Development of the Potential Energy

The first task is to develop convenient expressions for the gravi-tational potential energy of the interaction of a rigid body with a distant mass point. A rigid body can be thought of as made of a large number of mass elements, subject to rigid coordinate constraints. We have seen that the kinetic energy of a rigid body is conveniently expressed in terms of the moments of inertia of the body and the angular velocity vector, which in turn can be represented in terms of a suitable set of generalized coordinates.

The potential energy can be developed in a similar manner. We first represent the potential energy in terms of moments of the mass distribution and later introduce generalized coordinates as particular parameters of the potential energy.

The gravitational potential energy of a mass point and a rigid body (see figure 2.9) is the sum of the potential energy of the mass point with each mass element of the body:

−X

α

GM⁰mα

r_α (2.67)

whereM⁰is the mass of the external point mass,r_αis the distance between the point mass and the constituent mass element with index α, m_α is the mass of this constituent element, and G is the gravitational constant. Let R be the distance of the center of mass of the rigid body to the point mass; R is the magnitude of the vector ~x−X, where the external mass point has position~

r_α

~ξ_α

R X~ θ

m_α

M⁰

~x origin

Figure 2.9 The gravitational potential energy of a mass point and a rigid body is the sum of the gravitational potential energy of the mass point with each constituent mass element of the rigid body.

~

x, and the center of mass has position X. The vector from the~ center of mass to the constituent with index α is ξ~α, and has magnitudeξ_α. The distancer_αis then given by the law of cosines r_α² =R²+ξ²_α−2ξαRcosθα where θα is the angle between~x−X~ and ~ξ_α. The potential energy is then

−GM⁰X

α

mα

(R²+ξ²_α−2ξαRcosθα)^1/2. (2.68) This is complete, but we need to find a representation that does not mention each constituent.

Typically, the size of celestial bodies is small compared to the separation between them. We can make use of this to find a more compact representation of the potential energy. If we expand the potential energy in the small ratio ξ_α/Rwe find¹⁵

−GM⁰X

α

m_α1 R

X

l

ξ^l_α

R^lP_l(cosθ_α), (2.69)

15The Legendre polynomials P_l may be obtained by expanding (1 +y² − 2yx)^−1/2 as a power series iny. The coefficient ofy^l isP_l(x). The first few Legendre polynomials are:P₀(x) = 1, P₁(x) = x, P₂(x) = ³₂x²−¹₂, and so on. The rest satisfy the recurrence relation: lP_l(x) = (2l−1)xP_l−1(x)−(l− 1)P_l−2(x).

wherePlis thelth Legendre polynomial. Interchanging the order

Successive terms in this expansion of the potential energy typically decrease very rapidly because celestial bodies are small compared to the separation between them. We can compute an upper bound to the size of these terms by replacing each factor in the sum over α by an upper bound. The Legendre polynomials all have magnitudes less than one for arguments in the range−1 to 1. The distances ξα are all less than some maximum extent of the body ξ_max. The sum overm_α times these upper bounds is just the total massM times the upper bounds. Thus

kX

α

mαξ_α^l

R^lPl(cosθα)k ≤Mξ_max^l

R^l . (2.71)

We see that the upper bound on successive terms decreases by a factor ξ_max/R. Successive terms may be smaller still. For large bodies the gravitational force is strong enough to overcome the internal material strength of the body, so the body, over time, becomes nearly spherical. Successive terms in the expansion of the potential are measures of the deviation of the mass distribu-tion from a spherical mass distribudistribu-tion. Thus for large bodies the higher order terms are small because the bodies are nearly spherical.

Consider the first few terms in l. Forl= 0 the sum over αjust gives the total mass M of the rigid body. For l= 1 the sum over α is zero, as a consequence of choosing the origin of the ξ~_α to be the center of mass. For l = 2 we have to do a little more work.

The sum involves second moments of the mass distribution, and can be written in terms of moments of inertia of the rigid body:

X

where A, B, and C are the principal moments of inertia, and I is the moment of inertia of the rigid body about the line between the center of mass of the body to the external point mass. The momentI depends on the orientation of the rigid body relative to the line between the bodies. The contributions to the potential energy up tol= 2 are then¹⁶

−GM M⁰

R −GM⁰

2R³ (A+B+C−3I). (2.73)

Letα= cosθ_a,β = cosθ_b, andγ = cosθ_c be the direction cosines of the angles θ_a,θ_b and θ_c between the principal axes ˆa, ˆb, and ˆc and the line between the center of mass and the point mass.¹⁷ A little algebra shows I =α²A+β²B+γ²C. The potential energy is then

−GM M⁰

R −GM⁰

2R³ [(1−3α²)A+ (1−3β²)B+ (1−3γ²)C]. (2.74) This is a good first approximation to the potential energy of in-teraction for most situations in the solar system; if we intended to land on the moon we probably would want to take into account higher order terms in the expansion.

Exercise 2.14:

a. Fill in the details that show that the sum over consitutents in equa-tion (2.72) can be expressed as written in terms of moments of inertia.

In particular, show that X

α

m_αξ_αcosθ_α= 0, X

α

m_αξ_α² = 2(A+B+C),

and that X

α

m_αξ_α²(sinθ_α)²=I.

16This approximate representation of the potential energy is sometimes called MacCullagh’s formula.

17Watch out, we just reusedα. It was also used as the constituent index.

b. Show that if the principal moments of inertia of a rigid body areA, B, and C, then the moment of inertia about an axis that goes through the center of mass of the body with direction cosinesα,β, andγrelative to the principal axes is

I=α²A+β²B+γ²C.