# Hand Grenade in Solar Orbit

1. May 16, 2012

### jackpelham

Hi, folks. I'm new here and came looking for some expert input on this odd question that came up as a friend and I were considering a hypothetical.

Suppose a hand grenade were orbiting the sun---say, half way between the orbits of Mars and Jupiter. And to make things easy, suppose that the grenade orbiting in the same direction as these other planets, and at a speed generally consistent with that of planetary orbits.

What happens to the shrapnel if the grenade explodes? After a time, will we find a large field of scattered shrapnel all moving in the original direction and speed? Will some shrapnel end up on an opposite course from the original course? Will any shrapnel end up orbiting on a different plane than the original (pre-explosion) plane? Will some shrapnel leave the solar system? I realize these are only a few possible results.

If anyone can chime in with an answer, I'd sure appreciate it. And if a consensus can be reached, that's even better!

Thanks!

2. May 16, 2012

### Drakkith

Staff Emeritus
I haven't done the math, but this would depend on how fast the debris from the explosion moved at. If it were faster than the orbital velocity, then the shrapnel that was facing the trailing side would move backwards and fall into a new orbit. Shrapnel from the side facing the orbital direction could indeed leave the solar system if the new velocity was high enough. Overall the debris would assume slightly different orbits as they are now moving at different velocities relative to the Sun.

Now, seeing as how I don't know the velocity of shrapnel from a hand grenade, I cannot say how probable each event is.

3. May 16, 2012

### Filip Larsen

Welcome to PF!

The shrapnel will directly after the explosion end up in slightly different orbits than the grenade, with the pieces moving parallel to the orbit tangent (the orbital velocity vector) getting most change for the "money". If we assume the grenade is in near circular orbit at the distance of Jupiter from the Sun and the pieces after the explosion can move with at most 500 m/s relative to the original speed of the grenade, then the pieces can enter elliptical orbits that will their opposite point up to around 100 million km further in or out from the Sun, or up to a few degree of change in orbital inclination.

However, I would not expect any of the pieces to escape the solar system as that would require the pieces to get a kick of around 5.4 km/s from the explosion.

4. May 16, 2012

### jackpelham

Drakkith and Filip,

Thanks for your speedy and apt replies to this hypothetical. Another question, if I may.

Let us suppose that the unexploded grenade had been traveling along orbital Plane A, with the sun also on that plane. To make it easy, let's imagine that Plane as if represented by your table or desk top, with the sun in the middle of the table.

Do you suppose that after the explosion, provided that the force was sufficient, ANY piece of shrapnel might be seen traveling on orbital Plane B, where Plane B is PERPENDICULAR to Plane A?

That is, would there be any limit to the angular displacement between Plane A and Plane B that could result from such a scenario? It's probably not hard to imagine a displacement of a couple of degrees from orbital Plane A, but 90 degrees? Is that even possible?

5. May 16, 2012

### DaveC426913

No*. To see why, let's pretend you have an unlimited amount of energy in the explosion. No matter how much energy is put into the piece of shrapnel along axis B, it will always have potential energy along axis A (back toward the sun along the orbital plane).

Consider the equivalent in a static triangle ABC. You have a right triangle whose base AB is 1metre, and whose height BC is variable.
Qow many metres high would BC to be make AC vertical?
A: It would have to be infinitely high.

PHP:

C
/|
/ | ?
/  |
A---B
1m

Now:
A is sol,
B is Asteroid Belt,
C is the piece of shrapnel flying straight up.

B

* At least not without further interactions elsewhere/when in its orbit.

Last edited: May 16, 2012
6. May 16, 2012

### Steely Dan

I'm not sure what you mean by "potential energy along an axis," as potential energy is a scalar. Anyway, if the shrapnel leaves the grenade traveling directly perpendicular to the plane of the solar system, it must necessarily curve inwards towards the Sun as it leaves the solar system. If the shrapnel does not get the kick necessary to escape, then one would expect it to orbit the Sun in a plane perpendicular to the normal orbital plane of the planets. So the answer to the question is yes. I think Dave misread or misunderstood the question. If he was answering the question "is it possible for any piece of shrapnel to travel straight upwards indefinitely," then he is of course correct.

7. May 16, 2012

### DaveC426913

You can consider the x and y motion (AB and BC respectively in my diagram) separately. You are correct in that it will always be pulled toward the sun in the A plane, no matter how much it it moves in the B plane. Which is why:

No it won't. It still has a half billion kilometers to fall toward the sun in the AB direction no matter what you do in the BC direction. That will be a component of its orbit, meaning it cannot be perpendicular.

8. May 16, 2012

### Steely Dan

The point about gravitational potential energy being a scalar is relevant here. The shrapnel has no memory of what its previous potential energy was. The only thing relevant to the problem is the velocity of the shrapnel after the explosion. If that velocity is directly perpendicular to the solar system at less than the escape speed, then of course it will travel in an orbit perpendicular to the orbital plane of the planets. Suppose you had a gun sitting at the distance of Jupiter from the Sun, stationary with respect to the orbits of the planets but lying on the orbital plane. If that gun shot a bullet directly upwards from its present location at a speed lower than the escape speed, it would surely travel in the perpendicular plane as it orbits. Perhaps what you are commenting on is that if the grenade has some initial and final $x$ velocity (that is, in the plane of the planet orbits), then the resulting orbit is not perpendicular to the $xz$ plane, which of course is correct. My comment was that if it has only $y$ velocity after the explosion (and no $x$ velocity) then of course it will travel straight upwards, and afterward continue to curve inwards toward the Sun but in a plane perpendicular to the $xz$ plane (in the language we're using, that would be the $yz$ plane). Of course, this would not occur if the piece of shrapnel were fired straight upward in the frame of the grenade, but perhaps some piece of shrapnel would have a velocity equal in magnitude but opposite in direction to the orbital speed in that frame, as well as a vertical component. In that case, it could have a velocity only perpendicular to the plane of the planets (as viewed from a non-orbiting frame). (Of course, since apparently shrapnel doesn't actually travel at 10 km/s, this is not likely to occur)

9. May 16, 2012

### Filip Larsen

For an object in circular orbit to change orbital inclination by 90 degree by a single impulsive maneuver (like an explosion) requires a speed change (delta-V) around √2 ≈ 1.4 times the orbital speed. For instance, in a Jupiter-like orbit the orbital speed is around 13 km/s so there it will take a delta-V of around 18.4 km/s to change inclination by 90 degree, which by the way is far more than needed to simply escape the solar system from that orbit (which "only" is √2 - 1 ≈ 0.4 times the orbital speed).

I'll leave it up to you to consider if an exploding grenade can be expected to eject shrapnel with those speeds

10. May 16, 2012

### DaveC426913

No. While it initial trajectory is perpendicular, it is in orbit around the sun - which is a half billion miles to the "left". of its short-lived-perpendicular orbit.

11. May 16, 2012

### DaveC426913

Wait, sorry. There are two planes that are perpendicular to the plane of the Solar System.

It can be perpendicular to the plane of the Solar System (plane XY) in one plane (the blue orbit, in plane XZ) but not the other (the green orbit, in plane YZ). I've been assuming the OP mean the green YZ orbit.

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12. May 16, 2012

### D H

Staff Emeritus
Prior to the explosion your grenade (assuming a roughly circular orbit) will have an orbital velocity of 13 km/s (Jupiter) to 24 km/s (Mars). For example, the asteroid Ceres has an orbital velocity of about 18 km/s. I'll use that 18 km/s as the orbital speed of the unexploded grenade.

For an object in a circular orbit about the Sun, escape velocity is √2 times orbital velocity. So the grenade explosion would need to add 7.5 km/s to the shrapnel to have pieces escape the solar system. Grenade shrapnel is an order of magnitude slower than this, so with today's grenades nothing would escape.

To have this happen the shrapnel velocity relative to the unexploded grenade would have to be at least 18 km/s. That's way beyond the shrapnel velocity of today's grenades.

Ignoring this, to be exactly perpendicular, the explosion would have to exactly cancel the pre-explosion velocity and add some velocity normal to the pre-explosion velocity. Nothing is exact, particularly with explosions. It could be approximately perpendicular of course -- but that would still require a very powerful grenade.

Last edited: May 16, 2012
13. May 16, 2012

### DaveC426913

As long as it cancelled the motion in its current orbit (along the Y axis in my diagram), it could have any arbgirarily small component of motion along the Z axis and still be in orbit - albeit a highly eccentric sun-grazing orbit - in the XZ plane.

So, if it were given a velocity of, say, only (13.1km/s to 24.1 km/s), in just the right direction - that would be enough to put it in a perpendicular orbit. Which, granted, is still a pretty big explosion.

14. May 16, 2012

### Steely Dan

Yes, but D H's point was that it would need to exactly cancel that motion and not even a little more, or else it would have a non-perpendicular orbit going the other way. That is tough to manage for a normal grenade (the fact that normal shrapnel speeds are nowhere close to orbital speeds to begin with notwithstanding), but if we can rig the grenade the way we want to arbitrary accuracy, it's possible.

15. May 16, 2012

### DaveC426913

OP: considering my diagram in post 11, which perpendicular orbit did you mean?

16. May 16, 2012

### DaveC426913

Keep in mind the OP's qualification:
We don't need to set it just so. Assume an arbitrarily large collection of shrapnel pieces expanding spherically. Some of them will be moving just so.

Last edited: May 16, 2012
17. May 16, 2012

### Steely Dan

You can assume that if you want, but that's not how a real grenade works. One would need the grenade to split up into infinitesimally small pieces (not realistic) expanding perfectly spherically symmetrically (also not realistic) to guarantee this.

18. May 16, 2012

### DaveC426913

The point is, it satisfies the OP's question. Yes, some pieces can end up in that orbit.

19. May 16, 2012

### Steely Dan

Indeed. A more likely way to achieve this is to use a railgun that fires at exactly the right angle and speed, although if the OP is obsessed with grenades then it's going to be tough to get what he wants. Then again, we don't have railguns that can shoot at 10 km/s yet either, so apparently this is a pipe dream for now.

Anyway, serious digressions have occurred. I hope the OP is satisfied with the discussion.

20. May 19, 2012

### jackpelham

Yes, Steely Dan, I'm satisfied with the discussion. Thank you all very much for your input. After reading it all, the bottom line seems to be that the outcome of the explosion all depends upon the force of the explosion. With a low-powered explosion, I should expect to see all the shrapnel traveling more or less in a group along the path of the original orbit, and in a high-powered explosion, I could expect to see shrapnel in any orbit, or even leaving the solar system.

If anyone thinks I have mischaracterized the discussion in this summation, please correct me.

Thanks again!

Jack