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We all know Sun gravity, makes the space time curvature and space pushes the earth towards the Sun. But my question is "why the distance between the Sun and Earth is Same? Earth should have moved closer to Sun and obviously Earth should fall into the Sun."

Thanks.

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    $\begingroup$ did you mean to ask why the Earth-Sun distance remains unchanged? $\endgroup$
    – Walter
    Commented Dec 2, 2014 at 17:25
  • $\begingroup$ In space there is no friction. $\endgroup$
    – Wayfaring Stranger
    Commented Dec 2, 2014 at 17:35
  • $\begingroup$ Hi Wayfaring, In that case, A freely falling object should not reach the Earth Since there is no friction anywhere in space. $\endgroup$
    – Selva
    Commented Dec 2, 2014 at 18:31
  • $\begingroup$ Hi Walter, I hope you got my big picture of my question. Did you? $\endgroup$
    – Selva
    Commented Dec 2, 2014 at 18:32
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    $\begingroup$ "... and obviously Earth should fall into the Sun ..." -- What exactly is your basis for that? It's not at all obvious to me. I also have no idea why you think that the lack of friction implies that a freely falling object should not reach the Earth. A freely falling object may or may not reach the Earth, depending on its velocity (remember that velocity includes direction). $\endgroup$
    – Keith Thompson
    Commented Dec 2, 2014 at 19:45

2 Answers 2

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While the Sun and Earth attract each other, they cannot fall into each other because of angular momentum conservation. In a central field (where the force is acts in the direction of the distance vector and depends on distance only), the specific angular momentum vector $\boldsymbol{L}=\boldsymbol{r}\times\boldsymbol{v}$ is conserved ($\boldsymbol{r}$ is position and $\boldsymbol{v}$ the velocity). In particular $L=|\boldsymbol{L}|=rv_t$ (with $v_t$ the tangential velocity: the component of velocity perpendicular to the direction Earth-Sun). The specific orbital energy $E=\tfrac{1}{2}|\boldsymbol{v}|^2-GM/r$ is also conserved and must be negative for a bound orbit (such as Earth's). Combining these two we have $$ E = \tfrac{1}{2}v_r^2 + \frac{L^2}{2r^2} - \frac{GM}{r} $$ with $v_r$ the radial component of velocity. Since $v_r^2\ge0$, but $E<0$, not all radii are reachable. In particular there are two radii at which $v_r=0$: the apo- and peri-apse of the orbit.

In fact, for the Earth, the peri- and apo- apse are quite similar, and the orbit is nearly circular. The gravitational pull of the Sun is very nearly balanced by the centrifugal force due to the rotating orbit.

In GR, the picture doesn't really change much. Non-Newtonian (GR) effects are tiny in the Solar system and completely unimportant for Earth's orbit.

Note that the picture of pushing by ripples in space time (vs. pull from Sun's gravity) doesn't make a difference either. The point is that just because you're pushed some way doesn't imply that you're falling that way. All bodies also have something called inertia which makes them want to follow their path rather being pushed around. In case of a circular orbit (similar to Earth's around the Sun), gravity and inertia balance in such a way as to create a circular orbit: inertia want's to go straight, gravity want's to pull/push you into the Sun, but the result is something between those.

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  • $\begingroup$ Hi Walter, What you have answered is from Classical Mechanics perspective. But i really need a answer from General Relativity. Sorry if am an ignorant in this topic:( $\endgroup$
    – Selva
    Commented Dec 2, 2014 at 17:50
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    $\begingroup$ Walter's answer is the simplest (and most likely best) one you're going to get. There's generally no need to invoke GR for simple orbits unless you're talking about precession. One thing, though, Walter - did you mean 'centripetal' instead of 'centrifugal', or am I misinterpreting the usage? $\endgroup$
    – HDE 226868 ♦
    Commented Dec 2, 2014 at 21:17
  • $\begingroup$ I agree HDE. I took SUN and Earth as an example for this problem. I guess we can apply logic to all bodies in universe. am i correct? $\endgroup$
    – Selva
    Commented Dec 3, 2014 at 7:04
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    $\begingroup$ @HDE226868 Walter's usage is correct. The gravitational force would be the centripetal one here, and it is balanced by the centrifugal in an appropriate rotating frame, as Walter says. $\endgroup$
    – Stan Liou
    Commented Dec 3, 2014 at 8:47
  • $\begingroup$ Perhaps mentioning the spherically-symmetric GTR case would be useful for illustrative purposes, since it just adds a $-GML^2/(c^2m^2r^3)$ term to $E$, so morally speaking the idea is exactly the same even there (even if the terms have slightly different meaning). $\endgroup$
    – Stan Liou
    Commented Dec 3, 2014 at 9:00
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Technically, the distance between Sun and Earth changes throughout the year. The Earth moves around the Sun on an elliptical orbit. Its distance from the Sun changes between 152 million km (aphelion) and 147 million km (perihelion), a difference of 5 million kilometres, twice per year.

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