To illustrate what this means, let the big circular curve _BʹAʹCʹ_ in fig. 2 represent the earth’s surface, and imagine that a shot is fired horizontally from _A_, the top of an elevation _AAʹ_. If the shot were not pulled earthwards by gravitation, it would travel indefinitely along the line _AB_ out into space. If _AB_ is the distance it would travel in a second under these imaginary conditions, the end of a second’s actual flight does not find it at _B_, but at a point 16 feet nearer the earth, gravitation having pulled it down this 16 feet during its flight. For instance, if _BB′_ in fig. 2 should happen to be 16 feet, the shot would strike the earth at _Bʹ_ after a flight of precisely one second. As another example, let us suppose that the 16-foot fall below _B_ does not drag the shot down to earth but only to a point _b_, which is at precisely the same height above the earth’s surface as the point _A_ at which the shot started. If gravitation were not acting, so that the shot travelled along the line _AB_, its height above the earth would continually increase. Actually in the case we are now considering, gravitation pulls the shot down at just such a rate as to neutralise the increase of height which would otherwise occur, so that the height of the shot neither increases nor decreases; it neither flies off into space nor drops to earth, but continues to describe circles round the earth indefinitely. A simple geometrical calculation shews that for the distance _Bb_ to be 16 feet, the distance _AB_ travelled in one second must be 25,880 feet or 4·90 miles[2]. Thus if we could fire a shot horizontally with a speed of 4·90 miles a second, it would describe endless circles round the earth, the earth’s gravitational pull exactly neutralising the natural tendency of the shot to fly away along the straight line _AB_. [2] Let _C_ be the centre of the earth, and _bCD_ the diameter through _b_. Then _BA_² = _Bb_ × _BD_, where _Bb_ = 16 feet, and _BD_, which is 16 feet more than the earth’s diameter = 41,900,000 feet. From this we readily calculate that _BA_ = 25,880 feet. This calculation of course neglects the height of the hill _AAʹ_ by comparison with the earth’s diameter.