OA=r 1，OB=r2
Kepler's first law: the orbit of a satellite is elliptical and the earth is the focus.
Semimajor axis of ellipse a = (r 1+R2)/2.
The length of minor axis B2 = [(r1+R2)/2] 2-[(r1-R2)/2] 2 = r1R2.
Ellipse area: pi ab
Kepler's second law: the area swept by a satellite per unit time is equal. This is actually the conservation of angular momentum: p=v 1r 1=v2r2.
The swept area per unit time is: 1/2vr.
Kepler's third law: the ratio of the cube of the long axis of the orbit to the square of the period is constant. We can combine the first two laws with the conservation of energy to get the proportional constant (of course, we can also use it as a law to directly solve the circular orbit).
Conservation of energy: e =1/2mv12-GMM/r1=1/2mv22-GMM/R2.
p^2( 1/r 1^2- 1/r2^2)=2gm( 1/r 1- 1/r2)
p^2( 1/r 1+ 1/r2)=2gm
p^2 a=GMr 1r2
According to Kepler's first and second laws, the period is T=pi ab/( 1/2rv).
So t2 = pi2a2r1r2/(1/4p2).
=4pi^2 a^3/(GM)
So a 3/t 2 = GM/(4π2), which is the proportionality constant.
Direct calculation from circular orbit:
GMm/R^2=mR(2 pi/T)^2
A3/T2 = GM/(4π2), and the proportional constant is also obtained from a special case.
The following are specific stress states:
The magnitude of gravity is
F 1=GMm/r 1^2
F2=GMm/r2^2
If you mean the radius around centripetal acceleration, that's the distance to the center of the earth.
Centripetal acceleration is a 1 and a2 respectively, then
MB 1 = mv 1 2/r 1 = MP 2/r 1 3 = 2g MMR 2/(r 1+R2)/r 1 2 > f 1，000。
Ma1= MP 2/R23 = 2gmmr1/(r1+R2)/R22 < F2, the gravitational force is greater than the centripetal force, so the orbital radius at this point will be reduced.
Meanwhile, b1:a1= r23: r13.
After the second acceleration, it is a circular orbit, and gravity just provides centripetal force. ma2=F2=GMm/r2^2
A 1 < A2, and a1:a2 = 2r1/(r1+R2).
Regarding the radius of curvature of a point on an ellipse, this problem involves differentiation in universities. It can be proved that the radius of curvature of point B is a semi-major axis. The concrete proof is as follows: (Differential is required)
The elliptic equation x 2/a 2+y 2/b 2 =1,so XB 2+YY 'a 2 = 0.
The radius of curvature is defined as: R=ds/dθ, that is, the ratio of tangential angle change of arc length transformation.
θ = arctangent (x/y)
ds=( 1+y'^2)^0.5dx dθ=(x/y)'/[ 1+(x/y)^2]dx
Therefore, r = [1+(x/y) 2] [1+y' 2) 0.5]/(x/y)'
=y^2[ 1+(x/y)^2[ 1+(x/y b^2/a^2)^2)^0.5]/(-xx/y b^2/a^2+xy)
=(x^2+y^2)[y^2 a^4+x^2 b^4]^0.5/(x^2b^2+y^2a^2)
Replace the coordinates (a, 0) of point B with
R=a^2ab^2/a^2b^2=a
That is to say, the radius of curvature of point B is exactly the semi-major axis.
If centripetal acceleration you mean is v 2/ρ, ρ is the radius of curvature. This actually refers to the radial force of the object. In this problem, at point B, gravity has no component in the tangential direction, which naturally means that gravity completely provides centripetal force.
If the point is not on the two axes of the ellipse and the directions of force and motion are not perpendicular, there is tangential acceleration. The component of gravity perpendicular to the velocity direction provides centripetal force, and the radius of curvature can be solved by this conclusion, without starting from the above definition.