# Thread: plz help with this simple newton cannonballl problem

1. lets see if someone can help me solve this problem about the newton cannonball:

i shoot one cannonball at 1 m height from the center of the planet at 1 m/s, i check the cannonball makes a perfect circle

now i shoot the cannonball again at 2 m/s

what will be its height at the apogee?

2.

3. Originally Posted by luxtpm
lets see if someone can help me solve this problem about the newton cannonball:

i shoot one cannonball at 1 m height from the center of the planet at 1 m/s, i check the cannonball makes a perfect circle

now i shoot the cannonball again at 2 m/s

what will be its height at the apogee?
Infinite. The cannonball in the second instance exceeds escape velocity and enters a hyperbolic trajectory. For that body at that distance from the center any velocity equalling or exceeding m/s will be an escape trajectory.

4. thanks a lot of janus let me ask another question so it helps me clarify:

with the same data at what speed should i shoot the cannonball so it acquires 2 m height?

edit:

but how did you obtain that value for scape velocity?

5. Originally Posted by luxtpm
thanks a lot of janus let me ask another question so it helps me clarify:

with the same data at what speed should i shoot the cannonball so it acquires 2 m height?
1.1547 m/s
or
m/s

edit:

but how did you obtain that value for scape velocity?
Orbital velocity for a circular orbit is
Escape velocity is

Since a circular orbit at that distance is 1 m/s, the escape velocity is times larger.

6. Originally Posted by luxtpm
i shoot one cannonball at 1 m height from the center of the planet at 1 m/s, i check the cannonball makes a perfect circle
If you are talking about the Earth or any other planet I have ever heard about, then 1 m above the center of the planet is extremely deep inside it.

So I understand you are talking about a cannonball orbiting in an imaginary empty (I mean, vacuum inside) tunnel around the center of the planet.

If so, the formula relating the escape velocity to orbital velocity won't work.

Your orbit is around just the innermost 1m-radius ball of molten heavy metals, as the gravitational pull of the outer layers on your cannonball balances itself out. This is why the orbit velocity is so amazingly low. But to escape to outer space (presumably, through an immensely deep vertical mineshaft) you need to come to the surface and still have enough velocity to excape from there to infinity, against the gravitational pull of the whole planet.

On Earth, this means leaving the surface at about 11.2m/s. I don't have the time to find out what muzzle velocity you would need in your cannon positioned 1m above the center of the Earth, but it would sure have to be more than 1.41m/s.

Cheers, L.

7. no well i was assuming a particle planet

theres a formula to solve this problem but it can be solved too applying cons of energy and momentum

interesting problem thought the one youre considering

8. Originally Posted by Leszek Luchowski
Originally Posted by luxtpm
i shoot one cannonball at 1 m height from the center of the planet at 1 m/s, i check the cannonball makes a perfect circle
If you are talking about the Earth or any other planet I have ever heard about, then 1 m above the center of the planet is extremely deep inside it.

So I understand you are talking about a cannonball orbiting in an imaginary empty (I mean, vacuum inside) tunnel around the center of the planet.

If so, the formula relating the escape velocity to orbital velocity won't work.

Your orbit is around just the innermost 1m-radius ball of molten heavy metals, as the gravitational pull of the outer layers on your cannonball balances itself out. This is why the orbit velocity is so amazingly low. But to escape to outer space (presumably, through an immensely deep vertical mineshaft) you need to come to the surface and still have enough velocity to excape from there to infinity, against the gravitational pull of the whole planet.

On Earth, this means leaving the surface at about 11.2m/s. I don't have the time to find out what muzzle velocity you would need in your cannon positioned 1m above the center of the Earth, but it would sure have to be more than 1.41m/s.

Cheers, L.

I hadn't considered that he meant inside the body of a planet, and it looks like that isn't what he meant.

The escape velocity from the center of the Earth would be in the order of 17.7 km/sec.

9. Originally Posted by Leszek Luchowski
On Earth, this means leaving the surface at about 11.2m/s.
Ooops, sorry, of course I meant 11.2km/s, not 11.2m/s.

Cheers, L.

10. Originally Posted by luxtpm

interesting problem thought the one youre considering
There's a neat solution to it.

Assuming that our planet is of uniform density and our orbiting mass is a WIMP (weakly interacting massive particle) and only reacts gravitationally with the planet, we can set it up like this:

The gravitational potential of a mass m located at a distance r from the center of a sphere of radius a and mass M is ;

Thus the total energy of a particle moving at velocity v at a distance of r is

What we want to do is find r for our apapis given that we know v and r for the periapis by taking the total energy at perapis and apapis and equating them.

First perapis:
V=2 and r=1 so

For apapis we can use a short cut. Newtons shell theorem shows that a hollow sphere has no gravitational effect on anything inside of it. That means that we don't have to worry about any mass further away than the apapis. We can reduce the problem to one of a sphere with a radius equal to the apasis distance. This also means that "a" in the first equation will equal the apapis distance.

Thus we can find the total energy of the particle at apapis by treating as a particle at the surface of a sphere with a radius of a.

so:

equating the two equations:

first off we note that we can drop the 'm's

This still leaves us with M, a, and Va as unknowns.

M can be taken care of thus:

Our initial conditions were a 1 meter orbit at 1 m/s. Again making use of Newton's shell theorem, we can treat this as an object orbiting at the surface of a sphere of meter radius. Using the orbital velocity equation:

we can solve for M. This "M" (call it M1) is that of a 1 meter sphere, the "M" we need is the mass of a sphere of radius a (call it Ma).

If we increase the radius of a sphere by a factor of "a" we increase its volume(and mass) by a factor of a^3 so we can say that:

and we can substitute this for M in our equation:

this reduces to :

We can make things even simpler by noting that under our initial conditions,

reduces to

which means we can replace with 1 in our equation:

Gathering terms and reducing:

We still have two unknowns to deal with, but I've got to go for a bit so I'll show you how to deal with this in the next post.

11. Okay, I'm back.

Picking up where I left off, we had arrived at the equation:

This tells us tells us the velocity of the particle at apapsis if we know the distance and vice versa. however we know neither.

To go forward we need to invoke another conservation law; the conservation of angular momentum.

First, I need to mention something. Up to now I have been talking about perapsis and apapsis as if we were dealing with a typical elliptical orbit. However, this is not a typical orbit. Because the mass affecting the trajectory changes over the course of the orbit, instead of an elliptical orbit with the center of the planet at one of the foci, we get an oval orbit with the center of the planet at the center of the oval.

So the orbit is closest when the radial is aligned with the minor axis and furthest when it is aligned with the major axis. There are two minimum distance points and two maximum distance points to each orbit.

That being said we can deal with angular momentum.

Angular momentum can be found by

as along as direction of v is perpendicular to the radius vector. This is some thing that is true at all times for a circular path.

However, with our orbit, this is true only when the object is at the maximum or minimum distance. Which is fortunate, because these are just the points wr are concerned with.

So if we equate the angular momentum at minimum with that at maximum, we get:

Drop the 'm's

v1=2 and r=1 so

solve for Va

substituting into the equation from the last post:

Which gives us a single unknown to solve for.

There are two solutions to this problem:

a=1, which we already knew, as it is the solution for the minimum distance point from which we started.

and

a=2, which turns out to be our solution for the maximum distance of our orbit.

So our answer is 2 meters as the furthest distance of our orbit starting at 1 m and 2 m/s.
will equal 1 m/s

This assumes that is is perpendicular to the orbital radial at 1 m distance.

We can also solve for the situation where is along the radial. In this instance the mass will travel straight up away from the center of the planet until it stops and starts to fall back.

we can find this maximum height by using

and setting to 0

We get an answer of meters (2.236 meters)

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