Exercise: the cannon ball experiment (step 1.9)
This exercise is about finding the magnitude of the velocity that is necessary to put a cannonball in orbital motion around the Earth, at an altitude equal to h. To that effect, a cannonball is fired from a height h above the ground at a velocity v° parallel to the ground.
A few assumptions are made:
- The Earth is a perfect sphere of radius R = 6370 km.
- The ball is in free fall (the only external force acting on the ball is its weight).
- The gravitational field has a constant magnitude g and is pointing towards the center of the Earth.
The motion of the ball for small duration Δt is a combination of (see Figure):
- A horizontal motion of magnitude X.
- A vertical motion of magnitude Y.
1°/ a) Show that X = v° × Δt.
b) Show that X = (R+h) × sin(α)
c) Use the previous answers to write v° in terms of R, h, α, and Δt.
2°/ a) Show that after Δt, the ball will only remain at the same height above the ground if the following condition is met:
(R + h) × cos (α) + Y = R + h.
b) Use the second law of Newton to show that Y = ½ g (Δt)2.
c) Use the previous answers to write cos(α) in terms of R, h, g, and Δt.
If Δt is not too big (less than one minute), the distance travelled is very small compared to the circumference of the Earth. Therefore, we can assume that the angle α is very small relative to 1, in which case a very good approximation is: cos(α) = 1 – α2/2 and sin(α) = α.
3°/ Show that α² = g Δt² / (R + h).
4°/ a) Use the answers to 1) and 3) to show that vº = √[g(R+h)].
b) Deduce that, as long as h is much smaller than the Earth radius, v° depends only on g and the Earth radius R, and compute v° using g = 9.8m.s-2 .
5°/ Compare the value you found to the one mentioned in the video.