We could induce artificial gravity through centripetal acceleration. For example a ring-like structure in a spaceship could rotate about 1.34 rpm if the radius to the centre is 500 meters. This will give 1 g at the edges of the ring.

However we can also induce artificial gravity in the spaceship through its propulsive power and hence a constant acceleration at 1 g of the spaceship is required.

But what if the spaceship is accelerated at more than 1 g in order to achieve 75% the speed of light, even though the spaceship consists of the 500m radius ring-like structure that rotates at 1.34 rpm. What type of artificial gravity will the occupants feel? Will the excessive acceleration cancel out the centripetal effects in the ring? Is there an equation or formula that combine both linear acceleration of an object while the objects is also rotating?

  • $\begingroup$ Note that it only takes about one year at an acceleration of 1g to reach v = 0.75 c. $\endgroup$ – pela Mar 8 '17 at 11:27
  • $\begingroup$ 1 year, 1 month, and 5 days, to be more precise :) $\endgroup$ – pela Mar 8 '17 at 11:43

Since you mentioned high velocities (close to speed of light), you can not get an answer to this question in the framework of classical-mechanics and newtonian gravity.

For this we need General Relativity.
In this context, inertial acceleration is equivalent to gravitational acceleration, so you can get the same effect using centripetal acceleration as if it would be a gravitational field.
We could also use this assumption to simplify a bit the problem and use Special Relativity instead, since the mass of our object is so small as compared to a small moon.

Anyways, each parts of the ship will experience an overall acceleration so you are right to assume that "the excessive acceleration cancel out the centripetal effects in the ring", but it will depend on the overall "acceleration" at each point of the ring.
However, you can in no way cancel an acceleration of let's say $1km/s^2$ with a rotating ring of radius $500m$ - that is : the human body will get squeezed (and with $1km/s^2$ -as measured by someone on the ship- you'll need 62hours to get to 75% of speed of light and tremendous amount of whatever fuel you use).

Since including angular momentum is a bit complicated and would require us to use perfect (symmetrical) radius, we can use instead the assumption that our object is made out of molecules (or maybe even bigger parts) and ran a simulation on this molecules that make up our ship and assign to each of this molecules a four-velocity vector.
If the ship has angular momentum it will integrated as the velocity vector of this many molecules so that we need to use only "simpler" relativistic mechanics. (or if the speeds are small as compared to $c$ just newtonian mechanics).

  • 1
    $\begingroup$ You don't need GR. SR will do. Also, a minor correction: 62 hours at a = 1 km/s to reach 0.75 c is the classical calculation; the correct, special relativistic derivation yields 94½ hours. $\endgroup$ – pela Mar 8 '17 at 11:48
  • $\begingroup$ it depends who sees 1km/s^2 and who measures it. If i'am on the ship and I measure 1km/s^2 ... than that's how i'll see it $\endgroup$ – Mihai B. Mar 8 '17 at 11:56
  • $\begingroup$ i'think you were thinking of $ d U^\mu/ d\tau $ as measured by an outside observer $\endgroup$ – Mihai B. Mar 8 '17 at 11:58
  • $\begingroup$ Hmm… what speed would you then measure after 94½ hours? $\endgroup$ – pela Mar 8 '17 at 12:33
  • $\begingroup$ By who's clock ? :) $\endgroup$ – Mihai B. Mar 8 '17 at 12:35

Assuming the plane of the rotating ring is parallel to the direction of acceleration.

The artificial $g$ from rotating ring is always outward on the inner side of the ring. So, its direction keeps changing even in the reference frame of rest of the space ship.

The artificial $g$ due to $1g$ acceleration is always in opposite direction of the acceleration - i.e backward (or downward if you will).

So, if you do both together, then the person on the ring

Will feel $2g$ once per rotation, when s/he is on the rear end of the ring.

Will feel $0g$ once per rotation, when s/he is on the front end of the ring.

In between, the felt g will be a vector sum of the two.

It would not be hard to come up with the equations but I am not going to do that here.

It appears that the equation should be similar to that of a swing that 1) has a frequency of half the rpm you mentioned, 2) goes full 90 degrees 3) is on a planet with $2g$ gravity.

Assuming the plane of the rotating ring is perpendicular to the direction of acceleration - In this case, there will be a constant acceleration of square root of 2 times $g$ in conical shape backward & outward.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.