Irodov 1.213: Doubling a Hemisphere into a Sphere – Amit & Deepali Sinha Foundation

Problem Irodov 1.213

A particle of mass $m$ is transferred from the center of the flat circular face of a solid uniform hemisphere of mass $M$ and radius $R$ out to an infinite distance. Find the work done on the particle, during this displacement, by the gravitational force the hemisphere exerts on it.

The particle starts at the center of the hemisphere's flat face and is carried out to infinity along the axis. — The particle starts at the center of the hemisphere’s flat face and is carried out to infinity along the axis.

Given

$m$ , $M$ , $R$

Find

The work

A

done by the hemisphere’s gravity on

m

Solution

Step 1Double the hemisphere to reach a shape with a known potential

Join the given hemisphere to an identical mirror copy of itself across its flat face: the two together form one uniform solid sphere of radius $R$ and mass $2M$ , and the point in question, the center of the shared flat face, is exactly the center of that sphere. By the mirror symmetry, every point of the real hemisphere has a partner point in the mirror copy at the same distance from the center, so the two halves contribute equally to the potential there,

\varphi_{\text{sph}}(0) = \varphi_{\text{hemi}}(0) + \varphi_{\text{mirror}}(0) = 2\,\varphi_{\text{hemi}}(0).

Step 2Find the potential at the center of the full sphere

Build the sphere from thin concentric shells of radius $r'$ and thickness $dr'$ . The doubled body is uniform, so its density is the doubled mass over the sphere’s volume.

\rho = \dfrac{2M}{\frac{4}{3}\pi R^3} = \dfrac{3M}{2\pi R^3}

(1)

One shell carries the mass of its own surface area times its thickness.

dm' = \rho\,4\pi r'^2\,dr'

Every point of that shell is the same distance $r'$ from the center, so the shell contributes to the central potential as though its whole mass sat at that distance.

d\varphi = -\frac{\gamma\,dm'}{r'}

Substituting $dm'$ and integrating over every shell from the center out to the surface,

\begin{align*}\varphi_{\text{sph}}(0) &= -\int_0^R \frac{\gamma\,\rho\,4\pi r'^2}{r'}\,dr' \\ &= -4\pi\gamma\rho\int_0^R r'\,dr' \\ &= -2\pi\gamma\rho R^2\end{align*}

and substituting the density (1) for $\rho$ ,

\begin{align*}\varphi_{\text{sph}}(0) &= -2\pi\gamma R^2 \cdot \frac{3M}{2\pi R^3} \\ &= -\frac{3\gamma M}{R}\end{align*}

which, written with the doubled body’s own mass $2M$ on display, is

\varphi_{\text{sph}}(0) = -\dfrac{3\gamma(2M)}{2R}

(2)

Step 3Halve it for the real hemisphere, then find the work

By Step 1 the hemisphere alone supplies half of (2), which is the same as the sphere’s own expression with its own mass $M$ in place of the doubled body’s $2M$ .

\begin{align*}\varphi_{\text{hemi}}(0) &= \frac12\varphi_{\text{sph}}(0) \\ &= -\frac{3\gamma(2M)}{4R}\end{align*}

\varphi_{\text{hemi}}(0) = -\dfrac{3\gamma M}{2R}

(3)

The particle’s potential energy at the start is its mass times that potential, and infinitely far out the potential, and with it the energy, vanishes.

\begin{align*}U_i &= m\,\varphi_{\text{hemi}}(0) = -\frac{3\gamma mM}{2R} \\ U_f &= 0\end{align*}

The work done by the gravitational force over the transfer is the drop in potential energy between those two configurations.

A = U_i - U_f

A = -\dfrac{3}{2}\,\dfrac{\gamma mM}{R}

(4)

Check

The equality $\varphi_{\text{hemi}}(M) = \varphi_{\text{sph}}(2M)/2 = \varphi_{\text{sph}}(M)$ is forced, not coincidental: gravitational potential is linear in the source mass, so scaling every mass in the sphere by a factor scales the potential by that same factor at any fixed point, which is exactly the linearity Step 3 already relies on to convert the $2M$ sphere’s potential back to the real hemisphere’s. A genuinely independent check is on the dimensions of the boxed result: $\gamma mM/R$ carries units $(\text{N m}^2\text{kg}^{-2})(\text{kg})(\text{kg})/\text{m} = \text{N m} = \text{J}$ , an energy, and the value comes out negative, as work done by an attractive force pulling $m$ away to infinity must be.

Remark

The work is negative because gravity opposes the transfer: energy must be supplied to carry $m$ away, exactly $\frac32\gamma mM/R$ of it.