0
$\begingroup$

I'm familiar with the Newton's Cannonball thought experiment, but I've never really understood why the cannonball doesn't eventually spiral inwards towards earth, even if it takes 10^100 orbits. In the thought experiment, the cannonball is launched with a high enough velocity so that it falls to earth, but keeps missing so ends up at the same place.

But how is it possible to launch an object around the earth so it ends up at the same place? There is always a force acting on the cannonball which is perpendicular to its velocity, so even if the cannonball was given a very high velocity, surely it would have moved closer to the earth when it returns to the position from which it was launched, even if it was only by a fraction of a millimetre?

So over many many years does the cannonball spiral towards the earth? (ignoring the atmosphere and assuming the earth and cannonball are perfectly spherical)

$\endgroup$
0

2 Answers 2

1
$\begingroup$

There is always a force acting on the cannonball which is perpendicular to its velocity

The only way to change the magnitude of a velocity (i.e., the speed) is by applying a force with some nonzero component parallel to the velocity. This is expressed by Newton's second law.

A force applied perpendicular to the velocity changes its direction but not its magnitude. This is the essence of the Newton cannonball idea: the direction changes at the same angular rate as the Earth's surface beneath it, maintaining a permanent circular orbit.

$\endgroup$
7
  • $\begingroup$ Thanks for your answer, however I am still confused. Since even if the magnitude of velocity was to remain constant, why does this mean that the cannonball wouldn't end up at a lower position, since all that is needed is for this to happen is for the cannonball to gain a velocity component directed towards the centre of the earth, which is exactly what gravity will do. $\endgroup$ Commented May 9, 2023 at 0:22
  • $\begingroup$ As I wrote, in the case of a circular orbit, gravity changes the direction precisely as fast as the Earth's surface changes direction underneath. I would try making some sketches of the changing magnitude and direction of the velocity with a "flat" Earth and a spherical Earth. $\endgroup$ Commented May 9, 2023 at 0:27
  • $\begingroup$ @cookiecainsy It's basically always falling like you say, but it's moving so fast that it's continuously falling "over the edge" so it never hits the ground. Sort of imagine shooting off the edge of a radiused cliff such that it travels parallel to the curvature of the radius as it falls off the cliff. Now make the horizontal part before the cliff the same radius. Then make the vertical part of the cliff the same radius so it's a complete circle. $\endgroup$
    – DKNguyen
    Commented May 9, 2023 at 3:21
  • $\begingroup$ Chemomechanics said "A force applied perpendicular to the velocity changes its direction but not its magnitude". I'm finding that difficult to understand to be honest. If a point mass was moving at velocity 'v' and then a force is applied perpendicular to 'v' for some period of time, surely there will be a velocity vector perpendicular to 'v'. If you then added the 2 instantaneous velocity vectors (that are perpendicular to each other) , wouldn't one get a resultant vector that has a greater magnitude of 'v'? $\endgroup$
    – Dubious
    Commented May 9, 2023 at 12:34
  • $\begingroup$ I should have said a "resultant vector that has a greater magnitude than the original velocity 'v'. $\endgroup$
    – Dubious
    Commented May 9, 2023 at 12:42
0
$\begingroup$

Other than the fact that forces perpendicular to velocity always only ever provides centripetal acceleration and thus does not make it spiral inwards, as is in Chemomechanics's answer, you can also consider many other versions of similar answers.

You can also look at conservation laws. There are angular momentum conservation and energy conservation. Angular momentum conservation means that if it comes in closer, then it needs to speed up. But the force is always perpendicular to velocity, so it can neither speed up nor slow down, in the circular orbit, and so it cannot move closer or farther from the orbit centre.

Energy conservation says the same thing. Getting closer loses gravitational potential energy, which goes to kinetic energy, and vice versa.

Or you can look at it as a closed Keplerian orbit, and the eccentricity, or LRL vector, is conserved too. So it cannot change the shape of the orbit it is running along.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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