<p>The tides depend on the inverse cube of distance (<span class="math-container">$D$</span>) and the mass (<span class="math-container">$M$</span>) of the object causing the tides.
<span class="math-container">$$F_{\rm tide} \propto M D^{-3}.$$</span></p>
<p>The mass of an object depends on its density (<span class="math-container">$\rho$</span>) multiplied by the cube of its size (<span class="math-container">$R$</span>):
<span class="math-container">$$ F_{\rm tide} \propto \rho R^3 D^{-3}.$$</span></p>
<p>The ratio <span class="math-container">$R/D$</span> is proportional to the angular size of an object (<span class="math-container">$\theta$</span>). Thus
<span class="math-container">$$F_{\rm tide} \propto \rho \theta^3$$</span>
and two objects of similar density and angular size would exert the same tidal forces.</p>

CC BY-SA 4.0
2022-11-17T02:59:22.293Z
Thanks for this and the clarification above.
JohnHunt
CC BY-SA 4.0
2022-11-16T11:02:25.950Z
2022-11-16T11:02:25.950Z
ProfRob
<p>Exactly the same tides, yes.</p>
<p>The Sun is the same angular size as the Moon but about 400 times further away.</p>
<p>If the Sun were as dense as the Moon then its gravitational pull would be 400 times that of the Moon: <span class="math-container">$400$</span> times the diameter, so <span class="math-container">$400^3$</span> times the mass, divided by <span class="math-container">$400^2$</span> because of the inverse square law.</p>
<p><strong>But</strong> tides depend not on the gravitational pull but on the amount by which this pull diminishes over a distance equal to the Earth’s diameter. And the Earth’s diameter, <em>as a proportion of the distance to the Sun</em>, is <span class="math-container">$\frac1{400}$</span> of its diameter as a proportion of the distance to the Moon. So although the gravitational pull is <span class="math-container">$400$</span> times stronger, the tidal effects are <span class="math-container">$\frac{400}{400}$</span> times stronger - in other words, identical.</p>
<p>Since in reality solar tides are smaller than lunar tides, we can deduce that the Sun is less dense than the moon.</p>

CC BY-SA 4.0
2022-11-16T04:39:44.523Z
This was my thinking, but I wanted some confirmation.  I'll wait 24 hours as is normally recommended, but I think I will accept this answer.
JohnHunt
CC BY-SA 4.0
2022-11-15T22:20:49.257Z
2022-11-15T22:20:49.257Z
Martin Kochanski
<p>Someone else asked about various planets located at the orbital range of the Moon.  It made me wonder if an object with the same density as the Moon, at a distance which would present the same angular diameter of the Moon seen from the Earth, would have the same general gravitational effects of the Moon on the Earth, given that gravity follows an inverse square law and angular dimension follow an inverse linear law.  Would, for instance, a super-Earth of the Moon's density and angular diameter create the same tides as our Moon currently does?  We are, for this question, discarding our Moon for the scenario; there is only the super-Earth and the Earth.</p>
<p>It seems to me, intuitively (i.e. without any math, let's be honest, I'm still working on that bit), that such an object would have profound effects; the center of gravity of the system would, I would think, remain in proportion to the orbital distance, but be further out in absolute unit terms (kilometers), for instance.  In my head I picture an Earth orbit with wider oscillations, given the increased total mass of the system.  Yet I think the experienced pull from that super-Earth would be experienced on the ground at the same magnitude as our current Moon's gravitation.</p>
<p>How should I be thinking about this situation?  Thanks.</p>

CC BY-SA 4.0
2022-11-15T21:42:52.367Z
Angular diameter (under the small-angle approximation) isn't inverse-square, it's inverse linear.
notovny
CC BY-SA 4.0
2022-11-16T04:37:29.923Z
@ notovny; Corrected, thanks.
JohnHunt
<orbital-mechanics><gravitation>
Does the Inverse Square Law mean that the apparent diameter of an object of same mass has the same gravitational effect?
JohnHunt
# Does the Inverse Square Law mean that the apparent diameter of an object of same mass has the same gravitational effect?

(Asked by: username_0 on 2022-11-15)

Someone else asked about various planets located at the orbital range of the Moon. It made me wonder if an object with the same density as the Moon, at a distance which would present the same angular diameter of the Moon seen from the Earth, would have the same general gravitational effects of the Moon on the Earth, given that gravity follows an inverse square law and angular dimension follow an inverse linear law. Would, for instance, a super-Earth of the Moon's density and angular diameter create the same tides as our Moon currently does? We are, for this question, discarding our Moon for the scenario; there is only the super-Earth and the Earth.

It seems to me, intuitively (i.e. without any math, let's be honest, I'm still working on that bit), that such an object would have profound effects; the center of gravity of the system would, I would think, remain in proportion to the orbital distance, but be further out in absolute unit terms (kilometers), for instance. In my head I picture an Earth orbit with wider oscillations, given the increased total mass of the system. Yet I think the experienced pull from that super-Earth would be experienced on the ground at the same magnitude as our current Moon's gravitation.

How should I be thinking about this situation? Thanks.

**Tags:** orbital-mechanics, gravitation

> **Comments:**
> **username_1**: Angular diameter (under the small-angle approximation) isn't inverse-square, it's inverse linear.  
> **username_0**: @ username_1; Corrected, thanks.  

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## 2 answers

### Answer by username_3 on 2022-11-15. Score: 25 | Accepted answer

Exactly the same tides, yes.

The Sun is the same angular size as the Moon but about 400 times further away.

If the Sun were as dense as the Moon then its gravitational pull would be 400 times that of the Moon: $400$ times the diameter, so $400^3$ times the mass, divided by $400^2$ because of the inverse square law.

**But** tides depend not on the gravitational pull but on the amount by which this pull diminishes over a distance equal to the Earth’s diameter. And the Earth’s diameter, *as a proportion of the distance to the Sun*, is $\frac1{400}$ of its diameter as a proportion of the distance to the Moon. So although the gravitational pull is $400$ times stronger, the tidal effects are $\frac{400}{400}$ times stronger - in other words, identical.

Since in reality solar tides are smaller than lunar tides, we can deduce that the Sun is less dense than the moon.

> **Comments:**
> **username_0**: This was my thinking, but I wanted some confirmation.  I'll wait 24 hours as is normally recommended, but I think I will accept this answer.  

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### Answer by username_2 on 2022-11-16. Score: 12

The tides depend on the inverse cube of distance ($D$) and the mass ($M$) of the object causing the tides.
$$F\_{\rm tide} \propto M D^{-3}.$$

The mass of an object depends on its density ($\rho$) multiplied by the cube of its size ($R$):
$$ F\_{\rm tide} \propto \rho R^3 D^{-3}.$$

The ratio $R/D$ is proportional to the angular size of an object ($\theta$). Thus
$$F\_{\rm tide} \propto \rho \theta^3$$
and two objects of similar density and angular size would exert the same tidal forces.

> **Comments:**
> **username_0**: Thanks for this and the clarification above.  

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