I have a question regarding the Newtonian claim that a bullet fired at the exact escape velocity of Earth would indefinately follow an orbit around it, without ever touching the ground. The solution given is that, as the bullet travels, the Earths curvature causes the ground to 'continuously bend away', leaving the bullet no other option as to continuously free-fall. Then here's the question: if gravity is considered to be an acceleration (app. 10 m/s^2), wouldn't that bullet keep accelerating up to the point where it surpasses the escape velocity? Terminal velocity notwithstanding, since most of the thought experiments involved usually consider a 'perfect' sphere with no atmosphere ('perfect' vacuum)... After all, when simply dropping a bullet towards the ground, its speed is determined by the height from which it is dropped, the higher the 'drop-point', the longer the bullet is 'influenced' by gravity (ie, the longer the gravitational force can act upon it, and thus the higher its speed): a bullet dropped from 10 metres high (in a vacuum), takes 1 second to reach the ground, and if dropped from 30 metres, it'll take 2 seconds (since its speed would be 20 m/s during the second time-interval)... Or am I missing something here?

I was trying to figure this out, and tried to calculate how long it would take a projectile to cover a distance of 22 kms, given that it cannot travel faster then 11,2 km/s - which is the escape velocity at the surface of the Earth, and if it were travelling faster then that, it would leave the Earths gravitational pull and never hit its surface... So I calculated the 'extra curvature drop' to be an additional 38 metres (on top of the initial 10 metres; that was the initial gedanken experiment: drop a bullet from 10 metres high and at the same time fire a bullet from the same height: both bullets will hit the ground at the same time, according to Newton). So the fired bullet would need to drop 48 metres, but I can't see how it can do that in the 1 second it takes the dropped bullet to reach the ground, not without surpassing the escape velocity?

In case you're wondering, I got the 38 metres by taking the 22 km it would travel, divided it by the app. 6378.136 km of the Earths radius: that gives app. 0.0034492836 (this would be the cosinus if the Earth can be seen as a circle with radius equal to 1 - so 22 km would be 1/0.0034492836th part of that radius).

Sin(a) = sqrt(1 - cos(a)^2), so the difference in height due to the curvature should be app. 99.999405120 % of the radius, or app. 38 metres.
(I think )...