<p>Technically, the distance between Sun and Earth changes throughout the year. The <a href="https://en.wikipedia.org/wiki/Earth%27s_orbit" rel="nofollow noreferrer">Earth moves around the Sun</a> on an elliptical orbit. Its distance from the Sun <a href="https://en.wikipedia.org/wiki/Apsis" rel="nofollow noreferrer">changes</a> between 152 million km (aphelion) and 147 million km (perihelion), a difference of 5 million kilometres, twice per year.</p>

CC BY-SA 4.0
2014-12-04T13:49:28.067Z
2021-12-15T19:50:24.277Z
2021-12-15T19:50:24.277Z
iantresman
<p>While the Sun and Earth attract each other, they cannot fall into each other because of <strong>angular momentum conservation</strong>. 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 $E&lt;0$, not all radii are reachable. In particular there are two radii at which $v_r=0$: the apo- and peri-apse of the orbit.</p>

<p>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.</p>

<hr>

<p>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.</p>

<hr>

<p>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. </p>

CC BY-SA 3.0
2014-12-02T17:50:39.443Z
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:(
Selva
CC BY-SA 3.0
2014-12-02T21:17:12.963Z
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?
HDE 226868
CC BY-SA 3.0
2014-12-03T07:04:02.587Z
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?
Selva
CC BY-SA 3.0
2014-12-03T08:47:58.027Z
@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.
Stan Liou
CC BY-SA 3.0
2014-12-03T09:00:39.090Z
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).
Stan Liou
CC BY-SA 3.0
2014-12-03T09:09:37.277Z
@StanLiou Do you have a www reference for that?
Walter
CC BY-SA 3.0
2014-12-03T09:37:55.153Z
@Walter I derive [here](http://astronomy.stackexchange.com/a/7904/661) if you want a www reference specifically. Otherwise, it's in MTW §25.5, Hartle §9.3, Carroll §5.4, and other textbooks, up to some notational differences (e.g., $V_\text{eff}^{\scriptsize\text{MTW}} = \sqrt{2V_\text{eff}^{\scriptsize\text{Hartle}}+1}$). By the way, I didn't notice that you used $L$ as specific angular momentum when I made my previous comment, so the $m^2$ term would be unnecessary.
Stan Liou
CC BY-SA 3.0
2014-12-03T11:29:44.030Z
Thanks to everyone! You people really helped me in clearing my doubt! Regards,Selva
Selva
CC BY-SA 3.0
2014-12-02T17:34:43.867Z
2014-12-03T09:02:51.550Z
2014-12-03T09:02:51.550Z
Walter
<p>We all know Sun gravity, makes the space time curvature and space pushes the earth towards the Sun. But my question is &quot;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.&quot;</p>
<p>Thanks.</p>

CC BY-SA 3.0
2014-12-02T17:25:12.343Z
did you mean to ask why the Earth-Sun distance remains unchanged?
Walter
CC BY-SA 3.0
2014-12-02T17:35:04.773Z
In space there is no friction.
Wayfaring Stranger
CC BY-SA 3.0
2014-12-02T18:31:10.417Z
Hi Wayfaring, In that case, A freely falling object should not reach the Earth Since there is no friction anywhere in space.
Selva
CC BY-SA 3.0
2014-12-02T18:32:59.693Z
Hi Walter, I hope you got my big picture of my question. Did you?
Selva
CC BY-SA 3.0
2014-12-02T19:45:44.667Z
"... *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).
Keith Thompson
CC BY-SA 3.0
2014-12-03T06:59:43.593Z
Ok Keith . Let me rephrase my question. According to General Theory of Relativity, A Bigger Mass objects creates a ripple of Space-Time around it and that ripple pushes any  object in the Space-Time towards it . Due to this effect, the object feels no gravity(it's free falling, Please let me know if am wrong in my understanding).   In that Case, Earth is Pushed towards Sun by the Space-time ripple created by Sun's Mass and Earth keeps on escaping to form an orbit. Its the Push from Space and not the pull from Sun. What makes Earth to Escape the push and retain in the orbit?
Selva
CC BY-SA 3.0
2014-12-03T08:26:58.107Z
@SelvaKumar "Ripple"? There are no "ripples". The only things that could described such would be gravitational waves, but they are irrelevant here (indeed impossible for a spherically symmetric gravitational field).
Stan Liou
CC BY-SA 3.0
2014-12-03T09:03:42.060Z
Being pushed some way doesn't imply that you're necessarily falling that way. See also the update to my answer.
Walter
CC BY-SA 3.0
2014-12-03T11:29:22.353Z
Thanks to everyone! You people really helped me in clearing my doubt!
Selva
<gravity><general-relativity><space-time>
Why does the distance between Sun and Earth stay the same?
Selva
# Why does the distance between Sun and Earth stay the same?

(Asked by: username_0 on 2014-12-02)

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.

**Tags:** gravity, general-relativity, space-time

> **Comments:**
> **username_1**: did you mean to ask why the Earth-Sun distance remains unchanged?  
> **username_2**: In space there is no friction.  
> **username_0**: Hi Wayfaring, In that case, A freely falling object should not reach the Earth Since there is no friction anywhere in space.  
> **username_0**: Hi username_1, I hope you got my big picture of my question. Did you?  
> **username_3**: "... *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).  
> **username_0**: Ok Keith . Let me rephrase my question. According to General Theory of Relativity, A Bigger Mass objects creates a ripple of Space-Time around it and that ripple pushes any  object in the Space-Time towards it . Due to this effect, the object feels no gravity(it's free falling, Please let me know if am wrong in my understanding).   In that Case, Earth is Pushed towards Sun by the Space-time ripple created by Sun's Mass and Earth keeps on escaping to form an orbit. Its the Push from Space and not the pull from Sun. What makes Earth to Escape the push and retain in the orbit?  
> **username_4**: @username_0Kumar "Ripple"? There are no "ripples". The only things that could described such would be gravitational waves, but they are irrelevant here (indeed impossible for a spherically symmetric gravitational field).  
> **username_1**: Being pushed some way doesn't imply that you're necessarily falling that way. See also the update to my answer.  
> **username_0**: Thanks to everyone! You people really helped me in clearing my doubt!  

---

## 2 answers

### Answer by username_1 on 2014-12-02. Score: 7 | Accepted answer

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.

> **Comments:**
> **username_0**: Hi username_1,    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:(  
> **username_6**: username_1'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, username_1 - did you mean 'centripetal' instead of 'centrifugal', or am I misinterpreting the usage?  
> **username_0**: 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?  
> **username_4**: @HDE226868 username_1'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 username_1 says.  
> **username_4**: 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).  
> **username_1**: @StanLiou Do you have a www reference for that?  
> **username_4**: @username_1 I derive [here](http://astronomy.stackexchange.com/a/7904/661) if you want a www reference specifically. Otherwise, it's in MTW §25.5, Hartle §9.3, Carroll §5.4, and other textbooks, up to some notational differences (e.g., $V_\text{eff}^{\scriptsize\text{MTW}} = \sqrt{2V_\text{eff}^{\scriptsize\text{Hartle}}+1}$). By the way, I didn't notice that you used $L$ as specific angular momentum when I made my previous comment, so the $m^2$ term would be unnecessary.  
> **username_0**: Thanks to everyone! You people really helped me in clearing my doubt! Regards,username_0  

---

### Answer by username_5 on 2014-12-04. Score: 4

Technically, the distance between Sun and Earth changes throughout the year. The [Earth moves around the Sun](https://en.wikipedia.org/wiki/Earth%27s_orbit) on an elliptical orbit. Its distance from the Sun [changes](https://en.wikipedia.org/wiki/Apsis) between 152 million km (aphelion) and 147 million km (perihelion), a difference of 5 million kilometres, twice per year.


---

