For a man-made earth satellite with mass m in circular orbit, its distance to the ground is equal to the radius r of the earth, and the acceleration of gravity on the ground is g, Then () A. The linear velocity is root 2rg B. The period is 4 π times the root sign 2R / g C. The acceleration is 1 / 2 times G D. The angular velocity is 1 / 2 times the root sign g / 2R

For a man-made earth satellite with mass m in circular orbit, its distance to the ground is equal to the radius r of the earth, and the acceleration of gravity on the ground is g, Then () A. The linear velocity is root 2rg B. The period is 4 π times the root sign 2R / g C. The acceleration is 1 / 2 times G D. The angular velocity is 1 / 2 times the root sign g / 2R

GMM / (2R) ^ 2 = MV ^ 2 / 2R = M4 π ^ 2 2R / T ^ 2 = m ω ^ 2 2R can be obtained respectively: v = √ (GR / 2) a wrong t = 4 π √ (2R / g) b right a = g / 4 C wrong ω = 1 / 2 √ (g / 2R) d right BD

For a man-made earth satellite with mass m moving in a circular orbit, the distance from the ground is equal to the radius r of the earth, and the mass of the earth is m
(1) The expression of satellite velocity
(2) What is the period of satellite motion?

GmM/4R^2=mv^2/2R
So v = √ (GM / 2R)
T=2*π*2R/V=4πR/√（GM/2R）

For a man-made earth satellite with mass m moving uniformly in a circular orbit, the distance from the ground is equal to the radius r of the earth, and the acceleration of gravity on the ground is known to be g? (2) What is the cycle of the satellite in circular motion?

(1) Let the mass of the earth be m. for an object of mass M0 on the surface of the earth, the universal gravitation is approximately equal to the gravity, gmm0r2 = m0g ① The centripetal force provided by gravitation is GMM (R + R) 2 = mv2r + R ② The solution is: v = Gr2 ③ (2) from the formula of circular motion & nbsp; t = 2 π (R + R) V: T = 4 π 2rg A: (1) the linear velocity of the satellite in circular motion is Gr2, (2) the period of the satellite in circular motion is 4 π 2rg