Spherical Shell Binding Energy

Classical Mechanics Level 3

This problem was inspired by a comment from Legend of Physics

Consider a spherically symmetrical object with inner radius r 1 r_1 and outer radius r 2 r_2 , with uniform volume mass density ρ \rho . Let the total mass of this object be M M . If r 1 = 0 r_1 = 0 and r 2 ≠ 0 r_2 \neq 0 , this is a solid spherical ball.

In general, the gravitational binding energy of the object has the following form (where G G is the universal gravitational constant):

U = α G M 2 r 2 U = \alpha \frac{G M^2}{r_2}

It is well known that for a solid spherical ball ( r 1 = 0 ) (r_1 = 0) , α = 0.6 \alpha = 0.6 . What is the limit of α \alpha as r 1 r_1 approaches r 2 r_2 ? This is the result for a hollow spherical shell.


The answer is 0.5.

Steven Chase by Steven Chase , 37, USA

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2 solutions

Mark Hennings
Jan 26, 2020

The gravitational binding energy of a uniform object with mass M M and inner radius r 1 r_1 and outer radius r 2 r_2 is (building the solid one spherical shell at a time): U = ∫ r 1 r 2 G × 4 3 π ( r 3 − r 1 3 ) ρ × 4 π r 2 d r ρ × r − 1 = 16 3 G π 2 ρ 2 ∫ r ! r 2 r ( r 3 − r 1 3 ) d r U \; = \; \int_{r_1}^{r_2} G \times \tfrac43\pi(r^3 - r_1^3) \rho \times 4\pi r^2\,dr \rho \times r^{-1} \; = \; \tfrac{16}{3}G\pi^2\rho^2\int_{r_!}^{r_2} r(r^3 - r_1^3)\,dr where ρ \rho is the uniform density of the object, so that M = 4 3 π ( r 2 3 − r 1 2 ) ρ M \; = \; \tfrac43\pi(r_2^3 - r_1^2)\rho and hence U = 3 G M 2 ( r 2 3 − r 1 3 ) 2 ∫ r 1 r 2 r ( r 3 − r 1 3 ) d r = 3 G M 2 10 ( r 2 2 + r 2 r 1 + r 1 2 ) 2 ( 2 r 2 3 + 4 r 2 2 r 1 + 6 r 2 r 1 2 + 3 r 1 3 ) U \; = \; \frac{3GM^2}{(r_2^3-r_1^3)^2} \int_{r_1}^{r_2}r(r^3 - r_1^3)\,dr \; = \; \frac{3GM^2}{10(r_2^2 + r_2r_1 + r_1^2)^2}(2r_2^3 + 4r_2^2r_1 + 6r_2r_1^2 + 3r_1^3) after cancelling factors of r 2 − r 1 r_2-r_1 . Thus lim ⁡ r 1 → r 2 U = G M 2 2 r 2 \lim_{r_1 \to r_2}U \; = \; \frac{GM^2}{2r_2} making α = 1 2 \alpha = \boxed{\tfrac12} .

3 Helpful 3 Interesting 4 Brilliant 1 Confused

@Mark Hennings I can't understand your 1st step can you help me? Please

A Former Brilliant Member - 1 year, 4 months ago

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The gravitational potential of a solid with material from r 1 r_1 to r r is − G 4 3 π ( r 3 − r 1 3 ) ρ × r − 1 -G \tfrac43\pi(r^3 - r_1^3)\rho \times r^{-1} at the surface, and so the work done bringing a thin shell of radius r r and thickness d r dr up to that solid is d W = − G 4 3 π ( r 3 − r 1 3 ) ρ r − 1 × 4 π r 2 ρ d r dW = -G \tfrac43\pi(r^3 - r_1^3)\rho r^{-1} \times 4\pi r^2 \rho\,dr Adding these contributions up (integrating) and changing sign gives the formula for the binding energy U = − W U = -W .

Mark Hennings - 1 year, 4 months ago

There is a typo. In the expression for U U , the bracketed term in the numerator should be 2 r 2 3 + 4 r 2 2 r 1 + 6 r 2 r 1 2 + 3 r 1 3 2r_2^3+4r_2^2r_1+6r_2r_1^2+3r_1^3

A Former Brilliant Member - 1 year, 4 months ago

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Thanks. Fixed.

Mark Hennings - 1 year, 4 months ago

Let us subdivide the spherical layer into small elements, each of mass d M dM . The energy of interaction of each element with all others is d U = − G M d M r 2 dU=-\dfrac{GMdM}{r_2} . Integrating over all elements we obtain U = − G M 2 2 r 2 U=-\dfrac{GM^2}{2r_2} . So α = 0.5 α=\boxed {0.5}

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