There is one force only, gravity which pulls the moon in the direction of earth. The above formula gives the orbital velocity for an orbit at a distance $r$ from the center of the earth.Įssentially you are right, but your wording is slightly off. We used $g$ as the gravitational acceleration at the surface of the earth. We can rephrase the previous sentence as the ball should hit the orbit of radius $R$. The surface could have been a virtual one, i.e: any circular orbit. Have we used the fact that $R$ is the radius of the surface anywhere other than for using $g$? All we said was that the ball shouldn't hit the surface. Well, the surface of the earth needn't really be the surface of the earth. However, the above formula can be generalized easily. I had avoided deriving a formula for a general case where the object is a distance $r$ from the surface of the earth to avoid lengthening the answer. If you have studied gravitation, you'd immediately notice that the above formula gives the orbital velocity of objects orbiting the earth near the surface. Therefore, $x$ is the horizontal velocity. We had earlier assumed that $x$ is the distance traveled by the ball in one second. We can write the same quantity in terms of $g$ on the surface. ![]() We had considered $y$ as the distance the object would travel in one second. Solving for $x$ in terms of the other variables, you get, To be honest, in the limit where $y$ tends to zero, this is not an approximation. I could have made use of calculus to derive but it isn't really needed as it would unnecessarily complicate the answer.
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