How much does a swing-by change the orbit of the planet?

Question

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?

Answer 1

• 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:

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.}

Answer 2

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.

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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:

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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.