Question on velocity of an object orbiting a planet.

I’m pretty sure the answer is 1m but I wanted to know if it was more complex than that or that I was flat out wrong.

vf^2 = vi^2 + 2as where a is acceleration towards centre of earth s is distance.
I think we ought to find the pull exerted by the earth on that little ball by using newton gravitational law. From there we can find the acceleration thus the initial velocity when the ball enters the earth.
Then reuse the formula again using a=9.81 assuming gravity remains constant throughout the fall till it hits earth and use s=10

@mea05key Given the diameter of the Earth (and thus the distance between the centers of mass of the Earth and the ball), I would stipulate that a drop of that magnitude would make gravity effectively constant unless you go out to more decimal places than I would nless I was really bored.
Figure that G=~9.79m/sec^2 on top of Mt. Everest (~0.28% less than the 9.8226m/sec^2 at sea level) and that the difference between there and sea level is 8,850 meters, I think that we can ignore the change in gravity on a 1 meter drop in a classroom setting unless one of the students is really curious.

@quilm Gravity will actually pull the ball down 11 meters overall, so I would use that number. If it weren’t for the curvature of the Earth, the horizontal distance traveled would be much shorter.

@mea05key
Yeah I miss typed vf^2 = vi^2 + 2as *

What’s the equations Newton came up with for gravity? That states double the distance away gravity is half the force.

f = G[ m1 x m2/ r ^2]

@quilm Actually, it states that double the distance cuts it to one-fourth, or 1/(2^2) while tripling the distance cuts the force to one-ninth ( 1 / (3^2) ).

That is what the r^2 portion of the equation that @mea05key wrote above means.

FYI – The mean radius of Earth is 6371km. Add another 8.85km for the tallest mountain and you can see why gravity is fairly constant on the Earth’s surface and for object in low orbit, and why I say that 1m is pretty insignificant.