Problem Statement
Find the gravitational potential energy of a uniform sphere of mass M and radius R (self-energy).
Given Information
- All numerical data are stated in the problem above; symbols are defined as they appear.
Physical Concepts & Formulas
These problems use Newton’s law of gravitation together with the dynamics of orbital and central motion. Build sphere by adding shells. Energy to bring shell of mass dm from infinity: dU = $-GM_r \cdot dm/r$. Integrate over all shells.
- $F = G\dfrac{m_1 m_2}{r^2}$ — law of universal gravitation
- $v = \sqrt{\dfrac{GM}{r}}$ — circular orbital speed
- $T^2 = \dfrac{4\pi^2 r^3}{GM}$ — Kepler’s third law
Step-by-Step Solution
Step 1 — Identify the governing principle: We begin by recognising which physical law controls the situation and why it is the correct starting point for this problem.
$$At \text{rad}ius r, enclosed mass M_r = M(r/R)^{3}$$
Step 2 — Set up the relevant equations: Next we write down the equations that follow from that principle, introducing the symbols we will carry through the algebra.
$$Shell of thickness dr: dm = (3M/R^{3}r^{2}dr)$$
Step 3 — Apply the given conditions: We now substitute the specific conditions and constraints given in the problem so the equations describe this particular situation.
$$dU = -G \cdot M_r \cdot dm/r = -G \cdot M(r/R)^{3} \cdot (3M/R^{3}r^{2}dr/r = -3GM^{2r}^{4dr}/R^{6})$$
Step 4 — Solve for the required quantity: With the equations specialised, we isolate and solve for the unknown the problem asks us to find.
$$U = ∫_0ᴿ -3GM^{2r}^{4}/R^{6}dr = -3GM^{2}/(R^{6} \cdot R^{5}/5 = -3GM^{2}/(5R))$$
Worked Calculation
$$U = -3GM^{2}/(5R)$$
Answer
$$\boxed{U = -3GM^{2}/(5R)}$$
This is the quantity the problem asked for, expressed in terms of the given data: $U = -3GM^{2}/(5R)$.
Physical Interpretation
This is the gravitational self-energy — the energy released when a uniform sphere assembles from dispersed matter. It equals the energy needed to completely disperse the sphere. The magnitude of the answer is consistent with everyday physical experience for this class of problem in Irodov’s Part 1 — the result shows how the answer scales with the given quantities. If we doubled the dominant input, the boxed formula tells us exactly how the output would respond, and that scaling is the key physical insight this problem trains. Comparing the answer with the appropriate limiting cases (very small or very large values of the dominant parameter) recovers the familiar Newtonian or intuitive expectation, which is a useful sanity check.
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