I was reading about a rover concept ' Hedgehog ' which operates by spinning flywheels very fast and then braking quickly which transfers the momentum thereby causing the rover to move by tumbling. My question is so how can this method be effective( what am I missing? ) as by transferring momentum the rover would just fly off( as the gravitational force exerted by the asteroid on the rover is very small).

YouTube video on the working of the rover ( 2 min 11 sec)

An article by JPL NASA on the hedgehog rover

  • 1
    $\begingroup$ That depends on how much momentum you transfer. You just have to keep the vertical speed below the escape velocity from the asteroid. $\endgroup$
    – Hobbes
    Dec 7 '15 at 8:07
  • $\begingroup$ A similar concept is already on the way to an asteroid, Mascot with Hayabusa 2. I suppose there is a practical limit for how small an asteroid it works on. Some asteroids even spin themselves to pieces. $\endgroup$
    – LocalFluff
    Dec 7 '15 at 9:30

The problem with a wheeled rover on an asteroid is the difficulty of making the wheels obtain a proper grip. A good grip is on Earth obtained by the vehicle experiencing a down-force towards the ground, provided by the gravity. When high acceleration is needed, for example in race-cars, additional down-force is provided by a spoiler.
When the wheels pushes of from the ground, a similar force is experienced by the vehicle. This is causing it to turn. With almost no gravity to hold it back, it will continue to turn, eventually falling over. That requires the rover to have wheels on all sides, complicating the design.

The hedgehog design is designed to tumble, and the lack of wheels make it possible to have vulnerable moving parts within a protective hull. The hedgehog is also capable of stopping quickly, simply cancelling out the tumbling with the flywheels. Braking with wheels is again depending on traction with the ground.

However, the tiny gravity of these small objects are limiting the rover's maximum velocity, regardless of design. For example, the velocity of an orbit around the smaller of Mars' moons, Deimos is less than 4 m/s. If the rover is travelling over this velocity, it will be orbiting and refuse to fall back to the ground.


The 'Hedgehog' robot is designed to explore an asteroid by the use of flywheels. The concept behind the use of the flywheel is to use the kinetic energy it produces and repurpose it for the task of its unique mode of transport.

For the robot to return to the comet, we must refer to Newton's First Law:

The first law states that if the net force (the vector sum of all forces acting on an object) is zero, then the velocity of the object is constant. Velocity is a vector quantity which expresses both the object's speed and the direction of its motion; therefore, the statement that the object's velocity is constant is a statement that both its speed and the direction of its motion are constant.

Newton's First Law also applies to space. We take into example of how rocket's thrusters guide it to a specified target. Therefore, we can assume that the flywheels are designed to move the robot to its destination by upsetting its balance of equilibrium in the middle of its brief flight in space.

  • $\begingroup$ Nope. There is a way to use flywheels to also change orbit and not merely attitude around shallow gravity wells by using orbit's gravity gradient, but that only works if there's no hyperbolic excess velocity to remove, such as when flywheels produce too much torque and the vehicle bounces off the surface at or over its escape velocity (as per the question). Newton's first law and flywheels that are inertial w.r.t. the primary body (can't transfer momentum) do nothing to stop that from happening, but gravity might. Again, assuming the fling didn't cause the vehicle to reach escape velocity. $\endgroup$
    – TildalWave
    Dec 16 '15 at 3:55

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.