<p><a href="https://i.sstatic.net/TLtpE.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/TLtpE.png" alt="enter image description here" /></a></p>
<p>Here we go.  <em>Thanks for everyone's help!</em></p>

CC BY-SA 4.0
2021-10-23T05:20:45.997Z
It would be worth noting that in plotting such a trajectory the rotation of the Earth is significant. and practically, air resistance can't be neglected if you were to do this with a real cannonball. So the observed trajectory relative to a rotating observer in an atmosphere would be different.
James K
CC BY-SA 4.0
2021-10-19T14:53:00.333Z
2021-10-23T01:44:41.300Z
2021-10-23T01:44:41.300Z
Ray Andrews
<p>Kepler knew the answer to this one:</p>
<p>The cannon ball follows an elliptic trajectory with the centre of mass of the Earth-cannonball system at the focus of the ellipse.</p>
<p>And Newton was able to do the maths to deduce this from an inverse square law central force.
A modern derivation can be <a href="https://radio.astro.gla.ac.uk/a1dynamics/keplerproofs.pdf" rel="noreferrer">found for example in this worksheet for Glasgow University</a></p>

CC BY-SA 4.0
2021-10-18T00:33:50.940Z
Now that you mention it, it would have to be an ellipse or at the simplest a circle.
Ray Andrews
CC BY-SA 4.0
2021-10-17T16:49:49.600Z
2021-10-17T16:49:49.600Z
James K
<p>Everyone knows that if you fire a cannon, the cannon ball follows a parabolic trajectory.  But that's in an unvarying gravitational field.  What if the cannon ball flies so high that one must take the inverse square law into account?  I've been trying to work this out but I've got more than one possible solution.  Taking it to the next level there's the issue of the radius of the earth and thus the changing direction of the pull of gravity.</p>

CC BY-SA 4.0
2021-10-17T16:31:40.927Z
can you share your solutions with us?
planetmaker
CC BY-SA 4.0
2021-10-17T17:03:00.487Z
What about air resistance and the coriolis effect, which are more important?
ProfRob
CC BY-SA 4.0
2021-10-18T12:30:51.123Z
I’m voting to close this question because this is a physics question, not an astronomy question.
Carl Witthoft
CC BY-SA 4.0
2021-10-18T13:02:18.330Z
This question probably would belong on Physics SE but there probably is a duplicate somewhere.
WarpPrime
CC BY-SA 4.0
2021-10-18T13:12:53.313Z
Guys, I'm looking for something very, very simple.  All I want is to fire my cannon, but modify the equation 'y=x^2' such that the trajectory incorporates the inverse square law.  I'm on the surface of a planet of radius=1, I fire my canon at a 45 degree angle with sufficient powder that it reaches a height of radius=2, but I notice that the trajectory is not 'y=x^2', it is rather .... what is it?  Gravity at the apex of the trajectory is only 1/4 what it is a ground level so we are expecting a more 'rounded' trajectory.  Ignore the curvature of the planet.  What's the equation?
Ray Andrews
CC BY-SA 4.0
2021-10-18T13:13:57.463Z
BTW, this is an orbital mechanics question so seems to me to belong here, no?
Ray Andrews
<astrophysics>
What's the modification of a parabolic trajectory when accounting for variation in gravity?
Ray Andrews
# What's the modification of a parabolic trajectory when accounting for variation in gravity?

(Asked by: username_0 on 2021-10-17)

Everyone knows that if you fire a cannon, the cannon ball follows a parabolic trajectory. But that's in an unvarying gravitational field. What if the cannon ball flies so high that one must take the inverse square law into account? I've been trying to work this out but I've got more than one possible solution. Taking it to the next level there's the issue of the radius of the earth and thus the changing direction of the pull of gravity.

**Tags:** astrophysics
> **Comments:**
> **username_1**: can you share your solutions with us?  
> **username_2**: What about air resistance and the coriolis effect, which are more important?  
> **username_3**: I’m voting to close this question because this is a physics question, not an astronomy question.  
> **username_4**: This question probably would belong on Physics SE but there probably is a duplicate somewhere.  
> **username_0**: Guys, I'm looking for something very, very simple.  All I want is to fire my cannon, but modify the equation 'y=x^2' such that the trajectory incorporates the inverse square law.  I'm on the surface of a planet of radius=1, I fire my canon at a 45 degree angle with sufficient powder that it reaches a height of radius=2, but I notice that the trajectory is not 'y=x^2', it is rather .... what is it?  Gravity at the apex of the trajectory is only 1/4 what it is a ground level so we are expecting a more 'rounded' trajectory.  Ignore the curvature of the planet.  What's the equation?  
> **username_0**: BTW, this is an orbital mechanics question so seems to me to belong here, no?  

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## 2 answers

### Answer by username_5 on 2021-10-17. Score: 9 | Accepted answer

Kepler knew the answer to this one:

The cannon ball follows an elliptic trajectory with the centre of mass of the Earth-cannonball system at the focus of the ellipse.

And Newton was able to do the maths to deduce this from an inverse square law central force.
A modern derivation can be [found for example in this worksheet for Glasgow University](https://radio.astro.gla.ac.uk/a1dynamics/keplerproofs.pdf)

> **Comments:**
> **username_0**: Now that you mention it, it would have to be an ellipse or at the simplest a circle.  

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### Answer by username_0 on 2021-10-19. Score: 2

[![enter image description here](https://i.sstatic.net/TLtpE.png)](https://i.sstatic.net/TLtpE.png)

Here we go. *Thanks for everyone's help!*

> **Comments:**
> **username_5**: It would be worth noting that in plotting such a trajectory the rotation of the Earth is significant. and practically, air resistance can't be neglected if you were to do this with a real cannonball. So the observed trajectory relative to a rotating observer in an atmosphere would be different.  

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