The Voyager 1 lost speed gradually after gaining speed from gravity assist. Is the external thrust applied in the opposite direction to move closer to the planet, or does the spacecraft lose its momentum after the sudden rise in speed?
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47$\begingroup$ throw a ball up. Why does it slow down? $\endgroup$– njzk2Commented Mar 2, 2020 at 22:40
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$\begingroup$ Or in other words: Part of its kinetic energy is converted to gravitational potential energy. $\endgroup$– MichaelCommented Mar 5, 2020 at 11:01
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2 Answers
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It's the gravitational attraction of the Sun. Voyager was moving away from the Sun and was pulled back by its gravity. Since Voyager was not moving directly away from the Sun, it's trajectory also curved.
answered Mar 2, 2020 at 13:42
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3$\begingroup$ Note that, for simplicity's sake, Voyager is being affected by the Sun the same way a ball is being affected by the Earth if you throw it up (but not straight up in the air. It follows a parabolic trajectory in which it gradually slows down (vertically speaking), and were it to come to a complete (vertical) stop, it would then fall back down, i.e. towards the Earth (ball) or Sun (Voyager) again. $\endgroup$– FlaterCommented Mar 3, 2020 at 12:03
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3$\begingroup$ @Flater This is almost right. A ball only follows a parabolic trajectory if you ignore the curvature of the Earth, or fire it at exactly the right speed. The trajectory of Voyager is actually a hyperbola. Also note that, since the Sun's gravity gets weaker the further away from it you are, Voyager 1 will decellerate less and less as time passes, and never actually fall back. $\endgroup$ Commented Mar 3, 2020 at 17:25
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14$\begingroup$ In my defense, the ball throwing example was intended on a much smaller scale that one in which the Earth's curvature matters. Just exactly how far are you able to throw a ball!? $\endgroup$– FlaterCommented Mar 3, 2020 at 17:53
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$\begingroup$ Curvature doesn't play a role for the ball .. but ballistic influence due to atmospheric drag .. hence artillery calculates in atmospheric drag to hit a certain spot $\endgroup$– eagle275Commented Mar 5, 2020 at 12:40
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Additionally, the probe has just passed a large planetary body with its own gravity well. The probe has to climb up out of that planet's gravity well which costs momentum.
There is a net gain in momentum overall, the probe is leaving with more momentum than it entered the slingshot manoever, but the highest velocity is about where the probe is moving parallel to the planet's surface.
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$\begingroup$ It looks like the whole slingshot maneuver is instantaneous in the animation. $\endgroup$ Commented Mar 3, 2020 at 12:10
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2$\begingroup$ @user253751: It's definitely not instantaneous. Take a look a the timescale in the upper left of the animation. $\endgroup$ Commented Mar 3, 2020 at 12:46
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7$\begingroup$ @MichaelSeifert instantaneous in the animation. $\endgroup$ Commented Mar 3, 2020 at 12:50
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4$\begingroup$ @user253751 It seems to use about 3 frames. Enough that my brain can interpolate the animation into something meaningful, but --- as you say --- not enough to query any internal properties of the maneuver. $\endgroup$– jpaughCommented Mar 3, 2020 at 17:08