enter image description here

Let's I have a Newton's Cannon ball of 15 kg and want to fire so it will round the earth and back to me.

My query is which law can evaluate the speed that cannon ball will revolve round the sun ? any hints or answer will be appreciated.


closed as too localized by Brandon Enright, Emilio Pisanty, David Z Jun 4 '13 at 18:04

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Most people would use all three of Newton's laws plus Newton's law of gravity, as well as some kinematical facts about circular motion. However, because this is at the earth's surface it's possible to do it using a much more minimal approach. Basically, find the horizontal velocity such that the rate at which the ground curves out from under the cannonball matches the acceleration of the cannonball. $\endgroup$ – Ben Crowell Jun 4 '13 at 4:46

Newton's Law of Gravity + centripetal force: $$F_g=\frac{GMm}{r^2}$$ $$F_c=m\frac{v^2}{r}$$ Combine them and solve for what you need (usually you're given $r$ and asked to solve for $v$).

Those equations are for uniform circular motion (which is what you would encounter in an intro to physics course). More accurately, you would need the Lagrange function to derive the equation of motion for the cannon ball as a function of $\theta$, and in the process rediscover Kepler's laws of motion. It's a little lengthy, but you can find a derivation in Anderson's Intro to Flight, Chapter 8 (Astronautics).

  • $\begingroup$ What will be the r? $\endgroup$ – Physics_guy Jun 4 '13 at 15:10
  • $\begingroup$ What about the angle? $\endgroup$ – Physics_guy Jun 4 '13 at 15:12
  • $\begingroup$ @Physics_guy, $r$ is equal to the distance of the cannon ball to the center of mass of the earth (probably slightly higher then the radius of the earth). And when calculating the velocity $v$, also keep in mind the rotational speed of the earth (initial velocity). $\endgroup$ – fibonatic Jun 4 '13 at 15:19

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