Gravitation Potential Energy — Question 5

Question 5

The interior potential of a uniform planet must match the exterior value

A uniform spherical planet has mass $M$ and radius $R$. Its interior field magnitude is $g(r)=GMr/R^3$. Use $d\Phi/dr=g(r)$ and $\Phi(R)=-GM/R$.

1. Derive $\Phi(r)$ inside the planet.

2. Find the work to move a test mass from the center to the surface and then from the surface to infinity.

Question 5 - Solution

The interior potential of a uniform planet must match the exterior value

(a) Derivation and calculation

Integrate the specified radial derivative using the surface boundary value: $$\Phi(r)-\Phi(R)=\int_R^r\frac{GMs}{R^3}\,ds =\frac{GM}{2R^3}(r^2-R^2).$$ Since $\Phi(R)=-GM/R$, $$\boxed{\Phi(r)=-\frac{GM}{2R^3}(3R^2-r^2)}.$$

(b) Result and interpretation

The endpoint potentials are $$\Phi(0)=-\frac{3GM}{2R},\quad \Phi(R)=-\frac{GM}{R},\quad\Phi(\infty)=0.$$ Thus $$\boxed{W_{0\to R}=m[\Phi(R)-\Phi(0)]=\frac{GMm}{2R}},$$ $$\boxed{W_{R\to\infty}=\frac{GMm}{R}}=2W_{0\to R}.$$

Physical check

Differentiating the result recovers $d\Phi/dr=GMr/R^3$; the potential and its slope match the exterior at $R$.