We discuss the use of dimensional analysis and the quadrupolar emission hypothesis to determine the gravitational power radiated by a celestial body in a circular orbit. We then show how to derive the instantaneous power radiated in a general Keplerian orbit by approximating it locally by a circle. This derivation allows us to recover with good accuracy the nontrivial dependence given by general relativity relating the average radiated power to the eccentricity of an ellipse. The approach is understandable by undergraduate students.

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(
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,
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,
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4.
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,”
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7.
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Gravitational radiation from point masses in a Keplerian orbit
,”
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,
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440
(
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8.
A different discussion of the gravitational power is given in
Bernard F.
Schutz
,
Gravity from the Ground Up
(
Cambridge U. P.
,
Cambridge
,
2003
), p.
315
.
9.
If we had taken [T]β instead of [ω]β in the equality, the constant would have been multiplied by a very large factor (2π)66×104, and the dimensional analysis would not have yielded the right order of magnitude. The choice of ω1 instead of T for the characteristic time of evolution of harmonic phenomena is discussed for example in
Jean-Marc
Lévy-Leblond
and
Françoise
Balibar
,
Quantics: Rudiments of Quantum Physics
(
North-Holland
,
Amsterdam
,
1990
), Chap. 1.
10.
This dependence on ω4 is commonly associated in physics courses with Rayleigh scattering and the blue color of the sky.
11.
The equation for the orbit of the reduced mass is μd2r/dt2=Gm1m2/r3r, with μ=m1m2/(m1+m2). The magnitude of the acceleration is |a|=G(m1+m2)/r2.
12.
See, for example,
L. D.
Landau
and
E. M.
Lifschitz
,
Mechanics
, 3rd ed. (
Pergamon
,
Oxford
,
1988
), Vol.
1
;
Bradley W.
Carroll
and
Dale A.
Ostlie
,
An Introduction to Modern Astrophysics
, 2nd ed. (
Addison-Wesley
,
San Francisco
,
2007
).
13.
By using Eq. (13) and α=GMT, Pcirc also reads as Pcirc=325G4c5μ2MT3a5 or, to be consistent with the notation of Ref. 7: Pcirc=325G4c5m12m22(m1+m2)a5.
14.
For charges involved in harmonic motions (ω fixed) on elliptical trajectories, P¯ is proportional to a2+b2, where b2=a2(1e2); then f(e)=1e2/2 is a reducing factor.
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