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CALCULATING PLANETARY ORBITS ABOUT THE SUN
One of the earliest and most significant contributions of Newtonean mechanics was the verification of Kepler’s three laws of planetary motion. We want here to briefly go through the mathematics which allowed Newton to derive the properties of planetary motion about the sun. Our starting point is the following schematic-
We have here a planet of mass m moving in an orbit about the sun of much larger mass M. The polar coordinates used are the radial distance r between the centers of the two masses and θ the angle r makes with respect to the symmetry axis x. In terms of Newton’s second law and the universal law of gravitation one has the two governing equations-
( ^ 2 ) 2 and m ( r 2 r ) 0
r
GMm mr r
The second of these equations is just a conservation of angular momentum statement and is equivalent to saying –
h r^2 const.
This last result also says that the area swept out in unit time by a planet moving about the sun is h/2 (Kepler’s 2nd^ Law). Eliminating vθ from the first equation above we have-
3 2
2
r
GM
r
h r
Integrating once we get-
2
2 (^2) Const r
GM
r
h r
If we multiply this result by m we have the conservation of energy statement -
mE r
GMm v v m ( r ) 2
2 2
where E is the constant total energy of the planet per unit mass and –GMm/r the potential energy. Dividing this equation by m/2 and re-substituting for vθ yields-
E
r
h r
GM
v (^) r 2
2
2 2
Next letting u=1/r and noting that vr=-h(du/dθ), one finds-
( )^2 ^2 ( )^2
u d
du
, where α=GM/h^2 and β= sqrt[α^2 +(2E/h 2 )]. We can integrate this last result once more to get-
^2 ( )^2
u
du
Now recalling from the integral tables that-
2 2 cos^1 ( ) a
x a x
dx (^)
we find that the planet trajectory is given by the conic section-
1 e cos( )
r
where Δ=1/α^2 =h^2 /GM and e=β/α=sqrt[1-2h 2 |E|/(GM) 2 ] the eccentricity of the conic section defined by this last equation. The constant in the angle has been adjusted so as to make the x axis a symmetry axis and the near point(perigee) from the central mass M occur when θ=π.
When e<1, this trajectory will be an ellipse (Kepler’s 1st^ Law) whose eccentricity is given by-
p rp
GMm r
mh 2
2
2
This is equivalent to saying that the speed at perigee will be-
r p
GM
v
^
This last result is recognized as the escape velocity from mass M by a smaller mass m. For the case of escaping from the earth to infinity one has GM=gR^2 and rp =R the earth radius. Hence to escape from the earth’s surface will require an effective speed of-
v (^) 2 gR 2 9. 8066 6. 371 11. 2 km /sec=36.7 thousand ft/sec
To get out of the solar system will require a much higher speed than this because of the sun much larger mass. Note that the speed of a satellite in a near earth circular orbit will be less by a sqrt(2). In other words 7.9 km/sec or about 26 thousand foot per second for the earth.
When e>1 the trajectory will be in form of a hyperbola. Casting the r=Δ/[1-ecos(θ)] equation into Cartesian coordinates when e>1 produces a standard hyperbola-
2 2
2 2
2
e
a where b a b e
y e b
x b
and ‘a’ is the distance from the directrix , which crosses the negative part of the x axis at right angles, to the center of M. Here x=a+rcos(θ) , y=rsin(θ), and the eccentricity is e=r/[a+rcos(θ)].
