Irodov 1.206: A Point Mass Against a Rod – Amit & Deepali Sinha Foundation

Problem Irodov 1.206

Find the potential energy of the gravitational interaction in each of the following cases.

Two point masses $m_1$ and $m_2$ separated by a distance $r$ .
A point mass $m$ and a thin uniform straight rod of mass $M$ and length $l$ , with the point mass lying on the line containing the rod, a distance $a$ from the rod’s nearer end. Also find the force of the gravitational interaction between the point mass and the rod.

(a) Two point masses a distance r apart, pulling on each other with equal and opposite force F. — (a) Two point masses a distance $r$ apart, pulling on each other with equal and opposite force $F$ .

(b) Mass m a distance a from the near end of a uniform rod of mass M and length l, pulled toward it with force F. — (b) Mass $m$ a distance $a$ from the near end of a uniform rod of mass $M$ and length $l$ , pulled toward it with force $F$ .

Given

(a) $m_1$ , $m_2$ , $r$
(b) $m$ , $M$ , $l$ , $a$

Find

The potential energy of the interaction in each case; also the force in (b)

Solution

Step 1(a) Integrate the inverse-square force in from infinity

Take $U(\infty)=0$ . The force pulling $m_2$ toward $m_1$ at separation $r'$ has magnitude $\gamma m_1m_2/r'^2$ and acts to decrease $r'$ , so in the sign convention $F_{r'}=-dU/dr'$ (positive $F_{r'}$ meaning increasing $r'$ ), it reads

F_{r'} = -\frac{\gamma m_1 m_2}{r'^2}.

Integrating this force in from infinity gives the potential energy,

U(r) = -\int_\infty^r F_{r'}\,dr' = \int_\infty^r \frac{\gamma m_1 m_2}{r'^2}\,dr' = \gamma m_1 m_2\Big[-\frac{1}{r'}\Big]_\infty^r. \tag{1}

U = -\dfrac{\gamma m_1 m_2}{r}

(2)

Step 2(b) Set up the rod as a line of mass elements

The rod cut into elements: dm of width dx at distance x from the near end sits a distance a+x from m. — The rod cut into elements: $dm$ of width $dx$ at distance $x$ from the near end sits a distance $a+x$ from $m$ .

Measure $x$ from the near end of the rod, so an element $dm = (M/l)\,dx$ at position $x$ sits a distance $a+x$ from $m$ . Each element contributes the same inverse-square potential energy found in Part (a), so the rod’s total is the sum (integral) of those contributions,

U = -\int_0^l \frac{\gamma m\,dm}{a+x} = -\frac{\gamma m M}{l}\int_0^l \frac{dx}{a+x}. \tag{3}

Step 3Evaluate the integral and box the potential energy

\begin{align*} U &= -\frac{\gamma m M}{l}\Big[\ln(a+x)\Big]_0^l = -\frac{\gamma m M}{l}\big[\ln(a+l)-\ln a\big]. \tag{4}\end{align*}

U = -\gamma\,\dfrac{mM}{l}\,\ln\!\Big(1+\dfrac{l}{a}\Big)

(5)

Step 4Find the force from the same integral, taken over the separation

a

The force $m$ feels is $F = -dU/da$ at fixed $l$ and $M$ :

\begin{align*} F &= -\frac{d}{da}\left[-\gamma\frac{mM}{l}\ln\!\Big(1+\frac{l}{a}\Big)\right] = \gamma\frac{mM}{l}\cdot\frac{1}{1+l/a}\cdot\Big(-\frac{l}{a^2}\Big) \\ &= -\gamma\frac{mM}{a(a+l)}. \tag{6}\end{align*}

F = \dfrac{\gamma mM}{a(a+l)}

(7)

Check

As $a\to\infty$ (the rod shrinks to a point at the observer’s distance), $\ln(1+l/a)\to l/a$ , so $U \to -\gamma mM/l \cdot l/a = -\gamma mM/a$ : the two-point-mass result of Part (a) with separation $a$ , exactly as a rod must look from very far away.

Remark

The force could equally have been built by integrating $\gamma m\,dm/(a+x)^2$ directly over the rod; it gives the same $\gamma mM/[a(a+l)]$ , which is the check that $F=-dU/da$ was applied to the right variable.