$

$F = -\frac{d}{dx}(GMm(\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}}))$

$F = \frac{GMmx}{(x^2+R^2)^{3/2}}$

This expression is the resultant magnitude of the Force F.

PART B) The magnitude of loss in potential energy as the particle falls to the center

$U = GMm(\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}})$

According to conservation of energy,

$\frac{1}{2}mv^2 = GMm (\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}})$

$\therefore v = \sqrt{2GM(\frac{1}{R}-\frac{1}{x^2+R^2})}$