I need help with orbital speed? 11 points ASAP

If the distance of orbit (of a satellite orbiting earth ) is multiplied by 17 times, by how many times is the orbital speed v multiplied? Assume circular orbit.
I know v is the square root of (GM/r), and I tried to plug in the numbers (6.67*10^-11)(5.97*10^24)/17*(6.38*10^6)
However, I can't get it right. I'm sure I'm not putting in the right numbers,please help!!

A body stays in orbit when g = V^2/R = Ar the radial acceleration. And g = GM/R^2. So we have GM/R^2 = V^2/R and V^2 = GM/R. You got that part right, but here's the tricky part.

So here's how we compare the same factors at different values...ratios.

Let V^2 = GM/r and v^2 = GM/R = GM/17r then the ratio (v/V)^2 = GM/17r//GM/r = 1/17 so that v = V sqrt(1/17) = 0.242535625 V = .242 V. Thus, when the orbit goes from r to 17 r, the orbital speed is reduced to about 1/4 of the original speed. ANS.

Do you see the advantage of using ratios? G and M cancel out, we don't need them. And when the new orbit is 17 X r where r is the old orbit, even the r cancels. And we're left with 1/17 and v^2 = 1/17 V^2.

Call G = 1, M = 1, original r = 1 , then v = 1
Calculate with r = 17, then v = 0.242536
So , calculate original velocity * 0.242536 = new velocity