A planet with huge rotation speed but also huge gravity?

Question

Is it possible that a planet has a huge rotation speed and huge gravity so that at equator they mostly compensate for each other and the effective gravity is moderate (say, 1 g) but on poles the gravity is so huge that nobody can walk there?

Would spaceflight be as much possible from the surface of such planet as from Earth?

Answer 1

Physically I think such an object is possible. I would not know how a planet could gain such a high rotation velocity.

In the solar system massive planets tend to have a shorter rotation period than less massive planets (10 hours in the case of Jupiter).

Earth would have to spin with 11 km/s (escape velocity) for experiencing weightlessness at the equator resulting in a rotation period of 1 hour. The surface of Earth would be in geostationary orbit. A planet spinning with escape velocity though would be unstable and break apart.

Jupiter would have to spin 60 km/s for centrifugal force to equate g-force. To get 1 g at the equator rotation velocity would be something like 50 km/s or over 80% of its escape velocity resulting in a rotational period of 2 1/2 hours. That is only 4 times faster than Jupiter spins already.

Of course in reality Jupiter would lose all the hydrogen and helium by atmospheric escape, resulting in a smaller but denser planet, a sort of super-earth, so all parameters would change.

The form of a fast spinning planet would be ellipsoid rather than spherical.

We have flattened objects in the solar system, for example Haumea in the Kuiper belt. It rotates with an equatorial velocity of 400 m/s which is almost 60% of its escape velocity of 700 m/s.

Dwarf planets with rotation velocitiy approaching escape velocity obviously can form, although they are pretty rare. If and how a fast rotating super-earth - that is what you are asking for - can form I do not know.

"Would spaceflight be as much possible from the surface of such planet as from Earth?" Yes, from the equator.

"Is it possible that at the equator ... the effective gravity is moderate (say, 1 g) but on poles the gravity is so huge that nobody can walk there?"

A super-earth with a radius of 2 earth radii would have a surface gravity of over 3 g (says google) and would have to spin with close to 90% of its escape velocity to get down to 1 g at the equator, pretty close but - theoretically - still below break-up velocity. At the poles you would still experience 3 g, which means it would feel like you carry a bag on your shoulder twice the weight of your body. I do not think I could walk there.

Answer 2

Possible.

The acceleration of gravity exerted by a spherically symmetric distribution of mass on an object on its surface is given by the good old: $$g_{planet}=G\frac{M_{planet}}{R_{planet}^2}=\frac{4}{3}\pi G\rho_{planet}R_{planet}$$

And the centripetal acceleration at the equator is given by: $$a=R_{planet}(\frac{2\pi}{T})^2$$

So, it's quite simple really. We have 3 factors we can play with:

• Increasing the density of the planet $\rho_{planet}$ increases gravity everywhere
• Increasing the radius $R_{planet}$ increases gravity everywhere and increases the fictional centrifugal "force"
• decreasing the rotational period $T$ (faster rotation) increases centrifugal "force" (by a lot)

And the good news is that all those 3 factors can more or less be changed independently:

• size and density can be independent: Mercury and earth have similar densities but Earth is bigger
• size and rotational speed: earth and mars have almost the same length of day, but earth is bigger
• density and rotational speed: Jupiter is twice as dense as Saturn but they have almost the same rotational period

conclusion: you can tune any of those 3 parameters to get what you want.

To talk numbers: If you want the apparent gravity to be n times stronger at the poles than at the equator, while maintaining 1g of apparent gravity at the equator, you can work out that you must satisfy the following relations: $$R_{planet}=\frac{ng}{\frac{4}{3}\pi G\rho_{planet}}\text{ and }R_{planet}(\frac{2\pi}{T})^2=(n-1)g $$

We're gonna want a pretty big planet (that helps in reducing the revolution speed that you will need). I will make it not too dense so that it gets a larger radius, I will take $\rho_{planet}=690kg/m^3$ (density of Saturn). I will make it so that you get 2g at the pole and 1g at the equator. With the formulae above, we can then calculate the required radius and revolution period: $$R_{planet}=76300km$$ $$T=6h30$$

That's a pretty fast rotational speed, but it doesn;t seem implausible (the shortest day in our solar system is about 10h. You're less than twice as fast, seems reasonable to me)

So with those conditions, you would be twice as heavy at the poles. Probably difficult to walk for an untrained human