<p><a href="http://en.wikipedia.org/wiki/Gravity_assist" rel="noreferrer">Gravity assists</a> such as this are a form of elastic collision. There's a bit of number crunching here (hopefully no mistakes!), so you'll want to be familiar with the basics of momentum, kinetic energy, and the conservation thereof.</p>

<blockquote>
  <p>Question: If Ceres (the largest known asteroid and nearly 500 km in diameter) used Earth to perform a gravity assist to increase its own velocity, by how much would this slow the Earth down, and how much larger would Earth's orbit become?</p>
</blockquote>

<p>The <a href="http://www.wolframalpha.com/input/?i=earth%20orbital%20speed" rel="noreferrer">orbital speed of Earth</a> around the sun is $U = 29.8~\mathrm{km~s}^{-1}$. So at a mass of $$M = 5.97\times 10^{24}~\mathrm{kg},$$</p>

<p>it has a kinetic energy of </p>

<p>$$K = 2.65\times 10^{33}~\mathrm{J}$$ and momentum $$P = 1.78\times 10^{29}~\mathrm{kg~m~s^{-1}}.$$</p>

<p>So let's say Ceres is performing a gravitational slingshot as in the simple diagram below. Ceres has a mass  $m = 9.47 \times 10^{20}~\mathrm{kg}$. It approaches Earth at velocity $v$, and after the slingshot its final velocity is (up to, for a low-mass object) a velocity of $2\times U+v$.</p>

<p><img src="https://i.sstatic.net/fQw5I.png" alt="enter image description here"></p>

<p>The total momentum of the system <em>must be conserved</em>. Ceres has changed direction and thus gained a significant amount of momentum in the leftwards direction: the same momentum that Earth must then lose. Kinetic energy is also conserved. So, we have a system of equations, where the subscripts <em>i</em> and <em>f</em> are initial and final momenta and velocities. M and U are the mass and velocity of Earth, m and v are that of Ceres.</p>

<p>$$MU_i^2 + mv_i^2 = MU_f^2 + mv_f^2$$</p>

<p>which says that the sum of the initial kinetic energies of the two objects must equal the sum of the final kinetic energy. We also have conservation of momentum:</p>

<p>$$MU_i + m\vec{v}_i = MU_f + m\vec{v}_f $$</p>

<p>Solving these equations, the solution is</p>

<p>$$v_f = \frac{(1-m/M)v_i + 2U_i}{1-m/M} $$</p>

<p>If Ceres approached Earth at $v_i = 30~\mathrm{km~s}^{-1}$, I get a solution of $v_f = 89.6~\mathrm{km~s}^{-1}$ - even for such a massive object, the $v_f \approx 2U+v$ approximation is extremely good. This means that Ceres' velocity has nearly been tripled by the gravity assist.</p>

<p>So, the final momentum of Earth is</p>

<p>$$MU_f = MU_i - mv_i - mv_f = 1.78 \times 10^{29}~ \mathrm{kg~m~s^{-1}} $$</p>

<blockquote>
  <p>In fact, Earth's linear momentum will only decrease by $mv_i + mv_f =
 1.13 \times 10^{23} ~\mathrm{kg~m~s^{-1}}$. From this change in momentum and Earth's mass, we find its orbital velocity decreases by
  $0.019~\mathrm{m~s}^{-1}$.</p>
  
  <p>Approximating a circular orbit (using $r=GM_{sun} / v^2$), Earth's
  orbit widens by 190 km. Sounds like a lot, but bear in mind that's 190
  km out of 150 million!</p>
</blockquote>

<p>Ceres is many orders of magnitude larger than any satellite that we could launch. So we could never practically use spacecraft to change our orbit significantly, and even an enormous near-miss asteroid would be of little consequence. But, <a href="http://arxiv.org/abs/astro-ph/0102126" rel="noreferrer">it hasn't stopped some from trying</a>!</p>

CC BY-SA 3.0
2013-11-06T22:37:19.280Z
I'm confused by the assertion in your answer that, if the Earth slows down, its orbit widens (which I assume means it moves further from the Sun).  That implies that, as the Earth loses energy, it will drift away from the Sun; rather than falling towards it (which was my understanding of Newtonian physics and gravity).  I'm obviously missing something.
db9dreamer
CC BY-SA 3.0
2013-11-06T23:03:46.537Z
@dav1dsm1th It's a manifestation of [Kepler's Third Law](http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Third_law). Another way of thinking about it is that as the Earth moves further from the Sun, it gains gravitational potential energy in exchange for kinetic energy.
Moriarty
CC BY-SA 3.0
2013-11-06T23:19:43.510Z
I'm going to have to do some more reading...  I can't get my head around the idea that the Earth could lose a significant amount of its kinetic energy (in a very unlikely encounter with a large body) and end up flying away from the Sun, rather than falling towards it. Thanks for the response.
db9dreamer
CC BY-SA 3.0
2013-11-17T08:32:30.990Z
The change in orbital eccentricity would depend on where the collision took place. As stated in my example, I assumed circular orbits to limit the scope of the answer. In reality our orbit is eccentric, and the changes to the lengths of the semimajor and semiminor axes of our orbit will depend on how close we are to perihelion and aphelion. If Earth loses momentum near perihelion, we will lose eccentricity. If we lose momentum near aphelion, we will gain eccentricity. At least, that's what Kerbal Space Program taught me :)
Moriarty
CC BY-SA 3.0
2013-12-03T06:13:57.453Z
@dav1dsm1th You must take Angular Momentum into account. With the same angular momentum, the closer you go to the center, the faster you spin or orbit. Look at ice sliders when turning around theirselves. When they get arms close to the body they spin faster, and they extend the arms in order to slow rotation down.
Envite
CC BY-SA 3.0
2013-12-04T11:33:18.793Z
@Envite but if the gravity assist has transferred angular momentum from Earth to Ceres - Earth has slowed down.  In my head this would cause Earth's orbit to decay towards the Sun.  I imagine an encounter where Earth becomes stationary.  At that point, in my head, Earth would not then suddenly accelerate away from the Sun - but fall towards it (like the skaters arms falling to her side when she stops spinning - rather than being torn off and flung into deep space).
db9dreamer
CC BY-SA 3.0
2013-12-04T12:05:32.777Z
@dav1dsm1th It is a common misconception. Slowing an orbiting body actually makes it go farther, not nearer the central body. Note that this refers to orbital slowing. If you slow down the Earth until it is truly stopped it will be infinitely far away from the Sun. Orbital mechanics does not consider a full stop to happen on place (the energy required would be prohibitive anyway). About skaters arms, they fall to side due to gravity towards floor, not due to gravity towards the skater. If there were no floor, arms would continue extended with no effort.
Envite
CC BY-SA 3.0
2013-12-04T12:39:52.907Z
@dav1dsm1th Here is something you can try at home, which is a very analogous situation to that which I describe in my answer. Hold something of a moderate weight in each hand, and sit in an office chair (with plenty of room around to spin). Spread your arms out, and spin yourself with your feet. Try this gently at first, and then faster - you can fall off doing this! Now pull your arms in - it will take some effort, but you'll spin more quickly. Spread your arms out - this will feel easier, but you'll spin more slowly.
Moriarty
CC BY-SA 3.0
2013-12-04T12:46:42.183Z
So when the space shuttle used to fire its retro rockets to slow itself down to allow it to return to Earth, NASA was doing it all wrong and should have been firing their booster to speed up to re-enter the atmosphere?
db9dreamer
CC BY-SA 3.0
2013-12-05T04:59:59.657Z
@dav1dsm1th good question. Imagine the Space Shuttle is in a circular orbit. To return to earth, they need to lower their periapsis. It's a matter of orbital mechanics that by firing thrusters in reverse (a retrograde burn), the orbit will become elliptical and the periapsis will lower itself into Earth's atmosphere. Perhaps the matter needs another proper question, rather than a prolonged comments discussion!
Moriarty
CC BY-SA 3.0
2013-12-05T13:00:41.050Z
I see where I've been going wrong.  Thanks for taking the time to reply.
db9dreamer
CC BY-SA 3.0
2013-11-01T07:10:13.600Z
2013-11-01T23:20:09.857Z
2013-11-01T23:20:09.857Z
Moriarty
<p>Any time a spacecraft comes in close proximity to a planet and if the spacecraft has the right angle then it is able to use the planet's velocity to move itself further into space.</p>

<p>According to Newton's 3rd law: every action has an equal reaction.</p>

<p>In this case when the spacecraft uses for example the Earth's gravity to speed up, Earth will move towards the spacecraft. Earth's orbital change will be very small because the spacecraft's mass is small compared to Earth's mass but what if a big asteroid comes in close proximity or what if we use Earth's gravity to catapult our spacecrafts and keep doing it for an extended period of time.</p>

<p>What could happen in this case? Could that have a dramatic impact to Earth's orbit?</p>

CC BY-SA 3.0
2017-10-30T09:25:17.467Z
Related XKCD:    https://what-if.xkcd.com/146/
userLTK
CC BY-SA 3.0
2013-11-01T06:08:52.503Z
I think *“impact”* describes it pretty well indeed…
e-sushi
<orbit><planet><earth>
Changes to Earth's orbit
kalpetros
# Changes to Earth's orbit

(Asked by: username_0 on 2013-11-01)

Any time a spacecraft comes in close proximity to a planet and if the spacecraft has the right angle then it is able to use the planet's velocity to move itself further into space.

According to Newton's 3rd law: every action has an equal reaction.

In this case when the spacecraft uses for example the Earth's gravity to speed up, Earth will move towards the spacecraft. Earth's orbital change will be very small because the spacecraft's mass is small compared to Earth's mass but what if a big asteroid comes in close proximity or what if we use Earth's gravity to catapult our spacecrafts and keep doing it for an extended period of time.

What could happen in this case? Could that have a dramatic impact to Earth's orbit?

**Tags:** orbit, planet, earth

> **Comments:**
> **username_2**: I think *“impact”* describes it pretty well indeed…  
> **username_1**: Related XKCD:    https://what-if.xkcd.com/146/  

---

## 1 answers

### Answer by username_3 on 2013-11-01. Score: 12

[Gravity assists](http://en.wikipedia.org/wiki/Gravity_assist) such as this are a form of elastic collision. There's a bit of number crunching here (hopefully no mistakes!), so you'll want to be familiar with the basics of momentum, kinetic energy, and the conservation thereof.

> Question: If Ceres (the largest known asteroid and nearly 500 km in diameter) used Earth to perform a gravity assist to increase its own velocity, by how much would this slow the Earth down, and how much larger would Earth's orbit become?

The [orbital speed of Earth](http://www.wolframalpha.com/input/?i=earth%20orbital%20speed) around the sun is $U = 29.8~\mathrm{km~s}^{-1}$. So at a mass of $$M = 5.97\times 10^{24}~\mathrm{kg},$$

it has a kinetic energy of

$$K = 2.65\times 10^{33}~\mathrm{J}$$ and momentum $$P = 1.78\times 10^{29}~\mathrm{kg~m~s^{-1}}.$$

So let's say Ceres is performing a gravitational slingshot as in the simple diagram below. Ceres has a mass $m = 9.47 \times 10^{20}~\mathrm{kg}$. It approaches Earth at velocity $v$, and after the slingshot its final velocity is (up to, for a low-mass object) a velocity of $2\times U+v$.

![enter image description here](https://i.sstatic.net/fQw5I.png)

The total momentum of the system *must be conserved*. Ceres has changed direction and thus gained a significant amount of momentum in the leftwards direction: the same momentum that Earth must then lose. Kinetic energy is also conserved. So, we have a system of equations, where the subscripts *i* and *f* are initial and final momenta and velocities. M and U are the mass and velocity of Earth, m and v are that of Ceres.

$$MU\_i^2 + mv\_i^2 = MU\_f^2 + mv\_f^2$$

which says that the sum of the initial kinetic energies of the two objects must equal the sum of the final kinetic energy. We also have conservation of momentum:

$$MU\_i + m\vec{v}\_i = MU\_f + m\vec{v}\_f $$

Solving these equations, the solution is

$$v\_f = \frac{(1-m/M)v\_i + 2U\_i}{1-m/M} $$

If Ceres approached Earth at $v\_i = 30~\mathrm{km~s}^{-1}$, I get a solution of $v\_f = 89.6~\mathrm{km~s}^{-1}$ - even for such a massive object, the $v\_f \approx 2U+v$ approximation is extremely good. This means that Ceres' velocity has nearly been tripled by the gravity assist.

So, the final momentum of Earth is

$$MU\_f = MU\_i - mv\_i - mv\_f = 1.78 \times 10^{29}~ \mathrm{kg~m~s^{-1}} $$

> In fact, Earth's linear momentum will only decrease by $mv\_i + mv\_f =
> 1.13 \times 10^{23} ~\mathrm{kg~m~s^{-1}}$. From this change in momentum and Earth's mass, we find its orbital velocity decreases by
> $0.019~\mathrm{m~s}^{-1}$.
>
> Approximating a circular orbit (using $r=GM\_{sun} / v^2$), Earth's
> orbit widens by 190 km. Sounds like a lot, but bear in mind that's 190
> km out of 150 million!

Ceres is many orders of magnitude larger than any satellite that we could launch. So we could never practically use spacecraft to change our orbit significantly, and even an enormous near-miss asteroid would be of little consequence. But, [it hasn't stopped some from trying](http://arxiv.org/abs/astro-ph/0102126)!

> **Comments:**
> **username_4**: I'm confused by the assertion in your answer that, if the Earth slows down, its orbit widens (which I assume means it moves further from the Sun).  That implies that, as the Earth loses energy, it will drift away from the Sun; rather than falling towards it (which was my understanding of Newtonian physics and gravity).  I'm obviously missing something.  
> **username_3**: @dav1dsm1th It's a manifestation of [Kepler's Third Law](http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Third_law). Another way of thinking about it is that as the Earth moves further from the Sun, it gains gravitational potential energy in exchange for kinetic energy.  
> **username_4**: I'm going to have to do some more reading...  I can't get my head around the idea that the Earth could lose a significant amount of its kinetic energy (in a very unlikely encounter with a large body) and end up flying away from the Sun, rather than falling towards it. Thanks for the response.  
> **username_3**: The change in orbital eccentricity would depend on where the collision took place. As stated in my example, I assumed circular orbits to limit the scope of the answer. In reality our orbit is eccentric, and the changes to the lengths of the semimajor and semiminor axes of our orbit will depend on how close we are to perihelion and aphelion. If Earth loses momentum near perihelion, we will lose eccentricity. If we lose momentum near aphelion, we will gain eccentricity. At least, that's what Kerbal Space Program taught me :)  
> **username_5**: @dav1dsm1th You must take Angular Momentum into account. With the same angular momentum, the closer you go to the center, the faster you spin or orbit. Look at ice sliders when turning around theirselves. When they get arms close to the body they spin faster, and they extend the arms in order to slow rotation down.  
> **username_4**: @username_5 but if the gravity assist has transferred angular momentum from Earth to Ceres - Earth has slowed down.  In my head this would cause Earth's orbit to decay towards the Sun.  I imagine an encounter where Earth becomes stationary.  At that point, in my head, Earth would not then suddenly accelerate away from the Sun - but fall towards it (like the skaters arms falling to her side when she stops spinning - rather than being torn off and flung into deep space).  
> **username_5**: @dav1dsm1th It is a common misconception. Slowing an orbiting body actually makes it go farther, not nearer the central body. Note that this refers to orbital slowing. If you slow down the Earth until it is truly stopped it will be infinitely far away from the Sun. Orbital mechanics does not consider a full stop to happen on place (the energy required would be prohibitive anyway). About skaters arms, they fall to side due to gravity towards floor, not due to gravity towards the skater. If there were no floor, arms would continue extended with no effort.  
> **username_3**: @dav1dsm1th Here is something you can try at home, which is a very analogous situation to that which I describe in my answer. Hold something of a moderate weight in each hand, and sit in an office chair (with plenty of room around to spin). Spread your arms out, and spin yourself with your feet. Try this gently at first, and then faster - you can fall off doing this! Now pull your arms in - it will take some effort, but you'll spin more quickly. Spread your arms out - this will feel easier, but you'll spin more slowly.  
> **username_4**: So when the space shuttle used to fire its retro rockets to slow itself down to allow it to return to Earth, NASA was doing it all wrong and should have been firing their booster to speed up to re-enter the atmosphere?  
> **username_3**: @dav1dsm1th good question. Imagine the Space Shuttle is in a circular orbit. To return to earth, they need to lower their periapsis. It's a matter of orbital mechanics that by firing thrusters in reverse (a retrograde burn), the orbit will become elliptical and the periapsis will lower itself into Earth's atmosphere. Perhaps the matter needs another proper question, rather than a prolonged comments discussion!  
> **username_4**: I see where I've been going wrong.  Thanks for taking the time to reply.  

---

