Question about orbits

[mrbc19]

This is my first time posting here, hi to all and thanks in advance for any replies.

 

I took highschool AP physics, and have read a little bit of new stuff like The elegant Universe, so I consider myself to have a basic uderstanding of physics.

 

My question is about earth orbits. I know the equation for a projectile orbiting the earth is a parabolic one (I hope I am correct here, otherwise my post is unecessary.) Now in newtonian mechanics, if you fire lets say a cannonball fast enough, it will not strike the earth but will orbit it in a circle. I am confused as to how the equation for a projectile can be parabolic, yet an earth orbit can be circular, isn't a parabolic equation never going to produce a circular orbit no matter how fast an object is going? If anyone can clear this up for me, that would be great, thanks.

[blakeja]

the equation for any projectile is derived from newtons equations of motion:

where distance in the x direction is given by : x=Voxt and distance in the y direction is given by y=Voyt-(1/2)gt^2. Combineing these 2 equations gives us:y=xtan(theta) - g/(2Vo^2cos^2(theta)) which isn't parabolic. This equation is on the surface of the earth but to change it to work in anywhere you just do a small change to: y=xtan(theta)- GMe/[2(r^2)(Vo^2)(cos^2(theta))] where Me is the mass of the earth, G is the gravitational Constant, r is the distance from the center of the earth and theta is the angle at which the object is fired. This equation isn't parabolic either

[JaKiri]

My question is about earth orbits. I know the equation for a projectile orbiting the earth is a parabolic one (I hope I am correct here, otherwise my post is unecessary.) Now in newtonian mechanics, if you fire lets say a cannonball fast enough, it will not strike the earth but will orbit it in a circle. I am confused as to how the equation for a projectile can be parabolic, yet an earth orbit can be circular, isn't a parabolic equation never going to produce a circular orbit no matter how fast an object is going? If anyone can clear this up for me, that would be great, thanks.

 

It's all to do with assumptions and modelling.

 

If you don't travel very far along the earth's surface, you can assume (without too much error) that the gravitational force is pulling in a constant direction, and at a constant rate. This basically assumes that the earth is flat.

 

If you remember that the earth is curved, then you have to deal with the fact that, not only does the magnitude of the gravitational attraction change, but so to does the direction.

 

http://img13.paintedover.com/uploads/13/physicsho.jpg

 

Obviously the 2nd fight path is... inaccurate, but the rest of it's about right.

[JaKiri]

the equation for any projectile is derived from newtons equations of motion:

where distance in the x direction is given by : x=Voxt and distance in the y direction is given by y=Voyt-(1/2)gt^2. Combineing these 2 equations gives us:y=xtan(theta) - g/(2Vo^2cos^2(theta)) which isn't parabolic. This equation is on the surface of the earth but to change it to work in anywhere you just do a small change to: y=xtan(theta)- GMe/[2(r^2)(Vo^2)(cos^2(theta))] where Me is the mass of the earth' date=' G is the gravitational Constant, r is the distance from the center of the earth and theta is the angle at which the object is fired. This equation isn't parabolic either [/quote']

 

I'm afraid that's incorrect. I don't know how you've got to where you've got mainly because your starting notation is... haphazard, although now I examine it more closely, I think you mean, by Vox, 'Variable of x'. That's something I've not seen before. You've also made some rather silly mistakes, although quite frankly some of that is indecipherable. For the benefit of people who want to know the answer, I'm going to work through the entire problem.

 

Now, lets start from our basic equations of motion, where x and y are horizontal and verticle motion respectively, V is initial speed, (theta) is the angle of attack and g is gravity, assumed to be constant.

 

 

(this is a thumbnail, click on it for enlargement)

 

Start with the x dimension (for anyone who cares, a full explanation of these equations will be given at the bottom)

 

1. d^2x/dt^2 (acceleration in the x dimension wrt time) = 0

 

2. dx/dt (velocity in the x dimension wrt time) = Vcos(theta)

 

3. x (displacement in the x dimension wrt time) = Vtcos(theta)

 

Same for y...

 

4. d^2y/dt^2 = -g

 

5. dy/dt = Vsin(theta) - gt

 

6. y = Vtsin(theta) - g(t^2)/2

 

From 1, it's fairly clear we can cancel down and get

 

t = x/Vcos(theta).

 

Subbing into 6...

 

y = Vsin(theta) * x/Vcos(theta) - (g/2) * (x/Vcos(theta))^2

 

7. y = x tan (theta) - g (x^2)/(V^2)(cos^2(theta))

 

Given that theta (and therefore cos^2 and tan theta) is a constant, V is a constant and g is a constant, this means that we have a parabolic equation.

 

Aside: proving the parabolic nature of that curve

 

Parabolic equations are equations of the form (x-h)^2 = 4p(y-k)

 

By using basic simultaneous equations, we find that, from the given equation above,

 

(x - V^2sin(theta)cos(theta)/2g)^2 = 4(-V^2cos^2(theta)/4g) * (y - V^2sin^2(theta)/g)

 

Which multiplies out to equation 7. The curve is parabolic.

 

Proof of statements 1-6.

 

Newton showed that the acceleration wrt time is the differentiation of the velocity wrt time is the differentiation of the displacement wrt time.

 

Given we have our accelerations (and any constants), we can find out the information we require.

 

Equation 1: Trivial. There is no acceleration in the x direction.

 

Equation 2: Int (0)dt = 0 + c, where c is an arbitrary constant.

 

We know what c is, however, because we can work out the initial velocity.

 

 

(Again, click to enlarge)

 

V is our initial speed, and is (therefore) in the initial direction of travel, which is angled at theta to the horizontal. By basic trigonometry, we can find that the x component of the velocity is equal to Vcos theta, and the y component is equal to Vsin theta.

 

Therefore, our x velocity = 0 + V cos theta = V cos theta.

 

Equation 3: Our x displacement, similarly, is Int (x velocity) dt.

 

Therefore: x = 0 t + V t cos theta + C (which again represents an arbitrary constant)

 

x = V t cos theta + C

 

Since we're starting at (0,0), our initial displacement must be 0, and therefore C is 0.

 

x = V t cos theta.

 

Similarly for y...

 

Equation 4: The acceleration in the y direction = - g

 

'g' becaues it's accelerating (in this model) only due to gravity, and '-g' because gravity pulls down; our positive direction is up, and therefore our acceleration is negative g.

 

Equation 5: The velocity in the y direction = Int accel (y) dt

 

= - g t + c

 

We have already found the initial value for the velocity in y (V sin theta) and can sub in.

 

Velocity (y) = V sin theta - g t

 

Equation 6: Displacement (y) = Int (velocity (y)) dt

 

= V t sin theta - g t^2 / 2 + C

 

Again, we're starting at (0,0), so C is 0.

 

Displacement (y) = V t sin theta - g t^2 / 2.

 

Definitions used: wrt means 'with respect to'.

'Theta' is the english spelling of a greek letter which is commonly used to denote an angle. It looks like this.

The d^2x/dt^2 (and similar) is standard differential notation, and is down to the aforementioned integral and differential link between acceleration, velocity and displacement.

g is the acceleration due to gravity, which, for the purposes of small scale interactions, is assumed to be a constant. g has a numerical value of about 9.81N/kg.

[swansont]

I'm afraid that's incorrect. I don't know how you've got to where you've got mainly because your starting notation is... haphazard, although now I examine it more closely, I think you mean, by Vox, 'Variable of x'. That's something I've not seen before.

 

I assume Vox means initial velocity, x component, and the poster is unfamiliar with using either the "math" markups or the "sub" tags.

 

v_0x

[swansont]

One problem with this situation is that, ignoring air resistance, launching from a cannon into an elliptical orbit has the tendancy to make the projectile return to its place of launch, i.e. the orbit passes through the earth. Air resistance may modify this, but I wouldn't be surprised if you needed a course correction at some point to "circularize" the orbit.

[JaKiri]

The sheer number of possible interactions make the perfection of the circular orbit or the parabolic curve unlikely, although not impossible. Whether your model predicts that or not is dependent on how much you factor in, of course.

 

By and large we end up with elliptical or hyperbolic orbits.

[JaKiri]

I assume Vox means initial velocity' date=' x component, and the poster is unfamiliar with using either the "math" markups or the "sub" tags.

 

v[sub']0x[/sub]

 

Ah, yes, that's quite possible. It doesn't explain what's going on here: g/(2Vo^2cos^2(theta))

 

though.

[mrbc19]

Thanks for the replies, but I am still a little confused after that detailed response from jakiri. So are the eqautions for earth orbits parabolic or not? I think I understand that they are parabolic, so does that mean there are no purely circular orbits around the earth?

[Janus]

Thanks for the replies, but I am still a little confused after that detailed response from jakiri. So are the eqautions for earth orbits parabolic or not? I think I understand that they are parabolic, so does that mean there are no purely circular orbits around the earth?

 

All closed Earth orbits are elliptical (circle is just an ellipse with an eccentricity of 0).

 

If you fire from a cannon a projectile that strikes the Earth. it also follows the curve of an ellipse, it just can't complete the ellipse because the Earth gets in the way. Over short distances, this curve follows closely that of a parabola, so we can get away with assuming that it is a parabola and still get an accurate enought answer.

 

 

You can have parabolic orbits, but they would not be closed. Any object traveling at exactly escape velocity would follow such a path. if you fired a projectile at escape velocity, it would continue to fly away from the Earth following a parabola. (assuming that you didn't fire the cannon with a downward tilt.)

 

If the object is fired at greater than escape velocity, it will follow a hyperbola.

[JaKiri]

Thanks for the replies, but I am still a little confused after that detailed response from jakiri. So are the eqautions for earth orbits parabolic or not? I think I understand that they are parabolic, so does that mean there are no purely circular orbits around the earth?

 

That was dealing with Pure Newtonian stuff on the ground, as it were, and not taking anything else into account. A reply to a post in this thread, rather than the original post.

 

Projectiles fired from the earth which come straight back again (eg, throwing a rock, firing a bullet, whatever) can be estimated to follow a parabola, at a reasonable level of accuracy.

 

Earth orbits, as has been said, are ellipses; it is possible to have a circular orbit, but it's extremely unlikely.