Recent content by Dewgale

Given a particular semi-circle choice (with the properties I described in my previous post), we can divide the path (call it ##\gamma##) into the straight part (##\gamma_1##) along the real line, and the arc (##\gamma_2##). This would give (replacing ##-2\pi i \xi x## with just ##-i \xi x##, and...

That's true. If you consider it to be constant right up until that point, and then briefly change the force to stop the spring in place, then you'll be in equilibrium.

Well, the ##F## in ##F = -kx## is the force that causes the spring to move, the restoring force. So no, ##F_{app} \neq kx## except for at the exact situation where the spring is extended or compressed, and being held in that position by the applied force -- but there's nothing fundamental about...

Remember that ##x## is a real number in this case. You want a complex number, so consider the analogous case $$\int_\mathbb{C} \frac{1}{1+z^4} e^{-2\pi i \xi z}$$ where the integral follows a semi-circle that goes from ##-R\to R## along the real axis, then goes either up to ##R i## or down to...

Gneill, he was working from the geometric fact that if you draw a line from the two focii to a point on an ellipse, the angle between the tangent at that point and each line is the same, following from the link I'd posted in post #20. One can then easily derive the angle between the position...

Apologies, I wasn't thinking of the triangle correctly. As far as I can tell you're doing things correctly, but remember, you want half of the latus rectum, not the whole latus rectum - which is what that formula is for, so never mind about that. It still doesn't give the right answer, so I'll...

Your angle you derived is off - you want the conjugate angle to what you got. Remember, you want the angle betwen the Latus Rectum and the velocity, which you can see is not the angle in the top left of your right-angle triangle, but rather the bottom right (because it's 90-##\alpha##, where...

I just realized that you aren't given the distance from the sun to P (I thought you had been, my apologies) - so I'm not sure how you found the angle without finding that distance. Did you derive the distance?

Some geometric properties of an ellipse would probably help. Using the info you already have, you should be able to easily find the angle: https://www.mathopenref.com/ellipsetangent.html

Theta is the angle between the velocity and the line from the sun to the planet, not from the center of the ellipse to the planet. So it's not always 90 degrees - one can clearly see at the first point that it is a different value.

Like I said earlier, you need Conservation of Angular Momentum I assume you haven't learned about cross product yet, but you should know that for a point particle, the angular momentum is given by ##L = m v r \sin(\theta)##, where m is the mass, v is the speed, r is the distance from the origin...

##T## is not time in Kepler's Laws -- it's period, i.e. the amount of time to complete a full orbit. It's inversely proportional to the average speed over the course of the orbit, which doesn't help you when analyzing a given portion of the orbit.

How did you find v^2 to be inversely proportional to R^3? Kepler's law probably isn't going to be of much use in this case.

Yep! Though you can probably just approximate the planet as a point. Gneill, is it not the case that, even after finding the tangent line to the ellipse at that point, one would need to use conservation of angular momentum to find the velocity at a later time? As such, I'm not sure saying it...

Remember, it's a torque that changes angular momentum, not a force. Is there any torque being applied in this scenario?