# Will an object orbiting a non-rotating mass with a layer of water on it cause two bulges of water on opposite sides of it? [duplicate]

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Imagine a non-rotating spherical mass with a sea on its surface (somewhat like the earth, except for the rotating part). Around this mass is circling an object with a much smaller mass, but nevertheless big enough to exert a gravitational force on the non-rotating spherical mass. Will, in this case, two bulges of water develop on opposite places of the mass with the water on it (the line between the two bulges aligned with the line connecting the two masses)?

Because the gravity of the mass circulating around the mass with water, you would suspect that all the water on the big mass is drawn towards one side, so no two bulges will develop. Or am I wrong? Does the fact that the big mass revolves around the CM of the two masses influence the movement of the water?

## marked as duplicate by sammy gerbil, ZeroTheHero, Jon Custer, Bill N, YashasApr 25 '17 at 3:09

• This is really no different from the earth's tides being caused by the moon. The fact that your "planet" is non-rotating doesn't really change anything. Just look up tides to learn why you get two bulges. – Nuclear Wang Apr 19 '17 at 12:46

The water on the close side is pulled more strongly than the solid "earth" which in turn is pulled more strongly than the water on the far side: the result is two bulges.

E.g see Fig 7.5 in Feynman. If the moon pulls the whole earth toward it, why doesn’t the earth fall right “up” to the moon? Because the earth does the same trick as the moon, it goes in a circle around a point which is inside the earth but not at its center. The moon does not just go around the earth, the earth and the moon both go around a central position, each falling toward this common position, as shown in Fig. 7–5. This motion around the common center is what balances the fall of each. So the earth is not going in a straight line either; it travels in a circle. The water on the far side is “unbalanced” because the moon’s attraction there is weaker than it is at the center of the earth, where it just balances the “centrifugal force.” The result of this imbalance is that the water rises up, away from the center of the earth. On the near side, the attraction from the moon is stronger, and the imbalance is in the opposite direction in space, but again away from the center of the earth. The net result is that we get two tidal bulges.

• good link, copied from it – anna v Apr 19 '17 at 13:01
• The water on the side of the moon experiences less force towards the center of the earth because on the side of the moon the water undergoes the gravitation of the earth minus the gravitation of the moon, while on the opposite side the water undergoes the gravity of the earth plus the gravitation of the moon (if the earth wouldn't rotate). But the centrifugal force caused by the rotation of the earth more than compensates for the bigger gravitational force on the side of the earth opposite to the moon? – descheleschilder Apr 19 at 16:48 – descheleschilder Aug 14 '17 at 23:10

I tried it this way. Consider two equal spherical with masses M, covered with a layer of water, circling around each other at a distance $l$ (measured between their centers). They have radius $R$ and angular velocity $\omega$.

Now let's look at one of them (the same applies to the other). For a point on the outside, (at distance $\frac 1 2 {l+R}$ from their CM) there are three forces:

$F_{cf}$, the centrifugal force directed away from the CM

$F_{gM}$, the gravitational force from the sphere we are looking at, directed towards the CM

$F'_{gM}$, the gravitational force from the other sphere, directed towards the CM

So for a test mass m:

$$F_{cf}=-\frac{mv^2}{{\frac 1 2}l+R}=-m{\omega}^2({\frac 1 2}l+R)$$

$$F_{gM}=m \frac{GM}{R^2}$$

$$F'_{gM}=m \frac{GM}{(l+R)^2}$$

Now we must find out if $F_{cf}$ pulls harder or less (or equal) on a test mass than (to) $F'_{gM}$.

We can write

$$M{\omega}^2({\frac 1 2}l)=\frac{M^2 G}{l^2}$$ so $$MG={\frac 1 2}{\omega}^2 l^3$$

Putting this in the expression for $F'_{gM}$ gives:

$$F'_{gM}=\frac 1 2 m{\omega}^2\frac{l^3}{(l+R)^2}$$

So we have to find out if ${\frac 1 2}l+R$ is smaller or bigger (or equal) than (to) ${\frac 1 2}{\frac{l^3}{(l+R)^2}}$

Now we can write:

$${\frac 1 2}{\frac{l^3}{(l+R)^2}}={\frac 1 2}{\frac{l^3}{l^2+2lR+R^2}}={\frac 1 2}{\frac l {1+{\frac {2R}{l}}+{\frac{R^2}{l^2}}}}$$

which is obviously less than ${\frac 1 2}l +R$.

So we can conclude that for a mass $m$, on the outside of the spherical mass, the outward directed centrifugal force is bigger than the gravitational force caused by the other mass. Which implies an outward bulge develops.

We can apply the same procedure for the inside (the side that faces the other mass M) of the mass M, and that will show that the gravitational force from the other mass is bigger than the centrifugal force, which implies an inward bulge (towards the other mass) develops. I took two equal masses for simplicity, but you can do the same for two different spherical masses, circling around their CM. The result will remain the same (two bulges in opposite directions if both are covered with water; so if the moon was covered by a sea, also on the moon two bulges, on the line connecting the earth and the moon, would appear).

Looking at one mass I only had to compare the outward centrifugal force with the gravitational force caused by the other mass, so I didn't have to care about the force caused by the mass we were looking at, because they are equal everywhere (in approximation) on the surface of the mass, as is the case for the centrifugal force caused by an eventual rotation of the masses.

I didn't calculate if the bulges were different in size, but I guess they are.