Orbiting concept (1 Viewer)

[Dumbledore]

what is the range of angle for launching a projectile into orbit?

if you say angle is irrelevent: read below; else don't bother reading it

i posted an earlier question similar where i wanted to find out the direction of escape velocity and was told direction is irrelevent, so i assumed since if you fire a projectile slightly below escape velocity, the direction concepts should be similar. but if a projectile was fired directly up, the gravitational force would be vertically down and the velocity vertically up, so there would be no way to introduce a horizontal component so the projectile could not go into orbit.

also: what should i do if i have more questions.. i feel bad for having 3 threads on the first page of this forum

 

[helper]

0 to 89.9999999999999999999999999999999999999999999999999999999999999

It could be slightly below zero as well.

 

[Dumbledore]

ok change of question: what is the maximum angle of projection of a projectile so that it will clear the side of the earth on its way down, this gets annoying as the magnitude of the gravity force change as you get further from the earth and the direction of the force is always to the centre of the earth, not just down.

this is just for interest; i think it may be past hsc lvl physics

 

[Miss Neutrino]

ok change of question: what is the maximum angle of projection of a projectile so that it will clear the side of the earth on its way down, this gets annoying as the magnitude of the gravity force change as you get further from the earth and the direction of the force is always to the centre of the earth, not just down.

this is just for interest; i think it may be past hsc lvl physics

Hmm, it sounds a bit too complicated for an HSC question, unless they wanted you to just assume that g remained constant throughout?

Right now I'm just thinking you could work out the angle assuming g is constant and the range is Earth's radius, but it's 9pm and my brain isn't exactly functioning at its optimum level so forgive me if that doesn't make sense. I'm guessing if you did the above it would come out as some sort of an approximation, gross inaccuracies aside.

If you don't assume g is constant, it would be a pretty complicated calculation. I don't know though, I might think this through again tomorrow morning.

 

[k02033]

Goin to do this in polar coordinates

 

[k02033]

"for the transversal component..we use " should be"for the radial component.."

 

[Dumbledore]

whoa crazy stuff, just makes me want to do a physics degree more even though i only understand half of it.

just a few things i didn't get:
what does r bar and theta bar represent?
and how did u get the first 2 lines of working
a = ....
and F-> = (Gmm)/r^2 * r(hat?)
cause r(hat) or whatever u call it is the magnitude of the polar vector right? and i assume its supposed to create the relationship for the direction of the gravitational force(correct me if the assumption is wrong) since .../r^2 caters for the magnitude already.
so... how does the r(hat?) in the force equation work?

are u doing a physics degree btw?

 

[k02033]

 

[k02033]

and yes, doing a physics degree atm.

 

[Uncle]

and yes, doing a physics degree atm.

would you like to tell me how to calculate the gradient, divergence and curl of the gravitational field while you are at it please even though gravity is a conservative field?

 

[k02033]

too many pics here, goin to ans it in a new thread

http://community.boredofstudies.org/18/physics/212525/gravitational-field.html

 

[needFOOD]

but if a projectile was fired directly up, the gravitational force would be vertically down and the velocity vertically up, so there would be no way to introduce a horizontal component so the projectile could not go into orbit.

umm, i know your extra HSC was answered, but, you gotta remember that the earth is spinning, which in itself is a horizontal component. and also after reaching a certain velocity, the satellite or what ever is usually ejected from the rocket at various angles depending on the desired orbital 'shape' (for lack of a better word, sorry). so they really just need to give it the needed speed to orbit, then give it a push in the right direction... sorry if this is kinda a stupid answer >.<

 

[k02033]

umm, i know your extra HSC was answered, but, you gotta remember that the earth is spinning, which in itself is a horizontal component.

that horizontal velocity is the velocity of the object relative to the sun. what controls orbits around the earth, is the object's velocity relative to the earth. ie for object to achieve obit at Ro, the earth must "see" the object with tangential velocity Vo=sqrt(GM/Ro)

so if someone were to fire an object straight up from earth, a person standing on the sun will see that horizontal component (the trajectory of the object that the person on the sun sees is a parabola) but relative to earth it just falls back down. another words that horizontal velocity doesnt translate to any relative horizontal velocity (since object is fired from earth) which is needed for orbit.