B How long it takes the Earth to fall halfway to the Sun--ellipse method

[bruhh]

TL;DR

How long does it take for the Earth to fall half the distance to the Sun, without calculus?

There's a classic physics problem that is:

If Earth is orbiting the Sun at 1 au from and is suddenly stopped. How long does it take to fall into the Sun (neglecting the size of the Sun/Earth)?

I know that a clever way to solve this problem is by using degenerate ellipses and an object "falling" into the Sun is the equivalent of the object orbiting the Sun with very high eccentricity. Halving the semi-major axis of the orbit and using Kepler's Law gives: T=1AU/(4√2). However, I want to apply to strategy to another similar problem. This time:

If Earth is orbiting the Sun at 1 au from and is suddenly stopped. How long does it take for the Earth to travel half the distance to the Sun(neglecting the size of the Sun/Earth)?

If we were to use the result for T from the first problem and then simply apply Kepler's 2nd Law by looking at ratios of areas, we should be able to find this new time it takes to fall halfway to the Sun. However, I am having a problem visualizing what area of the ellipse the Earth covers when it travels half the distance to the Sun. I know about the calculus solution which is really bashy, so I would like some assistance with figuring out this more elegant method.

 

[Janus]

Towards the end of this article is an equation which allows you to calculate this.
https://en.wikipedia.org/wiki/Equations_for_a_falling_body

 

[Andrew Mason]

Use Kepler's third law to find the period of an elliptical orbit where the semi-major axis is the half the radius of the Earth's current orbit.

When stopped, the Earth is at the farthest point from the sun. Since the Earth would have no tangential speed, the orbit is the extreme elliptical orbit where the semi-minor axis is 0. In other words, and eccentricity of the ellipse is 1, putting the focii at the extrema of the ellipse. So the other extrema (also the other focus) is the centre of the sun (for practical purposes).

AM

 

[Andrew Mason]

As you will no doubt have realized, it is much easier to determine the time to fall to the sun rather than the time to fall halfway to the sun. But you can ball-park it by taking an ellipse with an eccentricity close to 1, which represents the orbit of the Earth with very little angular momentum relative to the centre of the sun. The area swept out by the Earth in the time it takes to reach the half way point is most of the area of one half of the ellipse. The remaining area of that half ellipse is a tiny fraction of the area swept out in the first half of the fall. Since the orbiting body sweeps out equal areas in equal times, you can see that it takes a small fraction of that time to fall the rest of the way. This means that the time it takes to fall half way is very close to the time it takes to fall the entire way.

You can verify this by figuring out the speed of the Earth at the half-way point using change in potential energy = - change in kinetic energy.

AM