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?

  • $\begingroup$ I think “impact” describes it pretty well indeed… $\endgroup$
    – e-sushi
    Commented Nov 1, 2013 at 6:08
  • $\begingroup$ Related XKCD: what-if.xkcd.com/146 $\endgroup$
    – userLTK
    Commented Oct 30, 2017 at 9:25

1 Answer 1


Gravity assists 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 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

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!

  • $\begingroup$ 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. $\endgroup$
    – db9dreamer
    Commented Nov 6, 2013 at 22:37
  • $\begingroup$ @dav1dsm1th It's a manifestation of Kepler's 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. $\endgroup$
    – Moriarty
    Commented Nov 6, 2013 at 23:03
  • $\begingroup$ 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. $\endgroup$
    – db9dreamer
    Commented Nov 6, 2013 at 23:19
  • 1
    $\begingroup$ If Ceres starts off moving away from the Sun and the orbital boost makes it move towards the Sun, then, to conserve momentum, the Earth's velocity away from the Sun may increase. Ceres gets a boost towards the Sun, the Earth gets a boost away from the Sun. It's this change in velocity that may result in a larger orbit. As a note, I think the Earth's semimajor axis increases, but so does the eccentricity of its orbit. $\endgroup$
    – user21
    Commented Nov 17, 2013 at 6:35
  • $\begingroup$ 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 :) $\endgroup$
    – Moriarty
    Commented Nov 17, 2013 at 8:32

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