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What would the gravity be like if Earth and the moon were touching?

enter image description here

Assuming the moon keeps its form and doesn't collapse on its own weight, would the area between it and the Earth (point B) have less gravity?

What about point A? Would gravity in that area stay the same?

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    $\begingroup$ Apply the gravity of a sphere and superposition principle. $\endgroup$
    – jinawee
    May 6, 2014 at 21:06

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The gravitational force would decrease at the point $B$ and increase at the point $A$. Actually this happens at all distances, and it's what causes the tides. Moving the Moon in until it touches the Earth is just an extreme case.

As long as the gravitational fields are weak (in this case weak means a lot less than a black hole) you can simply add the gravitational forces from the Earth and the Moon.

Earth-Moon

At point $B$ the total force is given by:

$$ F_B = F_E - F_{MB} $$

because the two forces point in different directions and oppose each other. So $F < F_E$ and the gravitational field is less than the field of the Earth alone. At point $A$ the total force is given by:

$$ F_A = F_E + F_{MA} $$

because the two forces point in the same direction and reinforce each other. So $F > F_E$ and the gravitational field is greater than the field of the Earth alone.

We can put numbers on this. I'll calculate the acceleration, which is the force per unit mass. At Earth's surface this is 1g, i.e. $9.81$ m/s$^2$, so $F_E = 9.81$ m/s$^2$. For a mass $M$ at a distance $r$ the acceleration is simply:

$$ a = \frac{GM}{r^2} \tag{1} $$

To calculate $F_{MB}$ we have to set $M$ equal to the mass of the Moon ($7.35 \times 10^{22}$ kg) and $r$ equal to the radius of the Moon ($1.74 \times 10^6$ m), and using equation (1) the acceleration is:

$$ a \approx 1.62 \text{m/s}^2 $$

To calculate $F_{MA}$ we set $r$ to the radius of the Moon plus the diameter of the Earth ($1.28 \times 10^7$ m), and using equation (1) the acceleration is:

$$ a \approx 0.023 \text{m/s}^2 $$

Combining these results and converting the accelerations to $g$ we get:

Point B: 0.835g
Point A: 1.002g
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I believe it would strip our atmosphere and create destructive ocean tides. It'd have to be orbiting very fast too or the earth would just pull it towards us.. which wouldn't be pretty.

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    $\begingroup$ I think the question assumes the moon is just sitting on the earth; like a terrifying - all crushing - marble, rather than orbiting. (Magic anti-crush soil and water are provided) $\endgroup$ May 7, 2014 at 10:51
  • $\begingroup$ Ah, well that's just terrifying! $\endgroup$ May 9, 2014 at 15:42
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The effect would be an extreme example of a tidal force. Your intuition has served you well.

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Another characteristic of the Earth/Moon system's orbit is that both the Earth and Moon actually orbit the shared bodies' center of gravity.

In the real Earth/Moon system, this center of gravity is located somewhere within the earth's mantle or outer core (as far as I can recall.)

In your imagined Earth/Moon system in which the two bodies touch, the shared center of gravity would be a good deal closer to the Earth's crust, but still in the mantle.

Also, the imagined Earth/Moon system as posed in the original question would have both bodies orbiting each other much faster than they do now. Once again, I'm not up to par on the math, but there is definitely a way to calculate it.

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    $\begingroup$ It would barely shift it by 98 km, around 1.5% of the radius of the Earth. wolframalpha.com/input/?i=mass+of+the+moon+*+%28radius+of+moon+%2B+radius+of+earth%29%2F%28mass+of+the+moon+%2B+mass+of+earth%29 $\endgroup$
    – Davidmh
    May 7, 2014 at 11:19

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