Let’s say the universe was empty and suddenly an astronaut and the sun appeared 2 light years apart. Using the reference frame of the Astronaut, would he be pulled towards the sun as soon as he can see it?

Q1. And from that moment, would his acceleration be constant until he smashed into the sun? What would that acceleration be and what would his top speed be?

I’m assuming here that gravity propagates at the speed of light to infinity (which is in line with mainstream theories as far as I’m aware). One of the things I’m interested in hearing thoughts on, is whether the gravity pull by the sun is just as strong 2 light years away, since there are no other gravitational forces at play in this scenario.

Also, let’s for argument’s sake say we consider this scenario from the reference frame of an observer located between the two objects - or slightly to one side so he doesn’t get hit by the astronaut ;-).

Q2. Would he observe the astronaut unaffected by the sun’s gravity for 1 light year before the gravitational waves reaches the astronaut? And what would then happen to the astronaut from the observer’s frame of reference, in terms of acceleration, speed, etc?

Here are my assumptions about the observer:

• The observer is not impacted by gravity • The observer is not using ‘eyes’ to observe, but rather a clever apparatus that can detect any object in the universe, including its speed and location (relative to the observer and also relative to each other). This deals with the problem of not being able to see the astronaut at the same time as the sun due to lack of light emission. • The method of observation is still subject to light speed, i.e. the apparatus will only detect an object once the particles or waves (travelling at the speed of light) emitted from that object has reached the apparatus

I realise time itself, as we know it (years, miles per second, etc), may be completely meaningless in this scenario – but try to entertain me. It is a hypothetical question after all (which perhaps means that there is no meaningful answer, turning the discussion into philosophy instead..)

  • $\begingroup$ Since the force of gravity varies as $1/r^{2}$, the gravity "pull" 2 light years away is not the same as 1 light year away, much less 1 mile away. $\endgroup$
    – Jon Custer
    Commented Feb 7, 2017 at 19:19
  • $\begingroup$ Post (v1) closed as a non-mainstream scenario: E.g a star cannot suddenly appear. $\endgroup$
    – Qmechanic
    Commented Feb 7, 2017 at 21:44
  • 2
    $\begingroup$ I believe these kind of "gedanken" experiments could help people understand physics better. Closing it as non-mainstream physics or personal theory seems extreme to me. The inconsistencies introduced by assuming the sudden unphysical appearance of an object are irrelevant to the argument. $\endgroup$
    – user126422
    Commented Feb 7, 2017 at 22:55
  • 1
    $\begingroup$ This isn't a question about challenging the correctness of accepted theories, but trying to understand the established ones. $\endgroup$ Commented Feb 8, 2017 at 0:26
  • $\begingroup$ Note: his impact speed will be around the escape speed from the Sun, which is around 500 km/s. I would be happy to explain it reason in an answer, if this question wouldn't be closed. Also I've heard there are (unaccepted, post-GR, but so or so mainstream) theories in which such a Universe wouldn't have gravity (but the focus of my answer would be the mainstream answer). $\endgroup$
    – peterh
    Commented Feb 8, 2017 at 1:27

1 Answer 1


1) Yes. Gravity propagates at the speed of light.

2) Acceleration would be continuous, but accelerate as the distance between the sun and astronaut diminishes. Acceleration and maximum speed depend on the mass of the sun.

3) As the previous comment states, gravitational pull is not the same at all distances. The closer you are, the more influence it has on you.

Tried to answer all of your questions simply. Let me know if I missed one; they were somewhat hard to spot.

Newton's Law of Gravitation would be a good read for you, but in general it states that the force of gravitational attraction is directly proportional to the product of the masses being considered (sun * astronaut (including whatever the astronaut is wearing). Call it G=SA

But it is also inversely proportional to the square of the distance between them.

Since we are dealing with two objects, we can simplify the calculation a bit and we get something like:

F(r) = mg(r)

F is the force applied on the astronaut due to the sun.

G is the gravitational constant: 6.67408(31)×10−11 m3⋅kg−1⋅s−2

R is the distance between the two objects

M is the astronaut's mass

Since you seem interested in velocity (measured in m/s2): V(r) = -G*m1/r

M1 is the sun's mass and for simplicity's sake we are assuming its mass is evenly distributed.

  • $\begingroup$ I think you should definitely include Newton's law of gravity, as to me, the OP does need the effect of $r $ explained explicitly. $\endgroup$
    – user140606
    Commented Feb 7, 2017 at 21:07
  • $\begingroup$ Oh, he meant Sol! I assumed he'd just used "sun" to mean any star. $\endgroup$ Commented Feb 8, 2017 at 1:27
  • $\begingroup$ @JakeWatrous, where did you get that value for velocity? $\endgroup$ Commented Feb 8, 2017 at 5:54
  • $\begingroup$ The 500km/s? I didn't. Peterh did. It's relatively easy to calculate, though. He assumed you meant our own sun, and plugged its mass into this equation: g = G*M/R^2, where g is the acceleration due to gravity, G is the universal gravitational constant, M is mass, and R is distance. $\endgroup$ Commented Feb 8, 2017 at 5:57
  • $\begingroup$ Thank you all for your comments and answers, they were very useful. I don't quite understand why this question was closed, I wasn't trying to propose outlandish theories or veer from mainstream physics, but rather try to understand established theories by using a thought experiment that my non-scientific brain might be able to understand better. Just like Albert Aspect pointed out, this could be useful to make physics more popular! As this was my first post however, I'll try to stick to the rules next time ;) Thank you @JakeWatrous JonCuster AlbertAspect peterh KyleKanos Countt010 $\endgroup$ Commented Feb 8, 2017 at 14:28

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