For circular orbit, because the gravity on the satellite just provides the centripetal force of satellite motion, it is convenient to solve the physical quantities such as speed, acceleration and period of satellite motion in circular orbit. But for elliptical orbits, it is relatively difficult to solve some problems. Here, several problems of satellite elliptical orbit are analyzed and explained one by one. First, the radius of curvature of any point on the ellipse. According to mathematical knowledge, the radius of curvature is given by formula 3222)xyryxxy? (For the convenience of derivation, the first and second derivatives of X and Y are substituted into the R expression with the help of parameter equations cosxa and sinyb (A and B of an ellipse are the semi-major axis and semi-minor axis of the ellipse respectively, and there are 322,222 SINCOS). At apogee and perigee, the parameter φ takes 0 and 0 respectively. Substitute and get (,0)a on the ellipse? The radius of curvature of these two points is the same, which is equal to 2ba, but not equal to ac? Or ac? Where c is the focal length of the ellipse. The requirement of this knowledge point for mathematical ability has exceeded the requirement of senior high school, but it is necessary to introduce its conclusion appropriately. Example 1: A satellite orbits the earth in an elliptical orbit. The distance from perigee to the center of the earth is C, and the distance from apogee to the center of the earth is D. If the speed of the satellite at perigee is cv, what is the speed dv of the satellite at apogee? Analysis: the radius of curvature of the orbit of the satellite in elliptical motion is the same at perigee and apogee, both equal to R. Then, the perigee has 22cvMmGmcr? Apogee 22dvMmGmdr? , the above two formulas are compared to get cdvdvc? , so 2dccvvd? . The student's error-prone solution is: the gravity of the satellite provides centripetal force. At perigee, there is 22cvMmGmcc? Apogee 22dvMmGmdd? , the above two formulas are compared to get cdVdVc? , get dccVVd? The above error is that the radius of curvature of the orbit of the satellite in elliptical motion is different at perigee and apogee, which are C and D respectively. Knowing the concept of ellipse curvature radius will not make this mistake. The second is the acceleration and centripetal acceleration of the satellite moving to any point in the elliptical orbit. According to Newton's second law, the acceleration of a satellite moving to any point in an elliptical orbit is given by the formula 2MmGmaR? Where r is the distance from the center of the earth sphere to the satellite, that is, the distance from the focus of the ellipse to the satellite. When the satellite moves in a circular orbit at a constant speed, all gravity is used to provide centripetal force. At this time, the acceleration of the satellite is centripetal acceleration, while the satellite moving in elliptical orbit is not used to provide centripetal force, and the centripetal acceleration will no longer be equal to the acceleration of the satellite moving in orbit. What is the centripetal force of a satellite moving at a certain point in orbit? Where r is the radius of curvature of the elliptical orbit where the point is located, centripetal acceleration nnFam? At apogee, the satellite is attracted by the earth? Where r is the distance between the satellite and the center of the earth. What is the centripetal force required for satellite movement at this time? ，rR？ And GnFF? At this time, the acceleration of the satellite is equal to centripetal acceleration, that is, naa? After that, the satellite makes centripetal motion near the earth under the action of gravity, which has two functions. On the one hand, it provides tangential force along the orbit and does positive work on the satellite, making the satellite speed higher and higher. On the other hand, it provides centripetal force and constantly changes the direction of the satellite. The tangential acceleration caused by gravity is a? And the normal acceleration, that is, centripetal acceleration na, is shown in figure 1. When we get to perigee, GnFF? ，naa？ After the satellite is far away from the earth, gravity also produces two effects. On the one hand, it provides tangential force along the orbit and does negative work on the satellite, making the satellite speed smaller and smaller. On the other hand, it provides centripetal force, constantly changing the direction of the satellite until apogee, and so on. During the whole movement, there are only two positions, perigee and apogee, GnFF? ，naa？ Other positions naa? .