It's a thought experiment which I've been running in my mind for long in order to understand tidal forces. Any suggestions/ideas are appreciated.

Assume a perfectly spherical planet with no atmosphere.
The planet is not in a translating motion, but it is rotating.
And we have a stationary rocket with infinite supply of fuel using which in jet propulsion, it is maintaining a fixed altitude over the planet. Jet propulsion is needed for counteracting the downward gravitational force.

The question is how the gravitational tidal forces will affect the rocket.
Will the rocket stay at its place forever while the planet keeps on rotating underneath it?
Or, will it start revolving the planet matching the period of the planet?
Will it start to have any sort of rotational motion of its own?

Note: this rocket has a considerable mass and size, and is orientable (we can tell which way it is facing). Further, we have millions of years at our disposal for this thought experiment.

  • $\begingroup$ "Jet propulsion is needed" - I don't really think that's essential to your question; what you're describing is just a spaceship that's in orbit. Now, could you say more about what aspect of tidal forces you're trying to figure out (what you're hoping understanding this scenario will clarify)? Tidal forces arise due to difference in gravitational pull on the near and the far side of the object, so you'll probably get more insight if, instead of a rocket, your spaceship was something huge and spherical (but orientable) - a Death Star type of thing. $\endgroup$ Jul 7, 2021 at 21:46
  • $\begingroup$ Have you studied frame dragging? en.wikipedia.org/wiki/Frame-dragging $\endgroup$ Jul 7, 2021 at 22:07
  • $\begingroup$ @FilipMilovanović I added jet propulsion to get rid of any initial rotation or revolution of the body (rocket), so that it is easier for me to visualize what exactly happens to it. Yes, this rocket has a great mass and is orientable. $\endgroup$
    – manisar
    Jul 8, 2021 at 0:43
  • $\begingroup$ @AdrianHoward In this question, I was not looking at any relativistic effects per se, but if anybody can throw light on it in addition to explaining the effects of tidal forces, that will be a bonus! I personally have a very vague understanding of frame-dragging at the moment. $\endgroup$
    – manisar
    Jul 8, 2021 at 0:45
  • $\begingroup$ Just to clarify, my remark that it's better if the spaceship is large in this scenario was not so much about its mass, but about its spatial extent, so that it's large enough that there's an appriciable difference in gravity it feels due to Earth on the far side (away from the Earth) vs the near side (facing the Earth), because that's what causes tidal forces. $\endgroup$ Jul 8, 2021 at 0:56

2 Answers 2


That's not what tidal "force" or tidal "drag" is. So, yes, the rocket will begin to rotate according to a distant observer, but that is because of a general relativistic effect called "frame dragging". The rocket, in its rest frame, remains stationary and doesn't rotate (around the planet). The problem is that spacetime is rotating around the planet relative to distant observers, so it appears to rotate according to them.

But that is general relativity and has nothing to do with tides.

If you want to understand tides, the 1st thing to understand is that it is not a "force". In Newtonian gravity, the field is completely described by the potential energy $\phi(r)$. A massive particle has energy:

$$ U(r) = m\phi(r) $$

So that couples a scalar mass to a scalar potential. The gradient of the energy leads to a vector force:

$$ F_i(r)=-m\frac{\partial\phi}{\partial x_i}$$

Linear motion results because energy now depends on a vector force coupled to a vector displacement:

$$ \Delta W =F_i\Delta x_i $$

With an extended mass distribution, the force on different parts of the distribution may be different, since $\nabla\phi$ is a function of position. This interaction is captured by coupling the quadrupole moment of the mass distribution to the tidal tensor:

$$ U_{\rm tide} = Q^{ij}\Phi_{ij} $$

The quadrupole tensor is the second moments of the distribution, with the trace (variance of the mass distribution) subtracted:

$$ Q_{ij}=\int\rho({\bf x})\big(3x_ix_j-||{\bf x}||^2\delta_{ij}\big)d^3{\bf x}$$

While the tidal tensor is related to the Hessian:

$$ \Phi_{ij}=J_{ij}-\frac 1 3 {\rm Tr}(J)$$


$$ J_{ij}=\frac{\partial^2\phi}{\partial x^i\partial x^j}$$

Note that in a vacuum: ${\rm Tr}(J)=0$. If mass is present, this term represents a volume shrinking stress. The remaining part is a volume preserving stress. In a radial field it stretches the object, as the part closer to Earth feels more gravity, and in free-fall (orbit), the top end is flung outward b/c centrifugal force. Meanwhile, in the 2 transverse directions, the fact that all forces point to a central point leads to compression with $\frac 1 2$ the magnitude of the stretching (but in 2 directions, so volume is preserved).

With that, your rocket is best modeled as 2 masses $m$ separated by a distance $d$. Align that along $z$ (vertical), and rotate it by $\theta$ in the $xz$ plane. The quadrupole moment is:

$$Q_{ij}=\frac{md^2}2\Big[ 3\left(\begin{array}{ccc} \sin^2{\theta}&0 &\sin\theta\cos\theta\\0&0 &0\\\sin\theta\cos\theta&0 &\cos^2\theta \end{array}\right)- \left(\begin{array}{ccc} 1&0 &0\\0&1 &0\\0&0 &1 \end{array}\right) \big]$$

$$Q_{ij}=\frac 3 2 md^2\left(\begin{array}{ccc} \sin^2{\theta}-\frac 1 3&0 &\sin\theta\cos\theta\\0&-\frac 1 3 &0\\\sin\theta\cos\theta&0 &\cos^2-\frac 1 3\theta \end{array}\right)$$

Meanwhile, a tedious calculation of the $1/r^2$ Hessian gives:

$$\Phi_{ij}= \frac{GM}{(x^2+y^2+z^2)^{\frac 5 2}} \left(\begin{array}{ccc} y^2+z^2-2x^2&-3xy &-3xz\\-3xy&x^2+z^2-2y^2 &-3xy\\ -3xz&-3xy &x^2+y^2-2z^2 \end{array}\right)$$

With that, you can show that your dumbbell shaped rocket wants to be vertically aligned, pointing up or down...it doesn't matter.

The configuration is stable. Tides don't apply a force, they apply a torque by coupling the quadrupole moment to the tensor gradient of the gravitational field.

In real world scenarios (like moons orbiting planets), one has to consider induced quadrupole moments. The same way an electric field induces a polarization in an otherwise spherically symmetric atom, the tidal tensor can deform a moon or planet. That induced Q-moment then couples to the tidal tensor, leading to various phenomena such has tidal locking.

  • $\begingroup$ Thanks, that's a very detailed answer and I'm in the process of absorbing it. If I've understood it correctly so far, if we talk about tidal effects only: 1. The rocket will be stationary and vertically (radially) aligned. In fact, if it was not aligned that way initially, it will soon. 2. It won't start to revolve the planet, nor rotate about any of its axis. If this is correct, this is contrary to what my incomplete understanding suggested me before I read your answer! $\endgroup$
    – manisar
    Jul 8, 2021 at 22:39

My answer supposes a planet with mass and rotation of the order of magnitude of the Earth, so that relativistic effects are small.

In this case, the situation is similar to an object on the planet surface without any friction. The only forces are gravity and Normal, both radial, so it is perfectly possible for the object to have a continuous range of tangential velocities, provided that it is smaller than the orbital velocity.

If we are at the planet equator, the velocity corresponding to the greatest Normal force is when an observer in the object sees the planet rotating with its angular velocity with respect to the distant stars. That means: the observer sees the stars static and the planet rotating behind.

For a rocket above the surface, the same principle applies. The engines must deliver the maximum force when the crew sees the planet rotating with its angular velocity with respect to the distant stars. The crew will feel this force as the normal force inside the ship.

We can even use this force as a criteria to verify if the planet is rotating or not, without the aid of the stars. It is not rotating if the maximum force corresponds to a situation when the rocket are always over the same spot of the planet.

The effect of the tidal force for this approximation is only a small correction in the compressive force acting on any object inside the ship. The gravity force is smaller far from the planet, so this compressive force is a bit smaller than if the gravitational field was contant.

  • $\begingroup$ Thanks for throwing light on some interesting aspects. I've read it over a few times, but am still not getting to the point where I can answer the questions that have troubled me: 1. Will the rocket stay at its place forever while the planet keeps on rotating underneath it? 2. Or, will it start revolving the planet (perhaps matching the period of the planet)? 3. Will it start to have any sort of rotational motion of its own? Or at least, will it try to orientate in a certain way? Any pointers in the direction of getting answers to these will be really helpful, kind sir. $\endgroup$
    – manisar
    Jul 10, 2021 at 17:45
  • $\begingroup$ The forces acting on the rocket (gravity and thrust) are both radial. If they have the same magnitude, the rocket stays forever at the same location, with the planet rotating behind. For any other case, the thrust have to be smaller than the gravity force. $\endgroup$ Jul 10, 2021 at 21:45

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