How could 99942 Apophis, in 2029, be captured and brought into a low Earth orbit?

Question

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)?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.)

Answer 5

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.

Answer 6

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.