Asatellite moves in a stable circular orbit with speed vo at a distance r from the center of a planet. for this satellite to move in a stable circular orbit a distance 2r from the center of the planet, the speed of the satellite must be? ? i said that f=ma, but m doesn't matter since it's constant. so, a0=a1. a=v^2/r. so v1^2/2r = v0^2/r. i ended up with v1 = v0sqrt(2). but that's not the answer. all the multiple choice answers have sqrt's, 2's, and v0's scattered around, but none are what i have. what did i do wrong? ?

Distance covered, s = 8ft - 4ft = 4 ftinitial velocity, u = 34 ft/sthe acrobat is moving upwards and therefore, a = -g = -9.81 m/s^2 = 32.19 ft/s^2applying equation of motion; s = ut -1/2gt^2then,4 = 34t -1/2*32.19t^24 = 34t - 16.095t^2rearranging; -16.095t^2 + 34t - 4 = 0solving the quadratic equation; t = {-34 +/- sqrt [34^2 - 4(-16.)]}/2(-16.095) = 1.06 +/- 0.93 = 0.13 or 1.99the acrobat will take 0.13 seconds to be at 8 ft above the ground. additionally, once at the maximum height, the acrobat will move down and be at the same spot (that is height) after 1.99 seconds moving down.

F= (mass)(acceleration) = mam = 0.25 kgvi = 16 m/st = 2 svf = 0 m/s (since it was put to stop)a=(vf-vi)/ta=(0-16)/2a = 8 m/s^2 (decelerating)f = ma = (0.25 kg)(8 m/s^2)f = 2 nhope this !