That tower question (3 Viewers)

[lolcakes52]

Okay bitchs let me explain.

So tower one is in the same frame of reference as the point of earth which is above. As far as we are concerned it stays above that point and it is motionless, non accelerating etc. So the ball drops straight down.

Now, tower number two. It is practically orbiting the earth, the velocity at the tip is changing every moment. THIS MEANS THAT IT IS A NON INERTIAL FRAME OF REFERENCE. SO WHEN the ball is dropped it isn't going to do whats expected, aka ball number one's action.
The ball at tower number two already has the altitude, velocity and orbital period of a satellite in a geostationary orbit; read as it is already in orbit.

lets pull an einstein thought experiment. Pretend these towers are not tower, they are satellites. Well straight away we see that the one above the north pole will plummet to its doom, with it's ball. But the other one lets go of it's ball and, what do you know, the balle just keeps on orbiting.

 

[lolcakes52]

For mass B the mass has a horizontal velocity same to that of the earth so RELATIVE to the sun an external observer will see it go in a parabolic motion around the earth before landing on the spot under the tower. It to has a constant vertical motion

It is a pity the principle of relativity doesn't apply to non inertial frames of reference, AKA circular motion.

 

[someth1ng]

talked to teacher who has been head of Marking for quite a bit:

This is what he had to said you needed to get full marks

earth orbits its axis at roughly 1700km/h at an easterly direction.
for mass A since it as the pole, the top of the axis it will gain the horizontal component of earths velocity. But it falls to the ground below the tower because it has a constant downward (vertical) accerlationa cting on it.
For mass B the mass has a horizontal velocity same to that of the earth so RELATIVE to the sun an external observer will see it go in a parabolic motion around the earth before landing on the spot under the tower. It to has a constant vertical motion

IT DOES NOT GO IN GEOSTATIONARY ORBIT BECAUSE MY TEACHER GAVE ME A SIMPLE EXAMPLE AND THAN LAUGHED.....

PLEASE NOTE: A GUY WHO JUMPED OF QUITE HIGH RECENTLY IN A DAREDEVIL STUNT DID NOT GO INTO EARTH ORBIT.... HE WENT STARIGHT DOWN.... I THINK THAT EVIDENCE IS ENOUGH THAT RELATIVE TO THE SUN THE MASS WILL GO A PARABOLIC DECAY WHERE IT FALLS TO THE GROUND BENEATH THE TOWER RELATIVE TO eARTH

Second one is definitely wrong, the ball WILL stay in orbit because the top of the tower has the same altitude and velocity as a geostationary orbit. Therefore, when the ball is dropped, it will maintain the orbit as it already has the component of altitude and orbital velocity required to maintain the stable orbit.

Go question his answer.

I don't understand why people are still saying things like parabolic paths when it will clearly maintain its orbit.

 

[Stealth_Mel0n]

I really don't think it has anything to do with frame of reference. The mass at the north pole has no orbital velocity due to the fact that the north pole doesn't spin like the rest of the earth. The one on the equator is already in geostationary orbit when you think about it, so when it's released nothing will happen, it will continue orbiting over the same point in earth with an orbital period of 24 hours. I think you guys all overthought it....

 

[manscux]

Second one is definitely wrong, the ball WILL stay in orbit because the top of the tower has the same altitude and velocity as a geostationary orbit. Therefore, when the ball is dropped, it will maintain the orbit as it already has the component of altitude and orbital velocity required to maintain the stable orbit.

Go question his answer.

I don't understand why people are still saying things like parabolic paths when it will clearly maintain its orbit.

Look at what NEWTON SAID in his thought experiment with a cannon ball...

If you have a high enough mountain and give a cannon ball a high enough velocity THAN IT's curvature will match the curvature of the earth... YOU MUST APPLY THAT INITIAL VELOCITY WHICH CANNOT BE PROVIDED SIMPLY BE EARTHS MOTION...... AN EXTRA INITIAL VELOCITY MUST BE PROVIDED

And secondly... every body is entitled to what they wrote.... if people wrote in the exam that the mass will travel in parabolic motion than thats what they are entitled to write on this thread.

 

[habitres]

The idea of geostationary orbit is that at that distance the period will be 24 hours.
The Tower B is in geostationary orbit.
Thus its period at the top is 24 hours
Thus if the tower disappeared and the ball remained (in other terms 'dropping' the ball), the ball will still have a period of 24 hours, at it was moving WITH the top of the tower before release.

It will remain in geostationary orbit.
There is no further reason to argue this topic.

 

[someth1ng]

Look at what NEWTON SAID in his thought experiment with a cannon ball...

If you have a high enough mountain and give a cannon ball a high enough velocity THAN IT's curvature will match the curvature of the earth... YOU MUST APPLY THAT INITIAL VELOCITY WHICH CANNOT BE PROVIDED SIMPLY BE EARTHS MOTION...... AN EXTRA INITIAL VELOCITY MUST BE PROVIDED

And secondly... every body is entitled to what they wrote.... if people wrote in the exam that the mass will travel in parabolic motion than thats what they are entitled to write on this thread.

The fact that it is on the tower means that it has the velocity to orbit the Earth at geostationary altitude - the tower had angular and hence tangential velocity and thus, the ball also had this velocity. Furthermore, newton's first law states that an object will continue in its state of motion unless enacted upon by a net force. .'. it will maintain the velocity.

 

[manscux]

The fact that it is on the tower means that it has the velocity to orbit the Earth at geostationary altitude - the tower had angular and hence tangential velocity and thus, the ball also had this velocity. Furthermore, newton's first law states that an object will continue in its state of motion unless enacted upon by a net force. .'. it will maintain the velocity.

Net force is gravity

But look at his though experiment... dO you agree or not that he talked about launching a cannon ball from a very high mountain (this can be assumed as the tower). If we give it a little velocity than by his though experiment the cannon ball will fall in a parabolic trajectory.... it is only when it is given enough energy that its curvature will match curvature of Earth..... this is exactly what our tower situation is

surely that is more than enough evidence that the mass will fall in a parabolic trajectory..... even Newtons supports this idea

 

[someth1ng]

Net force is gravity

But look at his though experiment... dO you agree or not that he talked about launching a cannon ball from a very high mountain (this can be assumed as the tower). If we give it a little velocity than by his though experiment the cannon ball will fall in a parabolic trajectory.... it is only when it is given enough energy that its curvature will match curvature of Earth..... this is exactly what our tower situation is

surely that is more than enough evidence that the mass will fall in a parabolic trajectory..... even Newtons supports this idea

Sure the net force is gravity but with the tangential velocity, it can be seen as centripetal acceleration and no, this situation is totally different. Have a look here.

"If the speed is the orbital speed at that altitude it will go on circling around the Earth along a fixed circular orbit just like the moon."

Clearly, you have the initial orbital speed due to the tower and hence, the ball will stay in orbit.

 

[manscux]

^ Cant really argue with that.... Luckily i did write that the the mass will go around in orbital motion before falling back to earth after a period of time

But thank you for clarification

 

[arrogant]

The mass at the equator would remain in orbit, the one on the pole would fall. That simple

 

[Zeroes]

talked to teacher who has been head of Marking for quite a bit:

This is what he had to said you needed to get full marks

earth orbits its axis at roughly 1700km/h at an easterly direction.
for mass A since it as the pole, the top of the axis it will gain the horizontal component of earths velocity. But it falls to the ground below the tower because it has a constant downward (vertical) accerlationa cting on it.
For mass B the mass has a horizontal velocity same to that of the earth so RELATIVE to the sun an external observer will see it go in a parabolic motion around the earth before landing on the spot under the tower. It to has a constant vertical motion

IT DOES NOT GO IN GEOSTATIONARY ORBIT BECAUSE MY TEACHER GAVE ME A SIMPLE EXAMPLE AND THAN LAUGHED.....

PLEASE NOTE: A GUY WHO JUMPED OF QUITE HIGH RECENTLY IN A DAREDEVIL STUNT DID NOT GO INTO EARTH ORBIT.... HE WENT STARIGHT DOWN.... I THINK THAT EVIDENCE IS ENOUGH THAT RELATIVE TO THE SUN THE MASS WILL GO A PARABOLIC DECAY WHERE IT FALLS TO THE GROUND BENEATH THE TOWER RELATIVE TO eARTH

fuck yes this is almost exactly what I wrote

THERE IS HOPE

but I'm still having serious doubts

 

[Demise]

Let's move on, it's over lol. Chances are they will accept numerous answers if we have accurate justification, since there are many things that could have been written and could have occurred, such as discussing that being at the pole/equator has different gravitational acceleration, that being at geostationary height that the object would go into orbit, that the objects would drop differently in regards to relative velocity, etc etc.

 

[someth1ng]

Let's move on, it's over lol. Chances are they will accept numerous answers if we have accurate justification, since there are many things that could have been written and could have occurred, such as discussing that being at the pole/equator has different gravitational acceleration, that being at geostationary height that the object would go into orbit, that the objects would drop differently in regards to relative velocity, etc etc.

That one is definitely wrong because it depends on the distance from the centre of mass. Therefore, g at equator and poles are equal.

 

[Demise]

That one is definitely wrong because it depends on the distance from the centre of mass. Therefore, g at equator and poles are equal.

No, at the equator it bulges out more therefore it's further from the centre when compared to the poles, which are closer to the centre.

 

[someth1ng]

No, at the equator it bulges out more therefore it's further from the centre when compared to the poles, which are closer to the centre.

That's only matters if you are on the surface of Earth. The distance of a geosynchronous from the centre of mass of Earth is always the same and hence, g will be equal.

 

[Demise]

That's only matters if you are on the surface of Earth. The distance of a geosynchronous from the centre of mass of Earth is always the same and hence, g will be equal.

Yeah I know, I was just talking in general. I mentioned that there would be the same force on both objects as the altitude is identical. I said others may see it like that.

 

[Demise]

I should've specified.

 

[jeffwu95]

if you assume uniform density of the earth and say that as there is more 'mass' at the equator then at the poles as the earth bulges then the value of g would be larger at the equator

 

[someth1ng]

if you assume uniform density of the earth and say that as there is more 'mass' at the equator then at the poles as the earth bulges then the value of g would be larger at the equator

It really only matters if you're standing on the surface of Earth. Otherwise, the distance from the Earth's centre of mass is the only important thing.