Answer:
Explanation:
From the given information:
a)
Let's have an imaginary view of the rod located at a given distance r from he the mass (m) of the sphere.
Then the equation for the potential energy as related to the small area of the dr of the rod can be computed as:
[tex]dU = -\dfrac{GMm}{L}*\dfrac{dr}{r}[/tex]
where,
G = gravitational constant
[tex]U = - \int^{x+L}_{x}Gm\dfrac{M}{L}*\dfrac{dr}{r}[/tex]
[tex]U = - \dfrac{GMm}{L}\int^{x+L}_{x}\dfrac{dr}{r}[/tex]
By taking the integral within the limit
[tex]U = - \dfrac{GMm}{L} \Big[In \ r\Big]^{x+L}_{x}[/tex]
[tex]\mathbf{\implies - \dfrac{GmM}{L} In \Big(\dfrac{{x+L}}{{x}}\Big)}[/tex]
b)
By using [tex]F= -\dfrac{dU}{dx}[/tex], the magnitude of the gravitational force can be determined as follows:
Here, we have:
[tex]F = -\dfrac{d}{dx} \Big [\dfrac{-GmM}{L}In(\dfrac{x+l}{x}) \Big ] \\ \\ = \dfrac{GmM}{L}\times \dfrac{x}{x+L}\times (0-\dfrac{L}{x^2}) \\ \\ By \ solving \\ \\ \mathbf{ =-\dfrac{GmM}{x(x+L)}}[/tex]
From above, the negative sign indicates an attractive force
