<p>"Why is it that the secondary feels more perturbations from the gravitational influence of the sun the further away it is from the primary?"</p>

<p>In brief, it is because the net perturbing acceleration on the secondary is just the (vector) difference between (a) the accelerative attraction towards the perturbing body experienced by the secondary, and 
(b) the accelerative attraction towards the perturbing body experienced by the primary.  </p>

<p>Thus, the closer the secondary is to the primary, the more nearly equal in size and direction are those two attractions towards the perturber, and the closer to zero is their vector difference.  Another result is that the the more similar are the changes in velocity in size and direction produced by the perturbing accelerations in the primary and the secondary, and the closer to zero is the resulting perturbation in their motions relative to each other.  </p>

<p>This has been known for a long time as a consequence of Newton's 6th corollary to the laws of motion: "If bodies, anyhow moved among themselves, are urged in the direction of parallel lines by equal accelerative forces, they will all continue to move among themselves after the same manner as if they had been urged by no such forces." </p>

<p>Because there can be immense variety in the possible trajectories of the primary and secondary apart from the perturbations, any detailed illustrations can speedily develop massive and intricate trigonometrical expressions. </p>

<p>But in all cases, including that of the Kozai-Lidov effect, the scale of the effect, however intricate its form, depends on the size of the net perturbing accelerations.</p>

<p>A highly simplified configuration can at least show by example how a net perturbing force, affecting the relative motion of primary and secondary, is almost directly proportional to the first power of the distance between primary and secondary -- though it also depends of course on further factors due to changes in angular configuration. </p>

<p>The diagram below indicates some highly simplified configurations. </p>

<p><a href="https://i.sstatic.net/Vs6kl.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Vs6kl.gif" alt="enter image description here"></a></p>

<p>Suppose first that the primary (E) and secondary (M) are for one instant in line with the perturbing body (S), with M at M1 between E and S.  Let s stand for distance ES, and d for distance EM (with d &lt;&lt; s).  Suppose also that the mass of E and M are negligibly small relative to the mass of S (though not negligible relative to each other). </p>

<p>With these approximations, the accelerative attractions of S on M and S on E are respectively $ k/(s-d)^2 $ and $ k/s^2 ,$ and the net perturbing acceleration on M is the difference $ ( k/(s-d)^2 - k/s^2 ) .$ </p>

<p>Putting $ s(1-d/s) $ for (s-d), and using the binomial expansion of $ 1/(1-d/s)^2 $ , one sees that the terms in $ k/s^2 $ cancel, leaving the net perturbing force as $ k/s^2 * (2d/s) $ , plus terms in higher powers of d/s, i.e in $ k d^2/s^4 $ and so on. </p>

<p>Where d is very much smaller than s, the higher-power terms in d/s can be neglected, and then the net perturbing acceleration on M at M1 in the chosen example-configuration is closely approximated by  $ +2 k d / s^3 $ , away from E and towards S.  </p>

<p>If instead the configuration has M at M2 so that E is in line between M and S, then the net perturbing acceleration on M clearly becomes $ ~ -2 k d / s^3 $ , i.e. away from E and away from S.   </p>

<p>If instead M is at M3, with the line EM3 at right-angles to ES, and if also the angle ESM3 can be treated as small enough that its cosine can be approximated as 1 and its sine as d/s , then it is easily found that the net perturbing acceleration on M at M3 is approximately $ k d/s^3 $ towards E, again neglecting higher powers of d/s. </p>

<p>If M is at an intermediate position M4, and D represents the angle ESM4, it can be seen by using a little more trigonometry that the net perturbing acceleration on M at M4, under the assumptions already made, has a component parallel to line ES of approximately $ +2 k d cos D / s^3 $, and a component perpendicular to line ES (acting always towards line ES) of approximately $ k d sin D / s^3 $. </p>

<p>All the components are proportional to the E-M separation d, to the extent that the omitted series of terms in higher powers of d/s can be treated as negligible as has been done here.   </p>

CC BY-SA 4.0
2018-08-12T22:30:17.547Z
2018-08-12T22:30:17.547Z
terry-s
<p>Why are mechanisms such as the Kozai-Lidov effect more prominent at at large semi-major axes?</p>

<p>If we had a binary system of a primary and secondary body, with the third perturber as the Sun. Why is it that the secondary feels more perturbations from the gravitational influence of the sun the further away it is from the primary?</p>

<planet><astrophysics><natural-satellites>
Why are gravitational perturbations stronger at larger semi-major axes?
Perturber
# Why are gravitational perturbations stronger at larger semi-major axes?

(Asked by: username_0 on 2018-08-11)

Why are mechanisms such as the Kozai-Lidov effect more prominent at at large semi-major axes?

If we had a binary system of a primary and secondary body, with the third perturber as the Sun. Why is it that the secondary feels more perturbations from the gravitational influence of the sun the further away it is from the primary?

**Tags:** planet, astrophysics, natural-satellites


---

## 1 answers

### Answer by username_1 on 2018-08-12. Score: 1

"Why is it that the secondary feels more perturbations from the gravitational influence of the sun the further away it is from the primary?"

In brief, it is because the net perturbing acceleration on the secondary is just the (vector) difference between (a) the accelerative attraction towards the perturbing body experienced by the secondary, and
(b) the accelerative attraction towards the perturbing body experienced by the primary.

Thus, the closer the secondary is to the primary, the more nearly equal in size and direction are those two attractions towards the perturber, and the closer to zero is their vector difference. Another result is that the the more similar are the changes in velocity in size and direction produced by the perturbing accelerations in the primary and the secondary, and the closer to zero is the resulting perturbation in their motions relative to each other.

This has been known for a long time as a consequence of Newton's 6th corollary to the laws of motion: "If bodies, anyhow moved among themselves, are urged in the direction of parallel lines by equal accelerative forces, they will all continue to move among themselves after the same manner as if they had been urged by no such forces."

Because there can be immense variety in the possible trajectories of the primary and secondary apart from the perturbations, any detailed illustrations can speedily develop massive and intricate trigonometrical expressions.

But in all cases, including that of the Kozai-Lidov effect, the scale of the effect, however intricate its form, depends on the size of the net perturbing accelerations.

A highly simplified configuration can at least show by example how a net perturbing force, affecting the relative motion of primary and secondary, is almost directly proportional to the first power of the distance between primary and secondary -- though it also depends of course on further factors due to changes in angular configuration.

The diagram below indicates some highly simplified configurations.

[![enter image description here](https://i.sstatic.net/Vs6kl.gif)](https://i.sstatic.net/Vs6kl.gif)

Suppose first that the primary (E) and secondary (M) are for one instant in line with the perturbing body (S), with M at M1 between E and S. Let s stand for distance ES, and d for distance EM (with d << s). Suppose also that the mass of E and M are negligibly small relative to the mass of S (though not negligible relative to each other).

With these approximations, the accelerative attractions of S on M and S on E are respectively $ k/(s-d)^2 $ and $ k/s^2 ,$ and the net perturbing acceleration on M is the difference $ ( k/(s-d)^2 - k/s^2 ) .$

Putting $ s(1-d/s) $ for (s-d), and using the binomial expansion of $ 1/(1-d/s)^2 $ , one sees that the terms in $ k/s^2 $ cancel, leaving the net perturbing force as $ k/s^2 \* (2d/s) $ , plus terms in higher powers of d/s, i.e in $ k d^2/s^4 $ and so on.

Where d is very much smaller than s, the higher-power terms in d/s can be neglected, and then the net perturbing acceleration on M at M1 in the chosen example-configuration is closely approximated by $ +2 k d / s^3 $ , away from E and towards S.

If instead the configuration has M at M2 so that E is in line between M and S, then the net perturbing acceleration on M clearly becomes $ ~ -2 k d / s^3 $ , i.e. away from E and away from S.

If instead M is at M3, with the line EM3 at right-angles to ES, and if also the angle ESM3 can be treated as small enough that its cosine can be approximated as 1 and its sine as d/s , then it is easily found that the net perturbing acceleration on M at M3 is approximately $ k d/s^3 $ towards E, again neglecting higher powers of d/s.

If M is at an intermediate position M4, and D represents the angle ESM4, it can be seen by using a little more trigonometry that the net perturbing acceleration on M at M4, under the assumptions already made, has a component parallel to line ES of approximately $ +2 k d cos D / s^3 $, and a component perpendicular to line ES (acting always towards line ES) of approximately $ k d sin D / s^3 $.

All the components are proportional to the E-M separation d, to the extent that the omitted series of terms in higher powers of d/s can be treated as negligible as has been done here.


---

