3
$\begingroup$

A friend of mine told me that one Voyager swing-by moved the orbit of Jupiter a foot closer to the sun. I understand that Jupiter was slowed by the gravity assist, but I figured it would be minuscule and basically unnoticeable. What is the true answer for this?

$\endgroup$
  • $\begingroup$ You've made me curious as to the mass of an object required for Voyager to shift it a full meter. $\endgroup$ – Magic Octopus Urn Jan 24 at 2:06
  • 1
    $\begingroup$ @MagicOctopusUrn $\approx 10^{14}$ times more, so in the order of $10^{16}$ kg. It is the mass of an around 20km moon/asteroid, like Lysithea. $\endgroup$ – peterh Jan 25 at 9:18
  • $\begingroup$ Hey just curious where does the term "swing-by" come from? I think it's clear to most people on here what you are referring to, but I've pretty much exclusively heard the term "flyby." $\endgroup$ – ben Jan 27 at 1:22
  • $\begingroup$ I don't know, it has several names. In German we would usually say Swing-By. As for wikipedia: In orbital mechanics and aerospace engineering, a gravitational slingshot, gravity assist maneuver, or swing-by [...] Its not a flyby, flyby does not use the object to increase speed necessarily $\endgroup$ – SinisterMJ Jan 27 at 16:27
  • 1
    $\begingroup$ JPL routinely uses the term "swingby" (no hyphen) for gravity assists. I'm not sure when that started. At least back in the early 70's. Possibly before. $\endgroup$ – Mark Adler Jan 27 at 16:41
14
$\begingroup$
  • The mass of the Voyager is $\approx$ 825 kg.
  • The mass of the Jupiter is $\approx 1.9 \cdot 10^{27}$ kg.

According this question, the Voyager accelerated from $\approx 10.2 \frac{km}{s}$ to $\approx 27.8 \frac{km}{s}$ by its Jupiter flyby at 1979.7.29:

enter image description here

Thus, the Voyager got $\approx 17.6 \frac{km}{s}$ velocity from the Jupiter by its gravitational slingshot maneuver.1

Step1

We use the Law of the Conservation of Momentum to calculate the change of the velocity of the Jupiter:

$$\Delta v_{Voyager} \cdot m_{Voyager} = \Delta v_{Jupiter} \cdot m_{Jupiter}$$

$$\Delta v_{Jupiter} = \frac{\Delta v_{Voyager} \cdot m_{Voyager}}{ m_{Jupiter} }$$

Substituting the values, we get $\Delta v_{Jupiter} = 7.642 \cdot 10^{-24} \frac{km}{s}$.

Step2

Kepler's Third Law says:

The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. (ref)

Formalized:

$$P^2 \propto a^3$$

Thus

$$P \propto a^\frac{3}{2}$$

Thus, a little change in the orbital velocity of the Jupiter causes the change of its 1.5 times in its orbital radius (more exactly, in the major axis of its elliptical orbit, but in the case of such little changes we can neglige this nuance).

According this source, the mean orbital velocity of the Jupiter is 13.1 $\frac{km}{s}$ and its mean orbital radius is $7.786\cdot10^8 km$. Thus, the change in the semi-major axis of the Jupiter's elliptical orbit is

$$\frac{7.64\cdot10^{-24}}{13.1}\cdot1.5\cdot 7.786\cdot10^8 km \approx 6.811 \cdot 10^{-16} km = \underline{\underline{6.811 \cdot 10^{-13} m}}$$

The size of a nucleus is $\approx 10^{-15}$ m. A foot is $\approx 0.3 m$.

1 Mark Adler mentions in comment, that this picture is not a very good one to estimate the speed gain. However, my this source says 16 $\frac{km}{s}$ which is not a significant deviation.

$\endgroup$
  • 1
    $\begingroup$ So, a foot then @peterh! $\endgroup$ – GdD Jan 22 at 9:09
  • $\begingroup$ @GdD It is billionths of the size of an atom. $\endgroup$ – peterh Jan 22 at 9:11
  • $\begingroup$ I know @peterh, I was being sarcastic. Hard to communicate in a comment. $\endgroup$ – GdD Jan 22 at 9:14
  • 1
    $\begingroup$ If Jupiter were that massive, it would be a red dwarf. $\endgroup$ – Mark Adler Jan 23 at 4:49
  • 1
    $\begingroup$ Also the velocity change of Voyager is a) based on a very poor plot of Voyager 2, and more significantly, b) the velocity change is not read from the top of that spike. The top of the spike is the maximum velocity at closest approach, which is then mostly lost on departure, with the momentum returned to Jupiter. You need to read the velocity change from the points before and after the swingby where the Sun dominates the trajectory. $\endgroup$ – Mark Adler Jan 27 at 17:00
8
$\begingroup$

I get that Voyager 1 brought Jupiter $6\times10^{-13}\,\mathrm{m}$ closer to the Sun. One hundredth the size of a hydrogen atom.

Then Voyager 2 did almost that again! Oh, the humanity.


The $825\,\mathrm{kg}$ (wet mass) Voyager 1 spacecraft was sped up $11\,\mathrm{km/s}$ by Jupiter, whose mass is $1.90\times 10^{27}\,\mathrm{kg}$. The speed up was calculated using the heliocentric specific energy of Voyager 1 before and after the swingby. By the conservation of momentum, Jupiter was then slowed down by about $5\times 10^{-21}\,\mathrm{m/s}$.

Jupiter's velocity around the Sun at that time was $12.8\,\mathrm{km/s}$. So it was slowed down by $4\times 10^{-25}$ of its velocity. By the relation of orbit velocity to radius, Jupiter's semi-major axis would have been reduced by that fraction times two. Multiplying by $5.2\,\mathrm{AU}$, you get $5.8\times10^{-13}\,\mathrm{m}$.

I fixed Voyagers' velocity change from a previous version of this answer. I should have looked at my own plot here. Here is that same plot for Voyager 1:

Voyager 1 heliocentric velocity


So why two times the fractional change in velocity? The relation between the semi-major axis $a$ and the velocity of an orbit $v$ at some point is the familiar equation:

$$-{\mu\over 2a}={v^2\over 2}-{\mu\over r}$$

The velocity change occurs in a short period of time over which $r$ can be considered to be a constant. We can then take the derivative:

$${\mu\over 2a^2}da=v dv$$

We can rewrite that a little as:

$${\mu\over 2a}{da\over a}=v^2{dv\over v}$$

If the orbit is circular, which Jupiter is close enough to, then:

$$v=\sqrt{\mu\over a}$$

So the $v^2$ and $\mu/a$ cancel giving:

$${da\over a}=2{dv\over v}$$

Therefore the resulting small fractional change in the semi-major axis is twice the small fractional change in velocity.

$\endgroup$
  • $\begingroup$ I like this answer most because you worked backwards from loss of velocity. $\endgroup$ – Magic Octopus Urn Jan 23 at 17:46
  • $\begingroup$ Isn't there some conflict between the two answers? In this answer, Voyager 1 (722Kg) gets 280m/s, Jupiter loses 10-22 m/s and this means ending up 4x10-15 closer. In peterh answer, Voyager 2 (825Kg) gets 17600m/s (!), Jupiter loses 10-23 m/s (gave more speed to Voyager 2 by losing less?) and this meant ending up 10-20 closer (lost one magnitude less and this means a smaller movement by 5 orders of mangitude - is this correct too?). $\endgroup$ – BlueCoder Jan 24 at 11:18
  • $\begingroup$ I've now read your comment in the other answer, the first inconsistency is probably due to Jupiter having 2 orders of magnitude more mass there (red dwarf!) :) $\endgroup$ – BlueCoder Jan 24 at 11:19
  • $\begingroup$ Marked this answer as correct, as the weight on Google says 10 to the power of 27, not 29. $\endgroup$ – SinisterMJ Jan 24 at 13:49
  • 1
    $\begingroup$ Not as competent as you seem to think. I too hastily pulled some HORIZONS data without plotting it, and I had completely missed the flyby date. Fixed now. $\endgroup$ – Mark Adler Jan 25 at 16:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.