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- Thread starter putongren
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Bandersnatch

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Have you tried looking up "escape velocity"?

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Simon Bridge

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Pedantic note: planets tend to be spherical rather than circular.

You are thinking of firing a cannot pointed along a great circle I guess... is the planet supposed to be rotating?

Or do you just want to see how the ballistic relations are modified in a circular coordinate system?

Less pedantic note - if you fire the cannon at 90deg elevation, then the ball falls back to ground unless the initial velocity is greater than the escape velocity.

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Simon Bridge, I am aware of the escape velocity.

Is it:

PE(int) + KE(int) = PE(final) + KE(final)

and then solve for escape velocity, assuming PE(final) = 0, and KE(int) = 0?

Simon, I guess I'm modifying the simple physics problem where we solve for distance and maximum height when given an initial force vector for the cannon ball. I modified the problem so the surface is not flat but curved and finite. I know how to obtain the equations that lead to solutions for the former problem. The new problem is a great deal harder.

My attempt at the problem:

We need to know the equation for the motion of the projectile in polar coordinates form, minus radius of the planet, and equate the other side of the equation to 0. So the equation of the problem would look something like:

0 = (term describing motion of cannon ball) - (radius of planet).

Is it:

PE(int) + KE(int) = PE(final) + KE(final)

and then solve for escape velocity, assuming PE(final) = 0, and KE(int) = 0?

Simon, I guess I'm modifying the simple physics problem where we solve for distance and maximum height when given an initial force vector for the cannon ball. I modified the problem so the surface is not flat but curved and finite. I know how to obtain the equations that lead to solutions for the former problem. The new problem is a great deal harder.

My attempt at the problem:

We need to know the equation for the motion of the projectile in polar coordinates form, minus radius of the planet, and equate the other side of the equation to 0. So the equation of the problem would look something like:

0 = (term describing motion of cannon ball) - (radius of planet).

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- #5

TumblingDice

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I've been trying to develop a equation for a ball that is shot out of a cannon and see if it would return to the surface of a circular planet.

Simon Bridge, I am aware of the escape velocity... I guess I'm modifying the simple physics problem where we solve for distance and maximum height when given an initial force vector for the cannon ball.

Are you trying to calculate IF the ball will return to the planet, or WHERE it will land? Simon and Bandersnatch mentioned escape velocity because if all you want to know is "will it fall back", that doesn't require launch direction or angle to calculate height and distance. Escape velocity will leave the planet in any direction, as long you don't aim at the ground.

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jbriggs444

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Where are you defining zero potential energy? A common convention is that PE is zero at infinity. Under that convention, PE is negative everywhere else.Is it:

PE(int) + KE(int) = PE(final) + KE(final)

and then solve for escape velocity, assuming PE(final) = 0, and KE(int) = 0?

Also, what are you taking as the "initial" and "final" states?

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Hey all, I'm trying to determine IF and if yes, WHERE would it hit the planet's surface. Wouldn't that necessitate the development to mathematic equations of the flight path? You're right in that if I'm just interested in the question of IF, I just need to compute the escape velocity.

jbriggs444:

Initial states for PE is the cannon ball still on the surface of the planet, the final state for PE is at infinity. Initial state for KE is 0 because it is the instant where the cannon ball leaves the cannon and gaining velocity. The final KE state is when the cannon ball is at an arbitrary large distance form the surface.

jbriggs444:

Initial states for PE is the cannon ball still on the surface of the planet, the final state for PE is at infinity. Initial state for KE is 0 because it is the instant where the cannon ball leaves the cannon and gaining velocity. The final KE state is when the cannon ball is at an arbitrary large distance form the surface.

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- #8

Nathanael

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When the ball leaves the cannon, it will already have velocity and will then be losing velocity. The initial velocity will not be zero.Initial states for PE is the cannon ball still on the surface of the planet, the final state for PE is at infinity. Initial state for KE is 0 because it is the instant where the cannon ball leaves the cannon and gaining velocity. The final KE state is when the cannon ball is at an arbitrary large distance form the surface.

Simon Bridge, I am aware of the escape velocity.

Is it:

PE(int) + KE(int) = PE(final) + KE(final)

and then solve for escape velocity, assuming PE(final) = 0, and KE(int) = 0?

The velocity needed to "escape" is not a single velocity, it's a range of velocities. (If you go faster than "escape velocity" you will still escape.)

So when people say "escape velocity" they are typically talking about the

So, when solving for the minimum escape velocity, what can you say about KE(final)?

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cjl

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As for what is required for the calculation of the trajectory? If you assume the planet is spherical and has no mass concentrations, it is simply orbital mechanics. Part of the orbit will be below the surface of the planet, but that doesn't actually change the calculations. Based on the initial velocity and angle, you can determine the orbital parameters. Once you know the orbit, you can find the two spots where it intersects the surface of the sphere. One of those points should be the location of the cannon, and the other will be the point of impact.

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Simon Bridge

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"initial force vector" is not a useful parameter, you want the initial velocity and position vectors... but I see what you mean.

You want to look up "central force problem"[1] - this gets you general orbital equations.

To do ballistics with the orbital equations, you have to include the radius of the body - using the initial position as the planet radius and the place the projectile comes to land is where the orbit next intersects that radius.

More realistically, you need the radial equation for atmospheric drag as well.

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[1] See: Tan S M, http://home.comcast.net/~szemengtan/ClassicalMechanics/SingleParticle.pdf [Broken] ch1.7 p1-12 for example.

You want to look up "central force problem"[1] - this gets you general orbital equations.

To do ballistics with the orbital equations, you have to include the radius of the body - using the initial position as the planet radius and the place the projectile comes to land is where the orbit next intersects that radius.

More realistically, you need the radial equation for atmospheric drag as well.

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[1] See: Tan S M, http://home.comcast.net/~szemengtan/ClassicalMechanics/SingleParticle.pdf [Broken] ch1.7 p1-12 for example.

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