<p>The Earth's orbital speed around the Sun is about 30 km/s.  So we can calculate the intersection of two spheres offset by 30 km.  Suppose we put the first sphere at the origin [0,0,0], and the second sphere at [30,0,0].  That is, the x-axis is aligned with the orbital velocity vector.</p>
<p>Then, from <a href="https://mathworld.wolfram.com/Sphere-SphereIntersection.html" rel="nofollow noreferrer">Wolfram</a>, the intersection of the spheres is a curve lying parallel to the z-y plane at a single x-coordinate (<span class="math-container">$x=d/2$</span>), described by <span class="math-container">$$y^2+z^2=r^2-d^2/4$$</span> where <span class="math-container">$d$</span> is the distance between the spheres, and <span class="math-container">$r$</span> is the radius of the spheres.  So, you can find specific points by varying <span class="math-container">$\theta$</span> between <span class="math-container">$0$</span> and <span class="math-container">$2\pi$</span> with:
<span class="math-container">$$x=d/2$$</span>
<span class="math-container">$$y=(r^2-d^2/4)\cos(\theta)$$</span>
<span class="math-container">$$z=(r^2-d^2/4)\sin(\theta)$$</span></p>
<p>In words instead of equations: only a subset of points on the Earth's surface will be in the same place in a Sun-centered coordinate system one second later.  These points will be on a plane perpendicular to the orbital velocity vector, and offset from the center of the Earth in the direction of the orbital velocity vector by 15 km.  One second later, a new set of points will occupy the same space.  The new set of points will also be in a plane perpendicular to the orbital velocity vector, but now offset opposite the direction of the orbital velocity vector by 15 km.  Here is a conceptual picture <strong>Not To Scale</strong>.</p>
<p><a href="https://i.sstatic.net/VD6yr.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/VD6yr.png" alt="enter image description here" /></a></p>
<p>Do you want to account for the revolution of the Earth during that time (a point on the equator moves about .46 km in a second)?  Then apply a rotation prior to calculating the latter set of points using a <a href="https://en.wikipedia.org/wiki/Rotation_matrix#Rotation_matrix_from_axis_and_angle" rel="nofollow noreferrer">rotation matrix from axis and angle</a> where your angle is <span class="math-container">$2\pi/86,400$</span> radians, and your axis is a unit vector specifying the orientation of the Earth's rotational axis from a Sun-centered coordinate system.  At summer solstice, this is about <span class="math-container">$[0,-\sin(23.5),\cos(23.5)]$</span>.</p>
<p><strong>Assumptions:</strong></p>
<ol>
<li>The Earth's orbital path doesn't curve over the order of a second.</li>
<li>The Earth is a perfect sphere.</li>
</ol>
<p><strong>Notes:</strong></p>
<ol>
<li>You could get a higher fidelity model using an orbital simulator like Universe Sandbox.</li>
<li>You could get a higher fidelity model by using <a href="https://en.wikipedia.org/wiki/DTED" rel="nofollow noreferrer">DTED data</a> instead of assuming the Earth is a sphere.  Then you might have to use a numerical engine to find point intersections.</li>
</ol>

CC BY-SA 4.0
2021-10-28T18:11:57.480Z
2021-10-28T18:45:18.440Z
2021-10-28T18:45:18.440Z
Connor Garcia
<p>In the following assumptions:</p>
<ul>
<li>Earth orbit is circular</li>
<li>Earth's surface is a sphere</li>
</ul>
<p>your task is to find the intersection of two spheres, which should be a circle if exists. The first sphere is the Earth now. The second one is the Earth 1 sec later.
The difference between these two spheres comes from the orbital motion and from rotation. Due to the Earth's rotational axis inclination to the orbital plane, you also have to keep in mind the day of the year (a 3D angle between the axis and the velocity changes).</p>
<p>In the next step, you may use any realistic model of the Earth's surface and orbit. Anyway, it will be only about finding intersections of 3D shapes.<br />
This way you can get all the points on Earth that have their counterparts as you stated them.</p>

CC BY-SA 4.0
2021-10-28T17:37:07.970Z
I agree with your assessment. Could you suggest some specific models that I could use to find these intersections? I’m not having much luck finding anything practical.
Erik
CC BY-SA 4.0
2021-10-28T15:55:56.050Z
2021-10-28T15:55:56.050Z
Mikhail Lisakov
<p>I’m hoping to find a method (probably some modeling software?) that would allow me to find pairs of locations on the Earth’s surface that occupy the same point in space relative to the sun, one second apart in time.</p>
<p>It would be best if this method is searchable by choosing a location and seeing at what times it would cross paths with another location one second later (with the sun as the frame of reference).</p>
<p>Any advice is greatly appreciated!</p>

CC BY-SA 4.0
2021-10-28T02:43:59.010Z
How far does a point on the Earth's surface move in 1 second? To a high degree of accuracy, that is how to answer your question.
JohnHoltz
CC BY-SA 4.0
2021-10-28T08:38:56.543Z
With the constraint that the points have to be at the Earth's *surface*, there will be only a few points that have a match at all.  For most places, the corresponding point will be above or below the surface.
gerrit
CC BY-SA 4.0
2021-10-28T13:22:50.843Z
@JohnHoltz Sure, but one must take into account rotation as well as orbit revolution.  or, for highest accuracy "wobble" due to Earth-moon interactions.
Carl Witthoft
CC BY-SA 4.0
2021-10-28T14:56:11.000Z
@CarlWitthoft You're right. When I wrote "high degree of accuracy", I was thinking that the Sun's change in RA and Dec is very small in 1 second of time, so that and other motions would be small compared to the Earth's rotation. Maybe I should have written "for a first approximation...".  I'm glad Erik is solving the problem and not me :-)
JohnHoltz
<the-sun><solar-system><earth><modeling>
2 points on Earth’s surface, 1 second apart in time
Erik
# 2 points on Earth’s surface, 1 second apart in time

(Asked by: username_0 on 2021-10-27)

I’m hoping to find a method (probably some modeling software?) that would allow me to find pairs of locations on the Earth’s surface that occupy the same point in space relative to the sun, one second apart in time.

It would be best if this method is searchable by choosing a location and seeing at what times it would cross paths with another location one second later (with the sun as the frame of reference).

Any advice is greatly appreciated!

**Tags:** the-sun, solar-system, earth, modeling

> **Comments:**
> **username_1**: How far does a point on the Earth's surface move in 1 second? To a high degree of accuracy, that is how to answer your question.  
> **username_2**: With the constraint that the points have to be at the Earth's *surface*, there will be only a few points that have a match at all.  For most places, the corresponding point will be above or below the surface.  
> **username_3**: @username_1 Sure, but one must take into account rotation as well as orbit revolution.  or, for highest accuracy "wobble" due to Earth-moon interactions.  
> **username_1**: @CarlWitthoft You're right. When I wrote "high degree of accuracy", I was thinking that the Sun's change in RA and Dec is very small in 1 second of time, so that and other motions would be small compared to the Earth's rotation. Maybe I should have written "for a first approximation...".  I'm glad username_0 is solving the problem and not me :-)  

---

## 2 answers

### Answer by username_4 on 2021-10-28. Score: 2 | Accepted answer

The Earth's orbital speed around the Sun is about 30 km/s. So we can calculate the intersection of two spheres offset by 30 km. Suppose we put the first sphere at the origin [0,0,0], and the second sphere at [30,0,0]. That is, the x-axis is aligned with the orbital velocity vector.

Then, from [Wolfram](https://mathworld.wolfram.com/Sphere-SphereIntersection.html), the intersection of the spheres is a curve lying parallel to the z-y plane at a single x-coordinate ($x=d/2$), described by $$y^2+z^2=r^2-d^2/4$$ where $d$ is the distance between the spheres, and $r$ is the radius of the spheres. So, you can find specific points by varying $\theta$ between $0$ and $2\pi$ with:
$$x=d/2$$
$$y=(r^2-d^2/4)\cos(\theta)$$
$$z=(r^2-d^2/4)\sin(\theta)$$

In words instead of equations: only a subset of points on the Earth's surface will be in the same place in a Sun-centered coordinate system one second later. These points will be on a plane perpendicular to the orbital velocity vector, and offset from the center of the Earth in the direction of the orbital velocity vector by 15 km. One second later, a new set of points will occupy the same space. The new set of points will also be in a plane perpendicular to the orbital velocity vector, but now offset opposite the direction of the orbital velocity vector by 15 km. Here is a conceptual picture **Not To Scale**.

[![enter image description here](https://i.sstatic.net/VD6yr.png)](https://i.sstatic.net/VD6yr.png)

Do you want to account for the revolution of the Earth during that time (a point on the equator moves about .46 km in a second)? Then apply a rotation prior to calculating the latter set of points using a [rotation matrix from axis and angle](https://en.wikipedia.org/wiki/Rotation_matrix#Rotation_matrix_from_axis_and_angle) where your angle is $2\pi/86,400$ radians, and your axis is a unit vector specifying the orientation of the Earth's rotational axis from a Sun-centered coordinate system. At summer solstice, this is about $[0,-\sin(23.5),\cos(23.5)]$.

**Assumptions:**

1. The Earth's orbital path doesn't curve over the order of a second.
2. The Earth is a perfect sphere.

**Notes:**

1. You could get a higher fidelity model using an orbital simulator like Universe Sandbox.
2. You could get a higher fidelity model by using [DTED data](https://en.wikipedia.org/wiki/DTED) instead of assuming the Earth is a sphere. Then you might have to use a numerical engine to find point intersections.


---

### Answer by username_5 on 2021-10-28. Score: 0

In the following assumptions:

* Earth orbit is circular
* Earth's surface is a sphere

your task is to find the intersection of two spheres, which should be a circle if exists. The first sphere is the Earth now. The second one is the Earth 1 sec later.
The difference between these two spheres comes from the orbital motion and from rotation. Due to the Earth's rotational axis inclination to the orbital plane, you also have to keep in mind the day of the year (a 3D angle between the axis and the velocity changes).

In the next step, you may use any realistic model of the Earth's surface and orbit. Anyway, it will be only about finding intersections of 3D shapes.  
This way you can get all the points on Earth that have their counterparts as you stated them.

> **Comments:**
> **username_0**: I agree with your assessment. Could you suggest some specific models that I could use to find these intersections? I’m not having much luck finding anything practical.  

---

