We first derive a general solution of a two-body problem subject to the universal gravitational attraction (Newton’s Law of Universal Gravitation), and then consider special cases of interest for our purpose.
1.4.1. Equation of Motion under Newton’s Law of Gravitation
Using the coordinates shown in Figure 1.2, Newton’s Law of Gravitation implies that the two bodies of masses M and m, and the radius vectors r ⃗_M and r ⃗_m , respectively, would exert the following forces of mutual gravitational attraction:
Figure 1.2. A two-body interaction coordinate system.
Two masses M and m at the radius vectors rM and rm , respectively.
Force on m: F ⃗_m = -m( r) ̈_(m )= -- GMm/r^2 r ⃗/r (1.5a)
Force …show more content…
Then in terms of the Cartesian coordinates (x,y,z), r = 〖〖(x〗^2+ y^2)〗^(1⁄2) ; and with unit vectors i ̂,j ̂,k ̂, along x-, y-, and z- axes, respectively,
r ⃗_o=cos∝ i ̂+sin〖∝j ̂ 〗 1.6a)
The unit vector r ⃗_p perpendicular to r ⃗_o , and in the direction of increasing ∝ is:
r ⃗_p=k ̂ x r ⃗_o= -sin〖∝i ̂ 〗+cos∝ j ̂ (1.6b)
Differentiating equation (1.2a) and (1.2b) with respect to time:
(r ⃗_o ) ̇=-sin∝ ∝ ̇i ̂+cos∝ ∝ ̇j ̂ = ∝ ̇ r ⃗_p (1.7a)
(r ⃗_p ) ̇=- cos∝ ∝ ̇i ̂-sin∝ ∝ ̇j ̂ = - ∝ ̇ r ⃗_o (1.7b)
Differentiating r ⃗ = rr ⃗_(o ), the velocity of m could be written as:
v ⃡= r ⃗ ̇=r ̇r ⃗_o+r(r ⃗_o ) …show more content…
Hence the equation of the orbit can be written as (Equation 5a).
r=p/(1+e cos〖(∝-∝_o 〗)) (1.11b)
where p = A^2/β (1.11c) and e=A^2/β B = p B (1.11d)
Equation (1.11d) is the polar equation of a conic section with ρ as the latur-rectum and e the eccentricity, with M as the focus, and the orbit of m being: (i) a circle with radius p for e=0 (ii) an ellipse for e<1 (iii) a parabola for e=1 (iv) a hyperbola for e>1
The cases of circular (e = 0) and elliptical (e<1) orbits are of most interest for a satellite orbiting around the earth. The semi major and semi minor axes of the ellipse, a and b, respectively, are given as: