Can We Stop a 10-km Asteroid Before It Hits Earth? | What If, Calculated #02

What if a 10-km asteroid were on a collision course with Earth?

We calculated the crater, the rock thrown out of it, and how Earth's and the Moon's motion would change. Then, using the real motion of the planets, we tried pushing it aside 10 years and 1 year ahead, and breaking it apart 1 day ahead.

The answer in 30 seconds

Earth's orbit around the Sun barely changes. That tells us about the motion of a planet, not the safety of its surface. We calculate the crater, falling rock and motion in space; we do not calculate shaking, heat, climate or harm to people.

The setup: one fictional collision, with specified inputs

The asteroid is a sphere of rock. Its impact site is a point in the Sahara chosen for this experiment, not a forecast.

Density tells us how much mass fits into a given volume. The ground here has no separate layers, mountains or water.

For the space experiments, we start from NASA JPL's ephemeris: a record of celestial positions and velocities. The planetary arrangement is from January 1, 2000, at 12:00 TDB, a time scale used in astronomy. “Hit Earth today” is the question we are exploring; the model does not identify an asteroid approaching Earth on the publication date.

If it hits: from mass to a crater and falling rock

How much energy does the asteroid carry?

Multiply a sphere's volume by its density to find its mass m. Use meters and meters per second in these equations. Superscripts and the ^ symbol indicate powers: d³ means diameter cubed, and d^0.78 means diameter raised to the power 0.78.

m = (π/6) × ρ × d³

m = (π/6) × 3,000 × 10,000³ ≈ 1.5708×10¹⁵ kg

Its kinetic energy, the energy of its motion, is:

E = m × v² / 2

E = 1.5708×10¹⁵ × 20,000² / 2 ≈ 3.1416×10²³ J

J stands for joules, the unit of energy. Speed is squared in this equation, so diameter alone does not specify the scale of an impact.

Why is the crater much wider than the asteroid?

We use the crater equations from Collins, Melosh and Marcus. They distinguish the initial cavity, called the transient crater, from the final crater after collapse. Let Dₜ be the initial cavity's diameter:

Dₜ = 1.161 × (ρ/ρₜ)^(1/3) × d^0.78 × v^0.44 × g^(−0.22) × (sin θ)^(1/3)

With our inputs, Dₜ is 66.30 km. To estimate the final diameter D, use the following equation, keeping all lengths in meters and setting D_c = 3,200 m:

D = 1.17 × Dₜ^1.13 / D_c^0.13

D_c is the transition scale between crater shapes. The result is 115.04 km, rounded to about 115 km. At 100 km/h, crossing that width would take 1 hour and 9 minutes. This is an estimate from a research model, not an exact reconstruction of the landscape.

What does “at least 17 cm” mean?

Rock thrown out of a crater is called ejecta. For the deposit thickness t, the model uses the initial crater diameter Dₜ and the surface distance r from the impact point. Both lengths go into this equation in meters:

t = Dₜ⁴ / (112 × r³)

At 1,000 km, r is 1,000,000 m. Substitution gives 0.1726 m, or about 17 cm. This is a lower-bound estimate: it excludes ground swept up by the falling rock and the extra volume of gaps between pieces.

Arrival times come from a separate calculation. We launch ejecta at 45° to the ground, find the speed needed to reach each distance, and follow its path under the gravity of a spherical Earth. This launch angle is a separate assumption from the asteroid's impact angle.

In this model, places outside the crater but inside the 1,000 km ring get a thicker deposit. Excluding the crater, that region covers about 3.125 million km², roughly the area of India. The 17 cm figure describes the distant edge of this region, not a uniform layer everywhere inside it.

Would the impact knock Earth out of its orbit?

Compare the masses:

Earth mass / asteroid mass = 5.9722×10²⁴ / 1.5708×10¹⁵ ≈ 3.8 billion

For this experiment, Earth and the asteroid merge at impact. We combine their momentum, meaning mass multiplied by velocity, then compare Earth's subsequent motion with the no-impact case.

The impact direction matters. Even in the direction that changes Earth's year the most, the change is only about 0.014 seconds, roughly one seventieth of a second. For our chosen Sahara collision, the year instead becomes about 0.00020 seconds shorter.

For the Moon, we measure the change in the size of its orbital ellipse, expressed by its semimajor axis, the ellipse's long radius. The largest change is about 3 m. This is not a statement that the instantaneous Earth–Moon distance jumps uniformly by 3 m. Also, the impact direction that maximizes this lunar change differs from the one that maximizes the change in Earth's year.

Deflection: change the meeting with Earth

What counts as a miss?

Our criterion is that the asteroid's lowest point passes more than 100 km above the surface. For a 10 km sphere, the relevant height is:

Bottom clearance = distance between centers − 6,371 km − 5 km

We move the Sun, eight planets and Moon under their mutual gravity. The asteroid responds to their gravity. Starting at the fictional impact, we calculate backward to construct its incoming path.

At the selected intervention time, we make a single change along or against the asteroid's direction of motion around the Sun. That velocity change is called delta-v, written Δv. We calculate forward again, narrow down the boundary between a hit and a miss, and choose whichever direction requires the smaller change.

An extra 8.5 mm/s changes its arrival time

Ten years before impact, the minimum is an increase of 8.519 mm/s, rounded to about 8.5 mm/s. This is an addition to the asteroid's speed, not its total speed. At the push, it is moving around the Sun at 25.94 km/s. The 20 km/s impact speed is measured relative to the ground and is a different quantity.

An 8.519 mm/s difference amounts to about 31 m over an hour. That is a way to picture the small initial change, not the method used to calculate the outcome a decade later. We follow the altered orbit under gravity.

In this case, speeding up enlarges the orbit slightly and increases the time needed to go around it. The asteroid reaches its closest approach about 7½ minutes late, after Earth has moved on. Its bottom passes at about 100 km altitude. One day before the original impact, the pushed and unpushed asteroids are about 6,000 km apart at the same instant.

Why does that tiny change take 1.9 million DARTs?

NASA's DART spacecraft hit Dimorphos in 2022. Research published in 2026 measured a roughly 0.15-second reduction in the Didymos–Dimorphos pair's orbital period around the Sun. That is distinct from the smaller asteroid's orbit around its companion.

We use DART's mass at impact, 579.4 kg, and impact speed, 6,144.9 m/s. Ejected material provides an additional recoil. The factor β, or beta, describes how much the transferred momentum exceeds that of the spacecraft alone. We set β = 2.0:

Momentum per DART p = β × 579.4 × 6,144.9 ≈ 7.1×10⁶ kg·m/s

Equivalent number of DARTs N = m × Δv / p

Convert Δv to m/s before substituting. For the ten-year case, N is 1.879×10⁶, or about 1.9 million. The asteroid is about 2.7 trillion times as massive as DART; giving it even a tiny speed change takes enormous momentum.

The one-year case needs about seven times the velocity change and momentum. These spacecraft counts are comparisons, not mission plans. Applying β = 1.7–2.3 gives about 1.6–2.2 million DART equivalents for the ten-year case. A 10 km asteroid would not necessarily transfer momentum with the same efficiency as DART's target.

Why can seven years work better than ten?

On this particular orbit, a push seven years ahead needs only about 6.05 mm/s. At that time the asteroid is closer to the Sun and moving faster, so a given velocity change can alter its orbit more strongly.

The lesson is that earlier detection generally reduces the required push. The asteroid's position along its orbit matters too. Seven years is not a universal optimum for incoming asteroids.

Breaking it apart: the fragments still need time to spread

One day before impact, we split the asteroid into 1,000 equal pieces, each about 1 km across. All receive the same outward speed, with randomly chosen directions.

Our energy benchmark is the 50-megaton Tsar Bomba, the largest human-made explosion. Using 4.184×10¹⁵ J per megaton of TNT gives about 2.09×10¹⁷ J.

We give the breakup every advantage: all that energy becomes motion spreading the fragments apart. Let Eₛ be this energy and u the scatter speed relative to the original asteroid. Since the pieces still have total mass m:

Eₛ = m × u² / 2, so u = √(2Eₛ/m)

The scatter speed is 16.3 m/s. We run the calculation three times with different fragment directions. All 1,000 pieces hit Earth in every run. They reach the planet before they spread far enough apart.

We then search for the energy at which at least 99% miss. The table reports the middle result, or median, of three runs.

At the one-day setting, the three runs leave 10, 3 and 17 fragments hitting Earth. The median is 10, which means 99% miss. It does not mean every fragment misses, or that every run reaches 99%. In the one-hour case, the scattering energy is about 40% of the original asteroid's impact energy.

How to read the results, and where the model stops

Deflection changes how the asteroid meets Earth. Breaking it apart requires the pieces to spread sufficiently before that meeting. Merely making smaller pieces barely reduces the energy reaching the planet.

These are deliberately simplified comparisons:

• The ground is uniform rock. We omit terrain, water, atmospheric slowing and breakup, and air resistance on ejecta.
• The crater coefficient is fixed at 1.161. The research gives a range, so 115 km is an estimate rather than an exact prediction.
• We omit asteroid rotation and sunlight pressure. A real deflection would also need a much wider safety margin than our 100 km boundary.
• The breakup model spends no energy on breaking the rock itself. All available energy goes into scattering already separated pieces, an exceptionally favorable assumption.

FAQ

Would humanity survive a 10 km asteroid impact?

These calculations cannot answer that. They cover the crater, falling rock and celestial motion, but not heat, shaking, tsunamis, climate or harm to people. A tiny change in Earth's orbit does not imply small consequences on the ground.

Could a few DART spacecraft deflect a smaller asteroid?

Under the same orbit, density and ten-year warning time, an asteroid 0.15 km across needs about six DART equivalents. This is a result for those conditions, not a general mission requirement. Momentum needed is proportional to mass; at fixed density, mass scales with the cube of diameter.

Is the Sahara impact a real prediction?

No. We chose the site and constructed the asteroid's path for this experiment. It is a fictional scenario for exploring the question, not a forecast involving a known incoming asteroid.

Try different conditions in the Web experiment

Open “Try it” on this page. The first challenge asks how much speed to add seven years before impact.

Start by changing the added speed and pressing “Push!” Once you make it miss, compare the ten-year and one-year cases. Then reduce the diameter to 0.15 km and watch the DART count change. The breakup and impact tabs let you vary the scattering energy and collision settings too.

Use the tables here as a starting point: which numbers change when you alter the warning time, and which change when you alter the asteroid's size?