What is the trajectory of a cannonball fired from the equator in space?

AI Thread Summary
When a cannonball is fired northward from the equator, its trajectory can be interpreted differently based on the frame of reference. In an inertial frame, the trajectory appears as a spiral due to Earth's rotation, while in a non-inertial frame, it seems like a straight line. The gravitational force is the primary influence on the cannonball's path, and without exceeding escape velocity, it will follow an elliptical orbit that intersects the Earth's surface. The Coriolis effect does not apply in the inertial frame of reference. Understanding these dynamics requires consideration of the two-body problem in orbital mechanics.
FermatPell
Messages
21
Reaction score
0

Homework Statement



Imagine that you are an observer in space (so you are in an inertial system), when the cannon (located on the equator) fires a cannonball in north direction. What does the trajectory of the cannonball look like from your perspective? Is it a straight line (that would mean that the cannonball is not affected by Earth's rotation) or something like a spiral?


Homework Equations




The Attempt at a Solution



I think that, since the only real force acting on the body is the gravitational force, the trajectory is like a spiral. I also know that no Coriolis force exists in my inertial frame of reference. Am I right?
 
Last edited:
Physics news on Phys.org
FermatPell said:
I think that, since the only real force acting on the body is the gravitational force, the trajectory is like a spiral. I also know that no Coriolis force exists in my inertial frame of reference. Am I right?

The question is not so clear for me. This is what I think-
If you consider the Earth to be an inertial frame of reference, then the trajectory of the ball would be a straight line.
If you consider the Earth to be a non-inertial frame of reference (take its rotation in account), then the trajectory of the ball would be similar to a spiral.

In both the above cases, you are in an inertial frame of reference, so there will not be any Coriolis force.
 
If you don't shoot the cannonball faster than escape velocity, the orbit will be an ellipse. Look up two-body problem. Of course the ellipse will intersect the surface of the Earth again at some point.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top