Math problem, last people I can think of

[Manao]

mpucoder : there are no loss of energy for the system Moon - Earth, so the rotation of both around the sun will not be disturbed. But the Moon will indeed be slightly closer to the Earth, and Earth's roation around itself will slightly be slower.

Kedirekin : It's possible. Orbits do not have to be circular, they can be elliptic. And such a situation could happen with the asteroid reaching it's perihelion / aphelion. At that point, it's motion vector would be perpendicular to the segment [Sun - Asteroid], and its speed would be the same as Earth's one, since distance from the Sun is the same.

[r6d2]

Originally posted by Kedirekin
If the asteroid is in a tangent to earth's orbit and moving at the same speed and direction, it is in the same orbit and can never collide. You are right! What I meant was that the asteroid was in course of collision along the tangent at the point of collision. The asteroid is going in a straight line. Sorry. Anyway, Manao got the idea. ;)

I really need a way to post pictures here. :o

Another problem based on the same idea:
What happens in the situation above if the asteroid is moving 10% faster than the Earth at the time of collision?

[Manao]

With the asteroid going faster than the Earth, the motion vector of Earth + Asteroid after collision will be (M + m*1.1) / (M + m) * v.

One of the keplerian laws says that angular speed is a constant of the movement of a planet around the Sun. Another says that the value a^3 / T^2, with a being the 'big' radius, and T the period of revolution is a constant for all planets around the Sun.

So angular speed ( radius * v when v is perpendicular to the radius ) has been raised, so the period has been reduced. So Earth will be revolving faster around the Sun. Moreover, since the period is lower, the 'big' radius of the elliptic orbit of Earth has also been reduced. Hence, if Earth's orbit was circular, it's now elliptic, with the crash point being its aphelion. So the new orbit will be inside the previous one.

[Kedirekin]

I have a feeling I'm going to get pedantic here.

It actually is not possible. If the asteroid's perihelion just touches earth's orbit, it cannot have the same velocity as the earth. It must have a higher velocity than the earth in order to move further away from the sun (barring collision of course).

Another way to put it, a body's orbit is completely described by it's velocity vector at any point in time. If it's velocity vector is the same as the earth's, then it's orbit is the same as the earth's.


PS. I don't mean to be a kill-joy. Ignoring the orbital mechanics, it's an interesting - slightly tricky - question. With the described conditions, Manao is of course correct; momentum is always preserved.

Here's another slightly tricky one.

Given two objects in the same orbit that are mutually gravitationally attracted, but separated by a distance significantly larger than their diameter, what will happen?

Note: this situation exists right here in the solar system.

[r6d2]

Originally posted by Manao
So angular speed ( radius * v when v is perpendicular to the radius ) has been raised, so the period has been reduced.Angular speed isn't v/r? It is more natural to think that if speed increases, the new orbit will have a bigger radius. Think for instance that the bigger the mass of the asteroid, the greater the momentum and thus the impact. Earth might get kicked to the outer solar system by a big asteoid, not towards the Sun.
Originally posted by Kedirekin
I have a feeling I'm going to get pedantic here.ac3 is getting used to it. :)
It actually is not possible. If the asteroid's perihelion just touches earth's orbit, it cannot have the same velocity as the earth.Consider then the asteroid is self-propelled and can compensate for the gravity of Sun. :)

Actually, if the orbit is elliptic, as Manao suggests, it is possible. I don't think the orbit is completely described by its velocity vector. This is true only for circular orbits, where speed is constant. I said the speed was constant, but to limit the scope of the problem you can consider it constant in the vecinity of Earth. We are not considering either the Earth's gravity effect on the asteroid.

[r6d2]

Originally posted by Kedirekin
what will happen?My first guess, based on Saturn's Rings, is "nothing". And the "significantly larger" part only reinforces that. But I'm sleepy. :)

[Kedirekin]

Agreed. If the asteriod is under thrust, the scenario is possible. Heck, any scenario is possible.

But I do stand by what I said. An free-falling asteriod with a perihelion the same distance from the sun as earth's orbit will have a higher velocity at perihelion than earth does. The velocity has to be higher, or the asteriod wouldn't move back outward again.

I can see all the diagrams and the machanics in my head (velocity vectors and changes over time, acceleration, translation between potential and kinetic energy), but I'm not eloquent enough to explain them with crystal clarity. I guess if you don't believe me, we'll just have to agree to disagree.

[Kedirekin]

In response to Manao's response to the 10% faster question.

The combined body will be moving faster, as Manao says, but the higher velocity immediately after impact will push the combined body outward - it will be moving at higher than orbital velocity. The point (or more accurately, the distance) of impact will become the perihelion.

As is so often true with orbital mechanics, thinking about it kind of twists your senses. In the words of one of my favorite authors -


"East takes you out, out takes you west, west takes you in, and in takes you east"
- Larry Niven

[Kedirekin]

Hint (or spoiler): Epimetheus and Janus

The universe is an amazing place.

[Manao]

r6d2 : angular speed was indeed v / r, and I said something wrong about the keplerian law, because in that case it's not the angular speed which is constant, but v * r. Which explains my second error, rightly pointed out by Kedirekin : in an elliptic orbit, speed decrease with the distance, and if angular speed was constant, it would be the contrary. So I inverted aphelion / perihelion.

Third mistake, it's indeed not possible for the Earth and the asteroid to meet with the same motion somewhere around the Sun.

Fourth, the period will be higher, not lower, because the 'big' radius will be higher, not lower.

[r6d2]

Originally posted by Kedirekin
Hint (or spoiler): Epimetheus and JanusBut these two don't fit into the "significantly larger" category, and they are not in the same orbit either. :confused: I'll give it more thought.

[duartix]

@Manao:Maybe I misunderstood the way the asteroid is meeting the Earth, No, Manao, it was me who missunderstood the colision, I thought it was a perpendicular colision...

[Kedirekin]

Okay, Epimetheus and Janus aren't in exactly the same orbit, but close enough as makes no never mind. Their orbits are so close that the body of one moon nearly (or maybe actually) overlaps the other's orbit.

And most of the time they are thousands or tens (or hundreds) of thousands of kilometers apart in their orbits, while their diameters are roughly 100 kilometers.

You can ignore Epimetheus and Janus if you want and deal with the ideal case. The mechanics is nearly the same, though in the ideal case the results are a little different. I wasn't getting at anything too intense or mathematical.

[r6d2]

Originally posted by Kedirekin
You can ignore Epimetheus and Janus if you want and deal with the ideal case.What I don't get is that if the two bodies share the orbit, they shoud have the same speed, as you pointed out above on the asteroid problem. So the "nothing" result I posted previously, meaning they stay like that forever, is the only one that makes sense... :confused:

[Kedirekin]

Sorry, I should have been clearer.

Assume two small moons 100 km in size are in orbit around a large planet. Further assume they are 10,000 km apart, traveling in the same orbit, and initially they have exactly the same velocity (tangential to the orbit at their respective positions, of course).

Taking into account that the moons are mutually gravitionally attracted to each other (in other words, don't ignore gravity), what will happen? Will the moons collide?

I fear I've already ruined the problem by giving you the spoiler.

[Kedirekin]

Here's another, perhaps more interesting way to state the same problem.

Assume you are floating in space (in a space suit) holding a rock. You are 500,000 km ahead of the earth in it's orbit, stationary with respect to the center of the earth (i.e. you are not in orbit around the earth).

You drop the rock and it very slowly begins to fall towards the earth - remember, you are stationary with respect to the earth.

Will the rock hit the ground?

[r6d2]

Just checking if I understood correctly. I'm in orbit around the Sun, at the same speed as the Earth and in the same orbit. The rock is with me but still at the same speed. I let the rock go. Is that it?

[sysKin]

Originally posted by Kedirekin
You drop the rock and it very slowly begins to fall towards the earth - remember, you are stationary with respect to the earth.

Will the rock hit the ground?As I understand, there is some engine that prevents you from falling.

OK, so why the rock wouldn't hit the Earth? It will.. if it wasn't for Earth's gravitation, the rock would be on orbit around the Sun, just like Earth currently is. Extra pull from Earth will make the rock accelerate exactly towards the centre of Earth, all the time, so eventually the rock will hit the Earth exactly in the centre (from your point of view).

Any extra forces present (Sun's pull) will make the rock and Earth accelerate, with exactly the same acceleration - so they don't matter.

:)
Radek

[RadicalEd]

Of course it won't hit the ground, it'll burn up in the atmosphere long before ;)

[Kedirekin]

@r6d2, yes, you've got it.

@sysKin, oh, that it were that simple. And yes, consider yourself magically suspended in space so you yourself aren't affected by earth's gravity.

@RadicalEd, I know your post is meant tongue-in-cheek, but actually I'm not sure the rock would get going fast enough to burn up. I know escape velocity is 11 km/sec, but I don't know if half a million km can be considered escape. I think I might be curious enough to check it out (later).