<p>While I think James K's answer is correct, it may help to add to the explanation.</p>
<p>Consider the traditional Newtonian definition of gravity:</p>
<p><span class="math-container">$$ F = G \frac{m_1 m_2}{r^2} $$</span></p>
<p>The two <span class="math-container">$m$</span> terms refer to the mass of two objects for which you want to know the attractive force. The <span class="math-container">$r$</span> term is more important to your question - it refers to the distance between the two objects. It's important to realise that it's a squared term, and on the bottom of the equation, meaning that the attractive force decays by the inverse square of the distance between the two objects. All this means is that objects need to be very close to have a significant attractive forces - what we experience as gravity.</p>
<p>Finally - <a href="https://i.sstatic.net/K6myY.png" rel="nofollow noreferrer">Here's what it looks like</a> if substitute in some numbers:</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th></th>
<th></th>
</tr>
</thead>
<tbody>
<tr>
<td><span class="math-container">$G$</span> = 6.67430E-11 <span class="math-container">$\frac{\mathrm{N \cdot m}^2}{\mathrm{kg}^2}$</span></td>
<td>Gravitational Constant</td>
</tr>
<tr>
<td><span class="math-container">$m_{\mathrm{you}}$</span> = 80 kg</td>
<td>Approx mass of observer</td>
</tr>
<tr>
<td><span class="math-container">$m_{\mathrm{Earth}}$</span> = 5.972E+24 kg</td>
<td>Approx mass of Earth</td>
</tr>
<tr>
<td><span class="math-container">$m_{\mathrm{Sun}}$</span> = 1.989E+30 kg</td>
<td>Approx mass of the Sun</td>
</tr>
<tr>
<td><span class="math-container">$r_{\mathrm{Sun}}$</span> = 149.6E+6 km</td>
<td>Approx distance to the centre of the Sun</td>
</tr>
<tr>
<td><span class="math-container">$r_{\mathrm{Earth}}$</span> = 6371 km</td>
<td>Approx distance to the centre of Earth</td>
</tr>
<tr>
<td><span class="math-container">$F_1 = \frac{G \cdot ( m_{\mathrm{you}} \cdot m_{\mathrm{Earth}} ) }{r_{\mathrm Earth}^2}$</span> = 785.598 N</td>
<td>Force between you and Earth</td>
</tr>
<tr>
<td><span class="math-container">$F_2 = \frac{G \cdot ( m_{\mathrm{you}} \cdot m_{\mathrm{Sun}} ) }{r_{\mathrm{Sun}}^2}$</span> = 0.475 N</td>
<td>Force between you and the Sun</td>
</tr>
<tr>
<td><span class="math-container">$ \frac{F_1}{F_2} $</span> = 1655.515</td>
<td>Factor difference between forces</td>
</tr>
</tbody>
</table>
</div>
<p>You'll notice that the force between you and the Sun is the equivalent of the force exerted by 50 grams of mass on Earth - so not much.</p>

CC BY-SA 4.0
2023-11-17T06:25:38.323Z
But we are in a rotating frame of reference, so to find the force we *feel* you would need to subtract the "centrifugal force",  which is 0.475 N
James K
CC BY-SA 4.0
2023-11-17T06:10:11.303Z
2023-11-18T02:24:38.597Z
2023-11-18T02:24:38.597Z
G.H.
<p>Firstly, the sun can't just &quot;disappear&quot;.  Even if it were converted by magic into &quot;pure energy&quot;, that energy can't go anywhere faster than the speed of light, and Energy has gravitation equal to mass (by E=mc²)</p>
<p>However, the Earth and all its denizens are in free-fall around the sun.  We are on a &quot;space station&quot; and so we are weightless with respect to the sun.
Imagine the Earth vanished and the Space Station floated off in a straight line. The astronauts would not notice any change in their weighlessness.</p>
<p>There are small tidal forces. The sun can cause a tidal acceleration of <span class="math-container">$5\times 10^{-7}\ ms^{-2}$</span>, more than a million times smaller than your weight on Earth. But these are not noticeable.</p>
<p>So you wouldn't directly notice if these forces suddenly ceased.</p>

CC BY-SA 4.0
2023-11-16T21:33:40.097Z
What about people on Earth's surface? If Earth had it's gravity magically turned off it seems it would feel like an elevator instantly dropping from under you. I'd certainly directly notice that! :)    this relates to the size/distance qualifier in the  body of question
BradV
CC BY-SA 4.0
2023-11-16T21:53:28.990Z
No, because people on the Earth's surface  are not in free fall about the Earth.  But the Earth (and everyone on it) are in free-fall about the sun.
James K
CC BY-SA 4.0
2023-11-17T04:28:35.870Z
The cycle of ocean tides would change enough to be noticeable to those who spend a lot of time on or near the ocean, right? Of course the lunar effect on tides would continue but I understand the sun contributes as well.
Todd Wilcox
CC BY-SA 4.0
2023-11-17T04:44:50.500Z
If the sun would disappear, tidal forces will be not the first thing we notice.
Rocket Nikita
CC BY-SA 4.0
2023-11-17T05:16:16.827Z
*"the sun can't just "disappear""* - Citation required
Valorum
CC BY-SA 4.0
2023-11-17T05:46:31.373Z
@BradV: "like an elevator instantly dropping from under you". Not instantly!  Earth is compressed by it's gravity somewhat (iron densities 7.8, but Earths center is ~12 despite being ~6000K). If gravity "turns off" Earth will decompress. Initially the ground will accelerate upward *exactly at 1g* due to decompression. It will take maybe seconds to notice anything, and minutes before gravity is substantially reduced.
Kevin Kostlan
CC BY-SA 4.0
2023-11-17T14:30:46.030Z
@Valorum In regards to the first sentence I imagine the sun being converted into energy which immediately recedes in the direction opposite of the Earth at the speed of light... (preferably some kind of energy which would not react with the solar wind and such to prevent reflections from destroying the Earth... in that regard perhaps a bit of inclination w.r.t. the ecliptic would be good.)
Michael
CC BY-SA 4.0
2023-11-17T17:26:10.917Z
If, with a magic wand, the Sun would disappear almost instantaneously -- say, within one nanosecond -- there would be a gravitational ripple propagating at light speed with a steep gradient. A ns equivalent is 33cm. One moment there is the Sun's gravitation, the next moment there isn't; your feet are pulled, your head isn't.  The difference creates tidal forces much larger than the ones caused by the inhomogeneous field, though still tiny. The Sun accelerates our feet at about 6mm/s^2, creating a tidal force of a few hundredth N which will be unnoticeable.
Peter - Reinstate Monica
CC BY-SA 4.0
2023-11-17T17:30:07.807Z
But it should be noted that objects close to the sun will be torn apart by the forces. On the Sun's "surface", the acceleration is >200m/s^2, so the tidal force within a 100kg body would be several kN.
Peter - Reinstate Monica
CC BY-SA 4.0
2023-11-18T07:49:52.743Z
@kostlan 1g's worth of sudden decompression is probably going to result in some exciting seismic consequences though!
G_B
CC BY-SA 4.0
2023-11-16T18:32:37.820Z
2023-11-16T20:46:29.693Z
2023-11-16T20:46:29.693Z
James K
<h3>No, the human body couldn't practically feel the sudden disappearance of a gravitational force</h3>
<p>Humans are able to feel because different accelerations applied to different parts of the body cause it to deform. For example you can feel the <a href="https://en.wikipedia.org/wiki/Jerk_(physics)" rel="nofollow noreferrer">jerk</a> as a car comes to a complete stop (non-zero acceleration to zero acceleration), because a force that was being applied only to your bottom is suddenly no longer applied, which changes the overall flex of your spine (among other deformations).</p>
<p>On the other hand, gravitational fields between a very large body and a very small body are practically uniform across the small body. So no matter how the field changes, the same acceleration is being applied to every part of the smaller body at any point in time. Therefore a human body wouldn't deform if the sun disapeared, so the human would not feel the change.</p>

CC BY-SA 4.0
2023-11-19T10:29:30.223Z
I'm afraid that's not how you feel jerk. Sure you can feel forces on your body too, but what you really sense is an inner-ear function, and it's inertial so it reports jerk relative to your current direction of motion. That's why those flight/driving simulators on big hydraulics never feel right - because the inertial component isn't there. And because it's dynamic, not static, the justification that you could remove all forces and have everything remain static is incorrect. Try removing the force on a swingball by cutting the string mid-swing and see if it changes the ball's travel. :)
Graham
CC BY-SA 4.0
2023-11-19T18:00:18.070Z
@Graham The inner ear can only feel acceleration because its structure deforms. Typically, the inner ear isn't being directly accelerated, it's being accelerated by its connection with the rest of the body; thus it deforms due to inertia. If the whole inner ear structure is accelerated uniformly, the acceleration will not be felt. As for why flight simulators feel different than real flight, it's because they don't reproduce all the same forces on the body as real flight. It's impossible to reproduce the same forces without reproducing the same flight path.
Vaelus
CC BY-SA 4.0
2023-11-17T14:59:01.447Z
2023-11-17T14:59:01.447Z
Vaelus
<p><strong>From 1,156,810 kilometers</strong> that is very near but still outside the Sun.</p>
<p>The formula to calculate the tidal force F exerted by the Sun on an object like a human is given by:</p>
<pre><code>F = 2 * G * M * r^2 / D^3
</code></pre>
<p>where G is the gravitational constant, M is the mass of the Sun, r is the the difference in distance between the two points on the object (assuming the height of the human, 2 m) and D is the distance between the astronaut and the Sun.</p>
<p>Further, we assume the human is able to sense a sudden change of the force equal to 1 power gram (nobody knows for sure but it you would feel a coin of this mass dropped on your palm).</p>
<p>This gives the distance of 1,156,810 km from the centre of the Sun that may be somewhat survivable with science fictional technologies. The Sun has the radius (not diameter) of 695,500 km.</p>
<p>Assuming a spaceflight (free fall), the astronaut would not feel the disappearance of the main gravity force.</p>

CC BY-SA 4.0
2023-11-17T08:57:49.893Z
2023-11-17T08:57:49.893Z
h22
<p>The text in the question body is a little at odds with the question itself, so I'm going to ignore anything specifically about the sun vanishing (since that involves a whole lot of extra baggage, and has been answered, and actually has a very small effect anyway).</p>
<p>Could we feel a sudden change in gravitation? Yes, but only indirectly. As far as our bodies are concerned, there's no such thing as gravity, but there <em>is</em> such a thing as weight (which is just gravity acting on a mass... like our body), involving things like prioperception and our inner ear.</p>
<p>We're quite good at detecting (large) changes in <em>that</em>. For experimental verification, use an elevator.</p>

CC BY-SA 4.0
2023-11-18T17:21:54.273Z
The case of an elevator is different than an orbiting body, because the elevator pushes on your feet only. If the elevator pushed on all parts of your body uniformly, including the structures in your inner ear, you would not feel it accelerating you.
Vaelus
CC BY-SA 4.0
2023-11-19T10:20:10.760Z
@Vaelus Not true - you've forgotten that inertia also exists, because this is not simply static forces in equilibrium. The jerk you experience from an elevator or car has nothing whatsoever to do with where the force is applied, and everything to do with the fact that there is a change of acceleration relative to the current direction of travel.
Graham
CC BY-SA 4.0
2023-11-21T17:41:55.700Z
@Graham I think that *is* true.  If the acceleration affected your body uniformly, there would be no changes for you to feel.   What we feel in an elevator is different forces acting through our body due to the force being applied or lessened at our feet.  That's the same reason we 'feel' the gravity of the earth when we're standing on it.
Jason Goemaat
CC BY-SA 4.0
2023-11-17T16:01:25.610Z
2023-11-17T16:01:25.610Z
Virtual Void
<p>For example, if the Sun suddenly disappeared, would we physically feel it 8 minutes later (not counting light/temperature), since propagates moves at the speed of light.</p>
<p>And if so, would the effect be noticeably different depending on size/distance?</p>
<p>Every answer I've seen so far is mostly about light and/or temperature, or our orbit. I am talking specifically about what the average person would feel in their body (like a drop or lurch or jolt).</p>

CC BY-SA 4.0
2023-11-16T16:54:01.597Z
Does this answer your question? [Sudden Effects of Sun Disappearing](https://astronomy.stackexchange.com/questions/18543/sudden-effects-of-sun-disappearing)
DarkDust
CC BY-SA 4.0
2023-11-17T08:38:05.417Z
@DarkDust Not really. It does touch on the most important parts of the sun dissapearing, but not if our bodies would be able to "feel" the gravity well effectively vanishing.
Parrotmaster
<gravity><gravitational-waves>
Could the human body feel the sudden disappearance or end of a gravitational force?
Parrotmaster
# Could the human body feel the sudden disappearance or end of a gravitational force?

(Asked by: username_0 on 2023-11-16)

For example, if the Sun suddenly disappeared, would we physically feel it 8 minutes later (not counting light/temperature), since propagates moves at the speed of light.

And if so, would the effect be noticeably different depending on size/distance?

Every answer I've seen so far is mostly about light and/or temperature, or our orbit. I am talking specifically about what the average person would feel in their body (like a drop or lurch or jolt).

**Tags:** gravity, gravitational-waves
> **Comments:**
> **username_1**: Does this answer your question? [Sudden Effects of Sun Disappearing](https://astronomy.stackexchange.com/questions/18543/sudden-effects-of-sun-disappearing)  
> **username_0**: @DarkDust Not really. It does touch on the most important parts of the sun dissapearing, but not if our bodies would be able to "feel" the gravity well effectively vanishing.  

---

## 5 answers

### Answer by username_2 on 2023-11-17. Score: 1 | Accepted answer

While I think James K's answer is correct, it may help to add to the explanation.

Consider the traditional Newtonian definition of gravity:

$$ F = G \frac{m\_1 m\_2}{r^2} $$

The two $m$ terms refer to the mass of two objects for which you want to know the attractive force. The $r$ term is more important to your question - it refers to the distance between the two objects. It's important to realise that it's a squared term, and on the bottom of the equation, meaning that the attractive force decays by the inverse square of the distance between the two objects. All this means is that objects need to be very close to have a significant attractive forces - what we experience as gravity.

Finally - [Here's what it looks like](https://i.sstatic.net/K6myY.png) if substitute in some numbers:

|  |  |
| --- | --- |
| $G$ = 6.67430E-11 $\frac{\mathrm{N \cdot m}^2}{\mathrm{kg}^2}$ | Gravitational Constant |
| $m\_{\mathrm{you}}$ = 80 kg | Approx mass of observer |
| $m\_{\mathrm{Earth}}$ = 5.972E+24 kg | Approx mass of Earth |
| $m\_{\mathrm{Sun}}$ = 1.989E+30 kg | Approx mass of the Sun |
| $r\_{\mathrm{Sun}}$ = 149.6E+6 km | Approx distance to the centre of the Sun |
| $r\_{\mathrm{Earth}}$ = 6371 km | Approx distance to the centre of Earth |
| $F\_1 = \frac{G \cdot ( m\_{\mathrm{you}} \cdot m\_{\mathrm{Earth}} ) }{r\_{\mathrm Earth}^2}$ = 785.598 N | Force between you and Earth |
| $F\_2 = \frac{G \cdot ( m\_{\mathrm{you}} \cdot m\_{\mathrm{Sun}} ) }{r\_{\mathrm{Sun}}^2}$ = 0.475 N | Force between you and the Sun |
| $ \frac{F\_1}{F\_2} $ = 1655.515 | Factor difference between forces |

You'll notice that the force between you and the Sun is the equivalent of the force exerted by 50 grams of mass on Earth - so not much.

> **Comments:**
> **username_3**: But we are in a rotating frame of reference, so to find the force we *feel* you would need to subtract the "centrifugal force",  which is 0.475 N  

---

### Answer by username_3 on 2023-11-16. Score: 16

Firstly, the sun can't just "disappear". Even if it were converted by magic into "pure energy", that energy can't go anywhere faster than the speed of light, and Energy has gravitation equal to mass (by E=mc²)

However, the Earth and all its denizens are in free-fall around the sun. We are on a "space station" and so we are weightless with respect to the sun.
Imagine the Earth vanished and the Space Station floated off in a straight line. The astronauts would not notice any change in their weighlessness.

There are small tidal forces. The sun can cause a tidal acceleration of $5\times 10^{-7}\ ms^{-2}$, more than a million times smaller than your weight on Earth. But these are not noticeable.

So you wouldn't directly notice if these forces suddenly ceased.

> **Comments:**
> **username_4**: What about people on Earth's surface? If Earth had it's gravity magically turned off it seems it would feel like an elevator instantly dropping from under you. I'd certainly directly notice that! :)    this relates to the size/distance qualifier in the  body of question  
> **username_3**: No, because people on the Earth's surface  are not in free fall about the Earth.  But the Earth (and everyone on it) are in free-fall about the sun.  
> **username_5**: The cycle of ocean tides would change enough to be noticeable to those who spend a lot of time on or near the ocean, right? Of course the lunar effect on tides would continue but I understand the sun contributes as well.  
> **username_6**: If the sun would disappear, tidal forces will be not the first thing we notice.  
> **username_7**: *"the sun can't just "disappear""* - Citation required  
> **username_8**: @BradV: "like an elevator instantly dropping from under you". Not instantly!  Earth is compressed by it's gravity somewhat (iron densities 7.8, but Earths center is ~12 despite being ~6000K). If gravity "turns off" Earth will decompress. Initially the ground will accelerate upward *exactly at 1g* due to decompression. It will take maybe seconds to notice anything, and minutes before gravity is substantially reduced.  
> **username_9**: @Valorum In regards to the first sentence I imagine the sun being converted into energy which immediately recedes in the direction opposite of the Earth at the speed of light... (preferably some kind of energy which would not react with the solar wind and such to prevent reflections from destroying the Earth... in that regard perhaps a bit of inclination w.r.t. the ecliptic would be good.)  
> **username_10**: If, with a magic wand, the Sun would disappear almost instantaneously -- say, within one nanosecond -- there would be a gravitational ripple propagating at light speed with a steep gradient. A ns equivalent is 33cm. One moment there is the Sun's gravitation, the next moment there isn't; your feet are pulled, your head isn't.  The difference creates tidal forces much larger than the ones caused by the inhomogeneous field, though still tiny. The Sun accelerates our feet at about 6mm/s^2, creating a tidal force of a few hundredth N which will be unnoticeable.  
> **username_10**: But it should be noted that objects close to the sun will be torn apart by the forces. On the Sun's "surface", the acceleration is >200m/s^2, so the tidal force within a 100kg body would be several kN.  
> **username_11**: @kostlan 1g's worth of sudden decompression is probably going to result in some exciting seismic consequences though!  

---

### Answer by username_12 on 2023-11-17. Score: 3

### No, the human body couldn't practically feel the sudden disappearance of a gravitational force

Humans are able to feel because different accelerations applied to different parts of the body cause it to deform. For example you can feel the [jerk](https://en.wikipedia.org/wiki/Jerk_(physics)) as a car comes to a complete stop (non-zero acceleration to zero acceleration), because a force that was being applied only to your bottom is suddenly no longer applied, which changes the overall flex of your spine (among other deformations).

On the other hand, gravitational fields between a very large body and a very small body are practically uniform across the small body. So no matter how the field changes, the same acceleration is being applied to every part of the smaller body at any point in time. Therefore a human body wouldn't deform if the sun disapeared, so the human would not feel the change.

> **Comments:**
> **username_13**: I'm afraid that's not how you feel jerk. Sure you can feel forces on your body too, but what you really sense is an inner-ear function, and it's inertial so it reports jerk relative to your current direction of motion. That's why those flight/driving simulators on big hydraulics never feel right - because the inertial component isn't there. And because it's dynamic, not static, the justification that you could remove all forces and have everything remain static is incorrect. Try removing the force on a swingball by cutting the string mid-swing and see if it changes the ball's travel. :)  
> **username_12**: @Graham The inner ear can only feel acceleration because its structure deforms. Typically, the inner ear isn't being directly accelerated, it's being accelerated by its connection with the rest of the body; thus it deforms due to inertia. If the whole inner ear structure is accelerated uniformly, the acceleration will not be felt. As for why flight simulators feel different than real flight, it's because they don't reproduce all the same forces on the body as real flight. It's impossible to reproduce the same forces without reproducing the same flight path.  

---

### Answer by username_14 on 2023-11-17. Score: 0

From 1,156,810 kilometers that is very near but still outside the Sun.
The formula to calculate the tidal force F exerted by the Sun on an object like a human is given by:
F = 2 * G * M * r^2 / D^3

where G is the gravitational constant, M is the mass of the Sun, r is the the difference in distance between the two points on the object (assuming the height of the human, 2 m) and D is the distance between the astronaut and the Sun.
Further, we assume the human is able to sense a sudden change of the force equal to 1 power gram (nobody knows for sure but it you would feel a coin of this mass dropped on your palm).
This gives the distance of 1,156,810 km from the centre of the Sun that may be somewhat survivable with science fictional technologies. The Sun has the radius (not diameter) of 695,500 km.
Assuming a spaceflight (free fall), the astronaut would not feel the disappearance of the main gravity force.


---

### Answer by username_15 on 2023-11-17. Score: 0

The text in the question body is a little at odds with the question itself, so I'm going to ignore anything specifically about the sun vanishing (since that involves a whole lot of extra baggage, and has been answered, and actually has a very small effect anyway).

Could we feel a sudden change in gravitation? Yes, but only indirectly. As far as our bodies are concerned, there's no such thing as gravity, but there *is* such a thing as weight (which is just gravity acting on a mass... like our body), involving things like prioperception and our inner ear.

We're quite good at detecting (large) changes in *that*. For experimental verification, use an elevator.

> **Comments:**
> **username_12**: The case of an elevator is different than an orbiting body, because the elevator pushes on your feet only. If the elevator pushed on all parts of your body uniformly, including the structures in your inner ear, you would not feel it accelerating you.  
> **username_13**: @Vaelus Not true - you've forgotten that inertia also exists, because this is not simply static forces in equilibrium. The jerk you experience from an elevator or car has nothing whatsoever to do with where the force is applied, and everything to do with the fact that there is a change of acceleration relative to the current direction of travel.  
> **username_16**: @Graham I think that *is* true.  If the acceleration affected your body uniformly, there would be no changes for you to feel.   What we feel in an elevator is different forces acting through our body due to the force being applied or lessened at our feet.  That's the same reason we 'feel' the gravity of the earth when we're standing on it.  

---

