We see the Sun about eight minutes after light left it. Presumably this means we are also experiencing the Sun's gravity 8 minutes after it 'left' the Sun.
So are we orbiting around the Sun we can see or around the Sun's actual position?
$\begingroup$
$\endgroup$
3
asked Dec 4, 2018 at 13:25
-
$\begingroup$ Please specify your question clearly $\endgroup$– reddotCommented Dec 4, 2018 at 13:29
-
$\begingroup$ Related: physics.stackexchange.com/q/62565/2451 , physics.stackexchange.com/q/5456/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Dec 4, 2018 at 13:34
-
$\begingroup$ @Sourabh - I'm trying to think of a way to formulate it differently. $\endgroup$– chasly - supports MonicaCommented Dec 4, 2018 at 13:49
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
We see the Sun about eight minutes after light left it. Presumably this means we are also experiencing the Sun's gravity 8 minutes after it 'left' the Sun.
No, not in the way you mean anyway.
I'm sure you've seen those diagrams where they illustrate gravity by placing a large mass like a shotput in the middle of a rubber sheet? Then you simulate Earth by rolling a golf ball or ball bearing across the surface?
It's important to understand what this is saying - the golf ball has no idea the shotput is there. All it knows is that the surface its running on is curved, and it's rolling along the path of least resistance on that surface.
Likewise, the Earth is not responding to the Sun, and for the most part is unaware it exists. It is simply travelling through 4D space along the path of least resistance. The Sun is not pulling on the Earth, it is pulling on space itself (spacetime) and creating the curvature.
The end result looks the same ultimately, but has a very different conceptual basis. Newton could never explain how gravity worked because he did see it as two objects pulling on each other, and there was no explanation of how one knew about the other. Field theories solve this problem - they're not pulling on each other, they're simply reacting to the shape of the field.
Now changes to that curvature do indeed travel at the speed of light, but that's not something that happens in this example - the Sun isn't jumping around a whole lot. So ultimately the field 8 minutes ago and right now are the same, or close enough for horseshoes anyway.
answered Dec 4, 2018 at 14:36
-
$\begingroup$ The Sun is however moving relative to the rest of the Milky Way and so Earth's orbit around it is not an ellipse. Presumably it's a spiral of some kind. As we are trailing around following the Sun, presumably we follow it to where it was 8 minutes ago. With the rubber sheet I imagine that 'light' gets to the golf ball very quickly but if the shot-put is moving across the sheet too quickly then the golf ball might not know soon enough and be left behind. $\endgroup$ Commented Dec 4, 2018 at 14:43
-
1$\begingroup$ Inertial movement is baked into the field. It's only non-inertial motion that causes changes that need to propagate. Think about how magnetism works, same basic underlying principle. $\endgroup$ Commented Dec 4, 2018 at 14:45
-
$\begingroup$ As long as the field is only a theorized construct it can be used to answer all sorts of questions. When you said “baked into the field” what’s the evidence for that other than it makes a theory work. The OP’s comment about multiple gravitational sources involved is not so easily dismissed. $\endgroup$– LambdaCommented Dec 4, 2018 at 17:16
-
$\begingroup$ That a field would have a different configuration due to movement for purely geometric reasons seems pretty obvious to me. But if you have a better answer, by all means,spawn one off. $\endgroup$ Commented Dec 4, 2018 at 19:24