Understanding Gravitational Potential Difference

The question asks us to find the gravitational potential difference between a point P located at a distance \(\frac{R}{2}\) from the center of a solid sphere and a point on its surface. The solid sphere has constant density \(\rho\), mass M, and radius R. Calculating the gravitational potential difference requires knowing the gravitational potential at both points.

Gravitational Potential Inside a Solid Sphere

For a solid sphere of uniform density, the gravitational potential at a distance \(r\) from the center, where \(r < R\), is given by the formula: $$V(r) = -\frac{GM}{2R^3}(3R^2 - r^2)$$ Here, \(G\) is the gravitational constant, M is the total mass of the solid sphere, and R is its radius. This formula describes the gravitational potential inside the solid sphere.

Gravitational Potential on the Surface of a Solid Sphere

The gravitational potential on the surface of a solid sphere (at \(r = R\)) is given by the standard formula for the potential outside a sphere (or on its surface), considering the mass is concentrated at the center for \(r \ge R\): $$V(R) = -\frac{GM}{R}$$ This gives the gravitational potential at the outer boundary of the solid sphere.

Calculating Potential at Point P (\(r = \frac{R}{2}\))

Point P is located at a distance \(r = \frac{R}{2}\) from the center, which is inside the solid sphere (\(\frac{R}{2} < R\)). We use the formula for the potential inside the solid sphere: $$V_P = V\left(\frac{R}{2}\right) = -\frac{GM}{2R^3}\left(3R^2 - \left(\frac{R}{2}\right)^2\right)$$ Now, let's simplify the expression: $$V_P = -\frac{GM}{2R^3}\left(3R^2 - \frac{R^2}{4}\right)$$ $$V_P = -\frac{GM}{2R^3}\left(\frac{12R^2 - R^2}{4}\right)$$ $$V_P = -\frac{GM}{2R^3}\left(\frac{11R^2}{4}\right)$$ $$V_P = -\frac{11GM}{8R}$$ So, the gravitational potential at point P is \(-\frac{11GM}{8R}\).

Calculating Potential at the Surface (\(r = R\))

The potential at the surface of the solid sphere is given by: $$V_{surface} = V(R) = -\frac{GM}{R}$$

Finding the Gravitational Potential Difference

We are asked to find the gravitational potential difference \(V_P - V_{surface}\). $$V_P - V_{surface} = V\left(\frac{R}{2}\right) - V(R)$$ Substitute the calculated values for \(V_P\) and \(V_{surface}\): $$V_P - V_{surface} = \left(-\frac{11GM}{8R}\right) - \left(-\frac{GM}{R}\right)$$ $$V_P - V_{surface} = -\frac{11GM}{8R} + \frac{GM}{R}$$ To combine these terms, we find a common denominator, which is \(8R\): $$V_P - V_{surface} = -\frac{11GM}{8R} + \frac{8GM}{8R}$$ $$V_P - V_{surface} = \frac{-11GM + 8GM}{8R}$$ $$V_P - V_{surface} = \frac{-3GM}{8R}$$ The gravitational potential difference between point P and the surface is \(-\frac{3GM}{8R}\).

Summary of Gravitational Potential Difference Calculation

By calculating the gravitational potential at the point inside the solid sphere (\(r=R/2\)) and subtracting the gravitational potential at the surface (\(r=R\)), we found the gravitational potential difference. The potential inside depends on the distance from the center using the specific formula for a solid sphere, while the potential on the surface uses the standard formula. The resulting gravitational potential difference is \(-\frac{3GM}{8R}\).