I'm currently attempting to plan out a mission where, in 2029, during Apophis's close approach to Earth, a rocket is sent up to intercept Apophis and decelerate it, making it orbit the Earth. At it's periapsis of the Earth, according to a few different sources, it may be anywhere from 31,900 to 37,720 kilometers high, travelling at roughly 7.433 km/s. What would a rocket need to look like in order to do this? How much delta-v would be necessary (and how do you calculate this)?

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    $\begingroup$ If you want to decelerate Apophis, you should know its mass. But in 2029 we may decelerate an object from the trajectory of Apophis into an Earth orbit with a mass of about 10 to 100 kg but not anything as heavy as Apophis. $\endgroup$
    – Uwe
    Commented May 16, 2022 at 16:04
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    $\begingroup$ @Uwe, according to a quick google search, Apophis's mass is about 26.99 billion kilograms. I know that's very heavy, but at the moment I'm more concerned about the actual math of this rather than the plausibility of actually doing it. $\endgroup$
    – An Axolotl
    Commented May 16, 2022 at 16:33
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    $\begingroup$ @AnAxolotl I also see that number (26.99 kg from google). It's ridiculous, for two reasons. One is that that value is ridiculously precise. The mass estimates of smallish asteroids that have not yet been visited by spacecraft are oftentimes off by a factor of three. The other reason is that that value is inconsistent with size estimates. That value corresponds to a density of about 1.3 g/cc, i.e., ice with a bit of intermingled rock. $\endgroup$ Commented May 16, 2022 at 19:09
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    $\begingroup$ Close approach is far less important than the difference in velocities. If you can find one that is closer in velocity to the earth at some point that will reduce the energy requirement substantially, even if it is farther away. $\endgroup$ Commented May 17, 2022 at 4:02
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    $\begingroup$ A more realistic approach, would be to land (technology of your choice) on Apophis in 2029, and then use the subsequent seven years until its 2036 near-Earth approach to SLOWLY adjust its course - course adjustment is pretty much a question of energy excised, duration of the adjustment and how far away you are - a small course correction a long way away will have a much stronger final effect than a huge correction while it is close to Earth. Low powered nuclear engines firing continually for seven years would have a much better chance of nudging it into a course that ends up as an Earth orbit. $\endgroup$
    – Torque
    Commented May 19, 2022 at 13:27

6 Answers 6


The edge of Earth's Hill Sphere is about 929000 km, so in order to capture Apophis, it needs to decelerate from a hyperbolic orbit to an elliptical one with an apogee of at most 929000 km. In 2029, Apophis will pass 38017 km from the surface of the Earth, which is equivalent to a perigee of 44388 km from the center of the Earth. Then the orbital velocity, at perigee, of a 44388 x 929000 km orbit would be

$$\sqrt{\mu\Bigg(\dfrac{2}{r}-\dfrac{1}{a}\Bigg)}=\sqrt{3.986\cdot10^{14}\Bigg(\dfrac{2}{44388000}-\dfrac{1}{929000000}\Bigg)}\approx 4187 \text{ m s}^{-1}$$

This means that 3246 m/s of delta-v must be applied in the retrograde direction to capture Apophis. As Apophis weighs $2.7\cdot10^{10}$ kg, here are a few propellant estimates for 3246 delta-v, assuming that the dry mass of propellant tanks and other infrastructure is negligible:

Engine Thrust (N) ISP (s) Propellant needed (kg) Deceleration time (y)
F-1 6909000000 304 $5.321\cdot10^{10}$ 0.728
NERVA XE PRIME 246663 841 $1.302\cdot10^{10}$ 13.8
VASIMR 5 18000 $4.97\cdot 10^7$ 55638.18

At this point, with the high-thrust, low-impulse engines, the mass of propellant is on the order of the mass of the asteroid itself, as @AntonHengst mentioned in a comment. So unless space mining is used to turn rock into fuel, the task is nearly impossible. In addition, the burn time is terribly long, which needs years of preparation. As of May 2022 the approach is less than 7 years away so not even the NERVA can slow it down fast enough. Unfortunately, we just don't have the time or resources to achieve this feat in 2029. But maybe our distant descendants can...

Edit: Assuming funding wasn't a problem, you would need even more fuel to put the asteroid-moving infrastructure onto Apophis. The deceleration time could be slowed by adding... you guessed it - moar engines.

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    $\begingroup$ Ah I see, that makes sense. Thank you so much for taking the time to write out such an in-depth response! $\endgroup$
    – An Axolotl
    Commented May 16, 2022 at 21:06
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    $\begingroup$ Would aerobraking be better? $\endgroup$
    – throx
    Commented May 17, 2022 at 2:23
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    $\begingroup$ @throx With an asteroid that size? Are you kidding? $\endgroup$ Commented May 17, 2022 at 12:19
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    $\begingroup$ Note that the mass of the propellant is on the order of the mass of the asteroid itself. Edited the propellant mass figures be in kg. $\endgroup$ Commented May 17, 2022 at 15:45
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    $\begingroup$ @throx - Too big. Gonna need lithobreaking. $\endgroup$
    – Eric G
    Commented May 17, 2022 at 16:43

In 2029? We can't. I doubt that that the next two closest approaches after 2029 (2036 and 2051) are within reach. We might be able to do it by 2116, or maybe even 2066. An object as large as Apophis will take many years, more likely many decades, of applied thrust to bring into a controlled Earth orbit.

We don't even know what we're up against. We'll eventually know what we're up against when the OSIRIS-REx vehicle, which will be renamed OSIRIS-APEX after it drops off the samples it captured from Bennu. OSIRIS-APEX is currently targeted to intercept Apophis in August 2029. That's four months or so after Apophis's closest approach to Earth. We need to know its mass (the mass estimates are all over the place) and we need to know whether it's a rubble pile versus more or less solid.


The Hill sphere delta-V is the minimum possible for Earth to capture Apophis from the Sun. To go all the way to LEO, as in the question title, you have to decelerate a lot more than that.

Note first that 7.4 km/s is a perfectly ordinary velocity for something already in a LEO orbit. In a circular orbit, $r = a$ everywhere, so $$v^2 = \mu \left( \frac{2}{a} - \frac{1}{a} \right) = \frac{\mu}{a} \ ,$$ and with $v=$ 7433 m/s we find the circle's radius $a = \mu/v^2 =$ 7215 km, which is 837 km above the equatorial radius of Earth. This means that if Apophis actually got that close at that speed, Earth would capture it from the Sun, all by itself, without us needing to do anything. The problem with going that fast much farther away is too much angular momentum, which we have to get rid of in order to reach LEO distances.

Now consider how fast an object at 44388 km would have to be going, in order for it to have apogee there and perigee at 7215 km, or whatever other final radius you like. Since the semimajor axis is the average of apogee and perigee, $$v_A = \sqrt{\mu \left( \frac{2}{r_A} - \frac{2}{r_A + r_P} \right)}\ ,$$ so we find $v_A=$ 1585 m/s, requiring 5848 m/s of $\Delta v$ to slow down Apophis enough for Earth's gravity to pull it in from its point of approach to a LEO altitude. However, when it gets to perigee, it will be going too fast for a circular orbit there (as it must, because without further intervention it will just climb back up to GEO again). Falling down the gravity well from GEO to LEO causes it to pick up quite a lot of speed.

When we leave gravity alone, it conserves angular momentum. In astrodynamics, we usually divide that by the mass, and talk about the "specific" angular momentum, $\mathbf{\vec{h}} = \mathbf{\vec{r}} \times \mathbf{\vec{v}}$. In the special case of apogee and perigee, that's just $r_A v_A = r_P v_P$, so $v_P = r_A v_A / r_P =$ 9749 m/s. To circularize that orbit at perigee, we need to slow it down by a further 2316 m/s.

This means the total $\Delta v$ we need is 8164 m/s, which is greater than the starting relative speed! Note, however, this is not the most efficient way to do the orbit adjust, but it does at least put you in the right ballpark (the factor of two difference between this and the minimal capture is probably in the noise of this scenario anyway). All questions of how to make the thrust happen I leave to others, except to note that of course you shouldn't do this with one thruster, if you can fit hundreds or thousands of them on the surface.

I just wanted to point out that all this work is necessary simply to change Apophis's motion to the same velocity it used to have before we interfered, except at 837 km above Earth rather than 38010 km above Earth.


Let's toss out some ballpark guesstimates, and go from there.

Using the same figures as other answers: Apophis is estimated to have a mass of $2.7 \cdot 10^{10}$ kg, and needs a Δv change of about 4187 m/s.

This works out to about ~57 megatons of TNT (237 petajoules) worth of kinetic energy. Note that this is not the same as the amount of energy required - we need to put in far more. But this gives a lower bound (at least assuming we aren't going gravity assist shenanigans - which we can't in this case).

This immediately rules out pretty much anything non-nuclear. The only thing even close to the scales required is an Orion-style drive. That is, roughly, repeatedly detonate nuclear bombs.

So. Let's use an optimistic Orion variant. Effective exhaust velocity of ~118 km/s. So. How much mass do we need to toss out the back to make 4187 m/s? Well, our mass ratio required is $e^{\Delta v / v_0} \approx 1.036$. So we need ~3.6 % of the mass of the asteroid worth of nuclear bombs... (n.b. you can't do the calculation this way for higher mass ratios, but it's close enough in this case.)

How many nuclear bombs is that? 3.6% of $2.7 \cdot 10^{10}$ kg comes out to... oh. About 28 million tons worth.

For comparison, the peak number of nuclear bombs in existence was 60-65 thousands, and has since drastically dropped. And nuclear bombs do not weigh half a million kg each.

We do not have 28 million tons of nuclear bombs. We don't have close to that. And that's even assuming that they could be converted to something that'd work for an Orion-style drive (which they likely couldn't be).

(There are many other issues, too - such as trying to actually get Apophis to hold together, getting the environmental approval for launching Orion, etc., etc.)

  • $\begingroup$ As an aside: you could potentially increase effective exhaust velocity with larger bombs - even that optimistic Orion variant is just using 15kT bombs, whereas you can get better efficiency with larger ones. But this gets even less plausible to do in 7 years. (The 'normal' reasons that Orion can't use larger bombs - namely max acceleration and not melting the pusher plate - don't really apply here.) $\endgroup$
    – TLW
    Commented May 19, 2022 at 23:58
  • $\begingroup$ You wouldn't need bombs. A NERVA engine had an exhaust velocity of roughly 5,000 km/s. Put a few of them on the asteroid and start converting it to superheated rocket exhaust. It should have a roughly similar mass requirement, but a lot less fissionable material. But it totally depends on what Apophis is made of and whether its a suitable nuclear engine ejection mass. $\endgroup$
    – Dan Hanson
    Commented May 21, 2022 at 2:32
  • $\begingroup$ @DanHanson - several things. A NERVA has ~_8_ km/s exhaust velocity, not 5,000 km/s. And you won't get anywhere near 8 km/s exhaust velocity running on rock (scales as the inverse square root of molar mass, to a first approximation. Same as ion engines). The tyranny of the rocket equation means that a NERVA would end up having to use nearly half the asteroid as reaction mass as a result. (8 km/s -> required mass ratio of ~1.7). $\endgroup$
    – TLW
    Commented May 21, 2022 at 3:32
  • $\begingroup$ NERVAs have relatively modest exhaust velocities (for nuclear-based designs) due to them being limited in maximum temperature by the need to not melt the reactor, whereas bombs have no such limitation. $\endgroup$
    – TLW
    Commented May 21, 2022 at 4:03

It depends on how much of Apophis you want to capture.

As others have stated, Apophis is estimated to have a mass of 2.7 $\cdot$ 1010 kg, which needs to experience a $\Delta v$ change of about 4187 m/s.

However, a rocket engine is nothing more than an accelerator for rocket fuel, which is mass to be expelled very quickly in the opposite direction. Don't bring your own fuel, use Apophis itself.

According to a cursory online search, the fastest baseball pitching machine can accelerate baseballs to speeds of just over 60 m/s. A baseball weighing about 0.15 kg. Assuming you can build a space-grade pitching machine and increase that speed tenfold and you manage to have solar panels efficient enough to keep a thousand of them going on the front-facing side of Apophis you can "capture Apophis" (a bit of it, anyway).

Tsiolkovsky tells us that $\Delta v = v_e \cdot ln(m_0 / m_f)$, where $m_f$ is the "dry mass", i.e. the bits of Apophis that remain captured in orbit:

$$m_f = m_0 / e^{(\Delta v / v_e)} \approx 3.8 \cdot 10^7 kg \approx 0.0014\ m_0$$

Seeing as this removes 150 kg/s, it will take slightly less than 6 years to complete the manoeuvre. I think we have enough time to plan this before it is too late.

The rest of Apophis's mass is violently spewed in 1.7 $\cdot$ 1011 baseball-sized chunks throughout the Solar System, eventually likely making any further space exploration close to impossible.

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    $\begingroup$ Heh, fun answer. It would take much less energy to take the 0.14% part of Apophis and decelerate it, leaving the rest of Apophis in one piece. But of course that's even further away from the question's goal. $\endgroup$
    – jpa
    Commented May 19, 2022 at 14:43
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    $\begingroup$ re: "eventually likely making any further space exploration close to impossible." Nowhere near. Assuming they were all within a 1au^3 region, something moving at 100km/s would expect to get hit on average once / m^2 of frontal area / 8 billion years. Space is, uh, not small. $\endgroup$
    – TLW
    Commented May 20, 2022 at 0:06
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    $\begingroup$ As @TLW said, Apophis isn't nearly big enough for this to have that effect on space travel. Earth gets hit by 15 thousand tons of meteoroids and micrometeoroids every year, that adds up to the mass of Apophis in less than 2 thousand years. Breaking up Apophis would be a drop in the bucket. $\endgroup$ Commented May 20, 2022 at 14:10
  • $\begingroup$ @TLW My thinking was that this would create a orbital zone around the Sun where close to a trillion of small baseballs would be perforating anything that crosses that particular orbit. Given that they only have slightly different dv they would be "close" to each other. But as long as you avoid the death corridor you'd be fine, yes. $\endgroup$
    – bitmask
    Commented May 20, 2022 at 17:10
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    $\begingroup$ The solar system is full of asteroids that are regularly impacting each other (en.wikipedia.org/wiki/…) or disintegrating due to the YORP effect. A mere trillion additional objects is not going to make a difference. $\endgroup$ Commented May 21, 2022 at 14:20

To bring Apophis in a low Earth orbit would result in a catastrophe. Low Earth orbits have a limited lifetime of years, decades or centuries depending on height caused by drag. An impact of Apophis may destroy civilisation. So only high orbits with unlimited lifetime may be considered.

But changing the trajectory of such a heavy object would not be possible anyway.


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