<p>The Earth doesn't <em>cover</em> the hole, the Earth <em>is</em> the hole. Gravity is an attractive force, so were you not standing on the Earth, the Earth's gravity would cause you to accelerate towards it. As you are standing on the Earth, you feel this acceleration as your weight, just as you would feel pressure if you were to push against a wall.</p>

<p>To take a more complex view, we would say that you are being attracted to the local center of gravity, which is located within the Earth. Take your bridges example, it doesn't matter that something obstructs you from reaching the center of gravity. You will feel the attraction of gravity regardless. If you jump while standing on a bridge, you would land just the same as if you jumped while standing on the ground (<em>though please try not to miss the bridge</em>). The graphic is meant to illustrate the strength of the force based on the mass of the attractive object and distance.</p>

<p>If we were to put you on a more massive object, which would have a steeper curvature, you would experience this as increased weight, which is why our lunar explorers bounced around on the moon instead of walking.</p>

CC BY-SA 3.0
2015-04-14T03:59:37.487Z
2015-04-15T02:31:11.123Z
2015-04-15T02:31:11.123Z
Mitch Goshorn
<p>Perhaps this is a silly question but I am trying to understand Einstein's general theory of relativity. </p>

<p>I would like to explain my question in more detail.</p>

<p>The greater the mass of an object, the greater the curvature of spacetime. </p>

<p>If I imagine the curvature of spacetime as a hole with a steep slope around it, I can picture that the closer I get to the center of the hole, the steeper the slope is and therefore the more I would be pushed into the center of the hole. I am not completely sure if that's a very good analogy but in general I think I get the picture of timespace curvature as depicted here: <a href="https://upload.wikimedia.org/wikipedia/commons/2/22/Spacetime_curvature.png" rel="nofollow noreferrer"><img src="https://upload.wikimedia.org/wikipedia/commons/2/22/Spacetime_curvature.png" alt="https://upload.wikimedia.org/wikipedia/commons/2/22/Spacetime_curvature.png"></a></p>

<p>Now, my question is why is it that the size of the hole matters if I am not standing on the edge (side) of the hole or in the hole but rather on an object that 'covers' the hole (Earth).  </p>

<p>Let's say if I stand on a bridge that covers a 20-meter-wide hole and on a bridge the covers 500-meter-wide hole. Wouldn't the gravitational force experienced be the same since the surface on both bridges is flat?</p>

CC BY-SA 3.0
2015-04-15T09:31:43.303Z
Why the down votes?
BustedSanta
<gravity><general-relativity><space-time>
Why does Earth have more gravitational force than the Moon if I stand on it?
BustedSanta
# Why does Earth have more gravitational force than the Moon if I stand on it?

(Asked by: username_0 on 2015-04-14)

Perhaps this is a silly question but I am trying to understand Einstein's general theory of relativity.

I would like to explain my question in more detail.

The greater the mass of an object, the greater the curvature of spacetime.

If I imagine the curvature of spacetime as a hole with a steep slope around it, I can picture that the closer I get to the center of the hole, the steeper the slope is and therefore the more I would be pushed into the center of the hole. I am not completely sure if that's a very good analogy but in general I think I get the picture of timespace curvature as depicted here: [![https://upload.wikimedia.org/wikipedia/commons/2/22/Spacetime_curvature.png](https://upload.wikimedia.org/wikipedia/commons/2/22/Spacetime_curvature.png)](https://upload.wikimedia.org/wikipedia/commons/2/22/Spacetime_curvature.png)

Now, my question is why is it that the size of the hole matters if I am not standing on the edge (side) of the hole or in the hole but rather on an object that 'covers' the hole (Earth).

Let's say if I stand on a bridge that covers a 20-meter-wide hole and on a bridge the covers 500-meter-wide hole. Wouldn't the gravitational force experienced be the same since the surface on both bridges is flat?

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

> **Comments:**
> **username_0**: Why the down votes?  

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## 1 answers

### Answer by username_1 on 2015-04-14. Score: 3 | Accepted answer

The Earth doesn't *cover* the hole, the Earth *is* the hole. Gravity is an attractive force, so were you not standing on the Earth, the Earth's gravity would cause you to accelerate towards it. As you are standing on the Earth, you feel this acceleration as your weight, just as you would feel pressure if you were to push against a wall.

To take a more complex view, we would say that you are being attracted to the local center of gravity, which is located within the Earth. Take your bridges example, it doesn't matter that something obstructs you from reaching the center of gravity. You will feel the attraction of gravity regardless. If you jump while standing on a bridge, you would land just the same as if you jumped while standing on the ground (*though please try not to miss the bridge*). The graphic is meant to illustrate the strength of the force based on the mass of the attractive object and distance.

If we were to put you on a more massive object, which would have a steeper curvature, you would experience this as increased weight, which is why our lunar explorers bounced around on the moon instead of walking.


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