# Trying to prove a student in my physics class wrong

#### Bonanza123d

##### Time Lord from Gallifrey
Hey guys,

We are early in my HS physics class. A student that is at the top of the class drew a theory that if you fire a cannon ball at a certain height off the ground, that you will hit the same spot on earth (without counting earth rotation) I will draw the diagram here.

Please excuse the wobbling line. It was a quick drawing
Is there a physics equation that can prove him wrong. I will be doing an experiment in orbiter to test it. The start point will be outside the atmosphere. Show me your solutions and diagrams if possible.

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#### Urwumpe

##### Not funny anymore
Donator
Sorry, but your student is perfectly right, that is nothing else but "Newton's cannonball".

It describes the situation of a circular orbit.

#### Loru

##### Retired Staff Member
Retired Staff
Donator
well.. He's right. - If you shoot it just under orbital velocity projectile will land in the same spot cannon is (minus height)

---------- Post added at 08:56 PM ---------- Previous post was at 08:56 PM ----------

:ninja: 'ed by Urwumpe

#### Rtyh-12

##### New member
You don't even need an equation. An orbiting body always describes a perfect ellipse (not counting external forces, thrust, nonspherical gravity, or other perturbations). Even if you throw a ball, its path is still an ellipse - it has a very high eccentricity, but, nevertheless, it's an ellipse. A side effect of this is that the orbiting body will always return to its starting point (or any other point along its orbit).

Edit: Urwumpe, I think you (or me) misunderstood the question. I think the student he's talking about said that the orbiting body would hit the surface right beneath the point where it was fired (the bottom of the mountain in the scenario of Newton's cannonball).

Last edited:

#### Izack

##### Non sequitur
I believe the question was meant thus: If you fire a projectile at a certain tangential velocity from a certain altitude above an alternate spherical Earth, it will land on exactly the spot beneath the point of firing.

In this case, unless I am mistaken, if the projectile is fired at slightly less than orbital velocity, it will be given a slighty less than perfectly circular orbit. The maximum distance it could travel without orbiting would be 180 degrees around the planet's surface, landing on the other side of the world. :shrug:

#### Rtyh-12

##### New member
I believe the question was meant thus: If you fire a projectile at a certain tangential velocity from a certain altitude above an alternate spherical Earth, it will land on exactly the spot beneath the point of firing.

That's how I understood it as well.

#### Urwumpe

##### Not funny anymore
Donator
Edit: Urwumpe, I think you (or me) misunderstood the question. I think the student he's talking about said that the orbiting body would hit the surface right beneath the point where it was fired (the bottom of the mountain in the scenario of Newton's cannonball).

I have not read anything like that there and it would be impossible unless you have a way to reduce the energy of the projectile during flight.

Elliptic orbits still have conservation of energy.

#### Izack

##### Non sequitur
That's how I understood it as well.

Yes, didn't see your post before replying. I suppose the student is still right, though. If the cannon imparts the projectile with no tangential velocity (it misfires) then the projectile will drop straight down and land on the surface, exactly where the student said it would. :lol:

#### martins

##### Orbiter Founder
Orbiter Founder
Yes, didn't see your post before replying. I suppose the student is still right, though. If the cannon imparts the projectile with no tangential velocity (it misfires) then the projectile will drop straight down and land on the surface, exactly where the student said it would. :lol:

A perfect example of lateral thinking (excuse the pun)! Since the velocity of the ball was not specified, a zero velocity should be admissable. It's still an orbit, although a degenerate one, and it hits the ground where predicted. Well done!

#### Hielor

##### Defender of Truth
Donator
Beta Tester
If you ignore the effects of atmosphere, this isn't possible--the cannonball will pass through its original point.

If you include atmosphere, that makes things a bit more complicated. You'd need the cannonball to start off sufficiently high that it's affected by a small amount of drag, but not much. I'm not sure if there's a "happy medium" point you can get to where the right amount of drag will happen.

##### Donator
Donator
Beta Tester
According to Kepler (and many others), a orbit must describe either an ellipse, a parabola or a hyperbola. Without adding or removing energy from the orbiting body it will never describe a spiral (which is what you are drawing there).

Then use the polar equation for an ellipse:

R(T) = sqrt(sin(T)^2*a^2 + cos(T)^2*b^2)

And tell him that unless a or b equal zero, which is the degenerate case, given a certain T(heta) there is only possible and defined R(adius).
You can't throw the cannon ball at Theta=0 and Radius = 6371+h and one revolution later have the ball at same Theta=0 but at Radius 6371.

Also, one thing you gotta love about math is that no lateral thinking is required. Degenerate case is right there in front of you!

#### Urwumpe

##### Not funny anymore
Donator
wasn't there some relation like:

R X V = const = h

... I really need to refresh my orbital mechanics again :idk:

#### C3PO

Donator

If the initial velocity is exactly orbital speed, you will have a circular orbit [Green]. (and hit the back of cannon )

If it's higher, you will get an elliptical orbit[Blue] (and still hit the cannon)

If it's lower[Red], you will either hit the surface in the first half of the orbit, or go into elliptic orbit. (and still hit the back of the cannon :lol

#### Face

##### Well-known member
Orbiter Contributor
Beta Tester
Is there a physics equation that can prove him wrong.

I think this is already a bad start. "Prove him wrong" in the title and the quote above sounds like your only goal in this dispute is to show that he is wrong.

Given the vague input on the problem I can see where you two can shout at each other for hours without coming to an end ("No, i meant it <such> and <such>!"; "No, you said ABOVE ground!").

I'd instead suggest to try to understand his point first, possibly read it back to him so you know you talk about the same scenario. Then try to explain how you see things, best without a "you are wrong, nana nana naaana".

my :2cents:

#### AlfalfaQc

##### Future Rocket Engineer
Yeah, I agree that the question is a bit fuzzy... From what I understand, the guy is right since the ball will always pass through the same point (from what I remember from my astrophysics class). But the ball won't hit the ground under the canon, so if it is what he thinks, then he is wrong, but more precision would be nice here.

The only way it could happen is if you consider the atmospheric drag, but then it becomes really complicated :hmm:

#### Bonanza123d

##### Time Lord from Gallifrey
So here it is how it played out. He said that a cannon that is fired laterally will hit the same point on earth that the cannon was fired from. Like I stated before, the start point would be from orbit. My question is how is it able to hit the same point without landing on the other side of earth due to the atmosphere.

EDIT: Can someone simulate this in orbiter?

#### Urwumpe

##### Not funny anymore
Donator
So here it is how it played out. He said that a cannon that is fired laterally will hit the same point on earth that the cannon was fired from. Like I stated before, the start point would be from orbit. My question is how is it able to hit the same point without landing on the other side of earth due to the atmosphere.

If you are outside the atmosphere, it doesn't matter.

#### C3PO

Donator
Can someone simulate this in orbiter?

It will only hit the same spot if you ignore all other forces/movements except for gravity. AFAIK you can't stop the Earth rotating in Orbiter without some serious tinkering.
You could try it on a body without atmosphere, and start on one of the poles (to avoid the rotation). But don't leave the sphere of influence.

#### Hielor

##### Defender of Truth
Donator
Beta Tester
It will only hit the same spot if you ignore all other forces/movements except for gravity. AFAIK you can't stop the Earth rotating in Orbiter without some serious tinkering.
You could try it on a body without atmosphere, and start on one of the poles (to avoid the rotation). But don't leave the sphere of influence.
Note that the original question isn't about hitting the same spot--it's about hitting the ground *under* the same spot (it was originally fired from altitude)>

#### statickid

##### CatDog from Deimos
Donator
Even if you do this on the moon it is very difficult with realistic settings such as light pressure and non-spherical gravity. I once tried to make a circular orbit on the moon about 50 meters altitude so I could witness the ultimate flyby of the base tower at orbital speed! however, the orbit took alot of maintenance and kept changing ever so slightly. eventually i crashed when i looked away for a second.

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