<p>Quite simply, the sun's gravity is pulling on the space probe at all times. As the probe moves away from the sun, the force of gravity pulls it back in, slowing it down. As the probe moves toward the sun, the force of gravity continues to pull, speeding it up. Any object orbiting the sun is <em>always</em> accelerating toward the sun - when that acceleration opposes motion, the object slows down, and when it's in the same direction as motion, the object speeds up.</p>

<p>It's no different from throwing a ball into the air and seeing it slow down as it rises, reverse direction, and then speed up again as it falls, except the primary gravitational body in that case is the earth and not the sun.</p>

CC BY-SA 4.0
2020-05-22T20:08:34.173Z
Or like a pendulum being pulled by the Earth. Closer to the Earth = fastest. Farthest from the earth = slowest. The difference between the ball and the pendulum is the pendulum "misses" the Earth when falling towards it like things in orbit.
DKNguyen
CC BY-SA 4.0
2020-05-22T13:28:48.460Z
2020-05-22T18:41:14.810Z
2020-05-22T18:41:14.810Z
Nuclear Hoagie
<p>The simplest explanation is that the satellite generally obeys <a href="https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Second_law_of_Kepler" rel="nofollow noreferrer">Kepler's second law of orbital mechanics</a>:</p>
<blockquote>
<p>A line joining a planet and the Sun sweeps out equal areas during equal intervals of time</p>
</blockquote>
<p>When the satellite is further away from the sun, the area swept out remains constant only because the satellite is moving more slowly.</p>
<p>The physical answer, as others have pointed out, is that the satellite trades out kinetic energy for gravitational potential energy as it travels &quot;up&quot; the Sun's gravity well. This is no different from gradually slowing as you coast up a hill on your bike (although you are climbing up the Earth's gravity well not the Sun's).</p>

CC BY-SA 4.0
2020-05-22T21:42:08.307Z
2020-06-17T18:45:35.523Z
2020-06-17T18:45:35.523Z
johnDanger
<blockquote>
  <p>Why does the “parker solar probe” lose speed as the distance from the sun increases?</p>
</blockquote>

<p>Because energy and angular momentum are individually conserved quantities in the two body problem. Except for where the Parker Space Probe has close fly-bys with Venus, the gravitationally interactions between the Parker Space Probe and the solar system are very closely modeled as a two body problem (the Sun and the probe), plus very small perturbations from the planets.</p>

<p>One way to express conservation of energy in the two body problem is the vis-viva equation, <span class="math-container">$$v^2 = \mu\left(\frac2r - \frac1a\right)$$</span></p>

<p>where</p>

<ul>
<li><span class="math-container">$\mu = G(M+m)$</span> is the sum of the central body's standard gravitational parameter and that of the orbiting body,</li>
<li><span class="math-container">$r$</span> is the distance between the two bodies,</li>
<li><span class="math-container">$a$</span> is the semi-major axis length (a constant), and</li>
<li><span class="math-container">$v$</span> is the magnitude of the velocity vector.</li>
</ul>

<p>Note that the mass of the Parker Space Probe is so much less than that of the Sun that one can drop the Parker Space Probe's mass from the expression <span class="math-container">$\mu = G(M+m)$</span>, resulting in <span class="math-container">$\mu = GM_{\text{sun}}$</span>.</p>

<p>Note that the only variable on the right hand side of the vis-viva equation is the radial distance. As radial distance increases the square magnitude of the velocity vector (and thus the magnitude of the velocity vector) decreases.</p>

<p>Without mathematics, conservation of energy dictates that the sum of an orbiting body's kinetic energy and gravitational potential energy must remain constant. As the orbiting body moves further from the central body, the orbiting body's potential energy increases, which means its kinetic energy must correspondingly decrease. This in turn means the orbiting body's velocity must decrease.</p>

CC BY-SA 4.0
2020-05-22T07:58:22.007Z
And how does this relate to the gravitational assistance used in Voyager missions? From what I understand, when moving away from the massive object, the probes should lose speed?
user1785960
CC BY-SA 4.0
2020-05-22T08:15:26.560Z
@user1785960 - They have lost speed, a whole lot of speed. But now that the Voyager probes are 123.9 astronomical units from the Sun (Voyager 2) and 149.3 astronomical units from the Sun (Voyager 1), the $\frac2r$ terms in the vis-viva equation are very small compared to the $\frac1a$ terms for both vehicles. While they do continue to lose speed, the loss is now very small for both of them.
David Hammen
CC BY-SA 4.0
2020-05-22T09:00:04.447Z
@user1785960 In fact this is how gravity assists work. If you pass by a planet, you first gain speed as you come closer and then lose the same amount as you move away. But this speed is with respect to the planet, which is itself moving. If you play around a bit with directions and velocities, you'll notice that there are even situations where losing speed with respect to the planet actually means gaining speed with respect to the sun. Now a gravity assist is just picking the right directions when approaching and leaving a planet to benefit from these effects.
mlk
CC BY-SA 4.0
2020-05-22T10:24:31.763Z
@user1785960 imagine very, very small rubber ball has a perfectly elastic collision with a very, very big moving ball. That's the same as gravity assist.
user11153
CC BY-SA 4.0
2020-05-23T11:55:43.443Z
To understand this more viscerally, I can highly recommend playing a bunch of Kerbal Space Program - it is dangerously good at teaching (some, not all) orbital mechanics.
Adam Barnes
CC BY-SA 4.0
2020-05-23T12:02:50.887Z
@AdamBarnes - Personally, I'm not a big fan of Kerbal Space Program. It makes too many simplifying assumptions for my taste. But nonetheless I do see the teaching value.
David Hammen
CC BY-SA 4.0
2020-05-23T23:57:27.997Z
@user11153 there is a slight difference – gravity is attractive (and works at distance), while rubber ball collision is repulsive and works only at the (short) time they are actually touching.
Paŭlo Ebermann
CC BY-SA 4.0
2020-05-22T01:17:36.790Z
2020-05-22T01:17:36.790Z
David Hammen
<p>Why does the Parker Solar Probe slow down as the distance from the Sun increases?</p>

<p><img src="https://upload.wikimedia.org/wikipedia/commons/6/66/Animation_of_Parker_Solar_Probe_trajectory.gif" alt="Image of probe&#39;s trajectory">
<br>
<sup>Image credit: Wikipedia user Phoenix777, <a href="https://creativecommons.org/licenses/by-sa/4.0/deed.en" rel="nofollow noreferrer">CC BY-SA 4.0</a></sup></p>

CC BY-SA 4.0
2020-05-25T03:21:43.207Z
Hey, folks. If you have an answer, please post it as an answer, rather than a comment. Thanks!
HDE 226868
<orbital-mechanics><space-probe><parker-solar-probe>
Why does the Parker Solar Probe slow down as the distance from the Sun increases?
user1785960
# Why does the Parker Solar Probe slow down as the distance from the Sun increases?

(Asked by: username_0 on 2020-05-21)

Why does the Parker Solar Probe slow down as the distance from the Sun increases?

![Image of probe's trajectory](https://upload.wikimedia.org/wikipedia/commons/6/66/Animation_of_Parker_Solar_Probe_trajectory.gif)
  
Image credit: Wikipedia user Phoenix777, [CC BY-SA 4.0](https://creativecommons.org/licenses/by-sa/4.0/deed.en)

**Tags:** orbital-mechanics, space-probe, parker-solar-probe
> **Comments:**
> **username_1**: Hey, folks. If you have an answer, please post it as an answer, rather than a comment. Thanks!  

---

## 3 answers

### Answer by username_5 on 2020-05-22. Score: 23 | Accepted answer

> Why does the “parker solar probe” lose speed as the distance from the sun increases?

Because energy and angular momentum are individually conserved quantities in the two body problem. Except for where the Parker Space Probe has close fly-bys with Venus, the gravitationally interactions between the Parker Space Probe and the solar system are very closely modeled as a two body problem (the Sun and the probe), plus very small perturbations from the planets.

One way to express conservation of energy in the two body problem is the vis-viva equation, $$v^2 = \mu\left(\frac2r - \frac1a\right)$$

where

* $\mu = G(M+m)$ is the sum of the central body's standard gravitational parameter and that of the orbiting body,
* $r$ is the distance between the two bodies,
* $a$ is the semi-major axis length (a constant), and
* $v$ is the magnitude of the velocity vector.

Note that the mass of the Parker Space Probe is so much less than that of the Sun that one can drop the Parker Space Probe's mass from the expression $\mu = G(M+m)$, resulting in $\mu = GM\_{\text{sun}}$.

Note that the only variable on the right hand side of the vis-viva equation is the radial distance. As radial distance increases the square magnitude of the velocity vector (and thus the magnitude of the velocity vector) decreases.

Without mathematics, conservation of energy dictates that the sum of an orbiting body's kinetic energy and gravitational potential energy must remain constant. As the orbiting body moves further from the central body, the orbiting body's potential energy increases, which means its kinetic energy must correspondingly decrease. This in turn means the orbiting body's velocity must decrease.

> **Comments:**
> **username_0**: And how does this relate to the gravitational assistance used in Voyager missions? From what I understand, when moving away from the massive object, the probes should lose speed?  
> **username_5**: @user1785960 - They have lost speed, a whole lot of speed. But now that the Voyager probes are 123.9 astronomical units from the Sun (Voyager 2) and 149.3 astronomical units from the Sun (Voyager 1), the $\frac2r$ terms in the vis-viva equation are very small compared to the $\frac1a$ terms for both vehicles. While they do continue to lose speed, the loss is now very small for both of them.  
> **username_6**: @user1785960 In fact this is how gravity assists work. If you pass by a planet, you first gain speed as you come closer and then lose the same amount as you move away. But this speed is with respect to the planet, which is itself moving. If you play around a bit with directions and velocities, you'll notice that there are even situations where losing speed with respect to the planet actually means gaining speed with respect to the sun. Now a gravity assist is just picking the right directions when approaching and leaving a planet to benefit from these effects.  
> **username_7**: @user1785960 imagine very, very small rubber ball has a perfectly elastic collision with a very, very big moving ball. That's the same as gravity assist.  
> **username_8**: To understand this more viscerally, I can highly recommend playing a bunch of Kerbal Space Program - it is dangerously good at teaching (some, not all) orbital mechanics.  
> **username_5**: @AdamBarnes - Personally, I'm not a big fan of Kerbal Space Program. It makes too many simplifying assumptions for my taste. But nonetheless I do see the teaching value.  
> **username_9**: @user11153 there is a slight difference – gravity is attractive (and works at distance), while rubber ball collision is repulsive and works only at the (short) time they are actually touching.  

---

### Answer by username_2 on 2020-05-22. Score: 15

Quite simply, the sun's gravity is pulling on the space probe at all times. As the probe moves away from the sun, the force of gravity pulls it back in, slowing it down. As the probe moves toward the sun, the force of gravity continues to pull, speeding it up. Any object orbiting the sun is *always* accelerating toward the sun - when that acceleration opposes motion, the object slows down, and when it's in the same direction as motion, the object speeds up.

It's no different from throwing a ball into the air and seeing it slow down as it rises, reverse direction, and then speed up again as it falls, except the primary gravitational body in that case is the earth and not the sun.

> **Comments:**
> **username_3**: Or like a pendulum being pulled by the Earth. Closer to the Earth = fastest. Farthest from the earth = slowest. The difference between the ball and the pendulum is the pendulum "misses" the Earth when falling towards it like things in orbit.  

---

### Answer by username_4 on 2020-05-22. Score: 5

The simplest explanation is that the satellite generally obeys [Kepler's second law of orbital mechanics](https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Second_law_of_Kepler):

> A line joining a planet and the Sun sweeps out equal areas during equal intervals of time

When the satellite is further away from the sun, the area swept out remains constant only because the satellite is moving more slowly.

The physical answer, as others have pointed out, is that the satellite trades out kinetic energy for gravitational potential energy as it travels "up" the Sun's gravity well. This is no different from gradually slowing as you coast up a hill on your bike (although you are climbing up the Earth's gravity well not the Sun's).


---

