1
$\begingroup$

An impact by a 10 kilometres asteroid on the Earth has historically caused an extinction-level event due to catastrophic damage to the biosphere. There is also the threat from comets coming into the inner Solar System. The impact speed of a long-period comet would likely be several times greater than that of a near-Earth asteroid, making its impact much more destructive...

Of course this is a paradoxical situation and question, but: suppose a perfectly spheric asteroid with radius of a few (hundred?) kilometers is on a collision course with the Earth.

If I were standing on its surface, could I deviate its direction just jumping on it in a direction perfectly perpendicular to the vector of velocity?

If it is possible, what is the angle of deflection, and the maximum size of an asteroid I could deviate?

Update

  • The asteroid has velocity and momentum in the direction of motion. On the normal direction momentum is zero and we know that it takes negligible/non zero energy to change the direcion of motion. Can we consider the action on the perpendicular as if the asteroid were at rest and treat it like a collision, with conservation of momentum etc?

  • If the abovesaid is correct, how do we measure the angle of deflection? In what way the push/jump can be calibrate to obtain a desired oucome?

$\endgroup$
1
3
$\begingroup$

The key here lies in two observations:

  1. If you can jump hard enough to escape the gravitational pull of the asteroid, this will permanently change the momentum of the asteroid; if you just jump and fall back down, nothing of lasting significance happens (conservation of momentum in the system)
  2. You want to change the angular momentum of the asteroid's orbit relative to the earth; so if you make a small change in the tangential velocity from a sufficiently long distance to earth, you can make a significant change in the angular momentum (which is velocity times perpendicular distance).

Let's do some math:

10 km diameter asteroid, density $5 kg/m^3$ has mass of $2.6\cdot 10^{15}kg$. If you have a mass of $100 kg$, and you make a big jump (say with a force of 2000N for one second - remember there is not much gravity and I'm going to assume Earth sent a strong guy to save the world...) the change in velocity of the asteroid is

$$\Delta v = \frac{F \Delta t}{m} = \frac{2000}{2.6 \cdot 10^{15}} = 7.6\cdot 10^{-13}m/s$$

If you wanted this to result in a deflection at Earth of at least 10,000 km (assuming that is enough of a "nudge", if done in the right direction, to get the asteroid to miss the earth; that is a big if - it is assuming the asteroid was not aiming straight at the earth, but the correct calculation needs to take account of the orbital dynamics a bit more carefully) then you need to give this nudge at a time

$$t = \frac{10^7}{7.6\cdot 10^{-13}} \approx 400 \text{ billion years}$$

Clearly this isn't going to work. We need a new plan to save the earth. "Dispatch war rocket Ajax".

$\endgroup$
3
  • $\begingroup$ Why momentum and ratio between masses? it takes negligible energy to change the path of a moving object, if you are bang on the perpendicular. It is not a collision between two masses, where momentum is relevant, right? $\endgroup$
    – bobie
    Sep 15 '14 at 13:58
  • $\begingroup$ Conservation of momentum: jumping off the asteroid is like a collision with time going backwards. As much momentum as you take away with you when jumping is the momentum given to the asteroid - and that's how you compute its velocity. You are right - the energy is (almost) all going to the person jumping. $\endgroup$
    – Floris
    Sep 15 '14 at 15:02
  • 1
    $\begingroup$ Good answer +1! One minor remark: Could you fix the exponent of the asteroids mass? There is a {} missing, and I don't have the reputation to propose the edit... $\endgroup$
    – Benedikt
    Sep 16 '14 at 11:18
2
$\begingroup$

The size of the asteroid does not matter: when you jump, the asteroid will move in the opposite direction.

Even if you jump up on earth, the earth moves backwards - but only infinitesimally so. Remember that $$F=ma$$ The mass of the earth is $5.97219 × 10^{24} kg$. If your mass is $100 kg$, then the earth's acceleration will be $6 × 10^{22}$ (almost 100,000 billion billion times) smaller than yours - in other words, not measurable.

You have a second problem though: when you fall back down you give a reverse push that cancels the first, and the net effect is zero. Of course this assumes the asteroid is small enough so you didn't jump completely away from it.

$\endgroup$
7
  • $\begingroup$ It would be helpful to estimate the escape velocity for, say, a Earth-killer asteroid of a couple km diameter :-) $\endgroup$ Sep 15 '14 at 11:41
  • $\begingroup$ @bobie - when you are "falling back", you are attracted to the asteroid (and it to you). During that time, you are deflecting the asteroid back to its old path (conservation of momentum). Only if you escape the asteroid can you truly change the course of history :-) . Thankfully the gravitational force is pretty small so escape velocity for a 10 km asteroid is miniscule. $\endgroup$
    – Floris
    Sep 15 '14 at 12:18
  • $\begingroup$ @Floris, two factors: a) when you jump you use your strength + weight, when you are up you attract only with weight (mass): b) when you push ++, whe you are up -, when you touch ground again you push again +, what is the balance? $\endgroup$
    – bobie
    Sep 15 '14 at 12:24
  • $\begingroup$ @bobie - the pull of gravity lasts for the time you go up (when you reach the top of the orbit both you and the meteorite stop moving outward) and the time you drop back (during which time you and the asteroid move towards each other). When you land, the two cancel out again. It's just conservation of momentum. $\endgroup$
    – Floris
    Sep 15 '14 at 12:35
  • 1
    $\begingroup$ @Carl For a comet of mass 3x10^12 kg (2km diameter) the escape velocity is 0.46 m/s, a slow walking pace. See wired.com/2014/08/comet-walk $\endgroup$
    – hdhondt
    Sep 16 '14 at 12:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.