# Does the Inverse Square Law mean that the apparent diameter of an object of same mass has the same gravitational effect?

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.

• Angular diameter (under the small-angle approximation) isn't inverse-square, it's inverse linear. Nov 15, 2022 at 21:42
• @ notovny; Corrected, thanks. Nov 16, 2022 at 4:37

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.

• 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. Nov 16, 2022 at 4:39

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.

• Thanks for this and the clarification above. Nov 17, 2022 at 2:59