The period of a man-made earth satellite orbiting the earth in a circular orbit may be () A. 60 minutes B, 80 minutes C. 180 minutes D, 25 hours Shouldn't it be more than 24 hours? I chose D, the answer is CD, why

The period of a man-made earth satellite orbiting the earth in a circular orbit may be () A. 60 minutes B, 80 minutes C. 180 minutes D, 25 hours Shouldn't it be more than 24 hours? I chose D, the answer is CD, why

At least 84 minutes, so CD
24 hours is a synchronous satellite cycle
By all possible satellites
The other 84 minutes are obtained by taking the radius of the orbit with the radius of the earth, which is the satellite on the surface of the earth
That is to say, the fastest period of a man-made satellite is 84 minutes and 24 hours, which is about 5.6 times the earth's radius from the earth. It does not mean that all satellites are synchronous satellites
The longer the orbit radius is, the longer the period is, so CD is correct

For a man-made earth satellite with mass m moving in a circular orbit, the distance from the satellite to the earth is equal to the radius r of the earth, and the acceleration of gravity on the ground is g, then the velocity of the satellite is?
mv^2/r=mg
G = root 2rg
Why not

What we should use here is the knowledge of gravitation, not the knowledge of circular motion
For objects near the earth's surface, if GM / R ^ 2 = g, we get GM = GR ^ 2
For artificial earth satellite, GM / (R + R) ^ 2 = V ^ 2 / (2R)
We get v = √ (GR / 2)

A man-made earth satellite with mass m is moving in a circular orbit. The distance from the satellite to the ground is equal to the radius r of the earth. It is known that the acceleration of gravity on the earth's surface is g and the constant of gravitation is g. ignoring the influence of the earth's rotation, we can find: (1) the mass of the earth; (2) the period of the satellite's circular motion around the earth

(1) Let the mass of the earth be m. since the mass of the earth's surface is M0, the universal gravitation is equal to the gravity of the object, so: gmm0r2 = m0g, the solution is: the mass of the earth M = gr2g. ① (2) the universal gravitation provides the centripetal force for the circular motion of the satellite, then: GMM (2R) 2 = m2r (2 π T) 2 ② simultaneous ① ② get: T = 2 π 8rg