E and e  (Energy and eccentricity)

In this section, we use the notation:  r’ = dr/dt;  θ’ = dθ/dt;  F (bold blue) is a vector;  r (bold plum) is a unit vector.

For an inverse square Central Force, we have  F = (K/r2)r     where for gravity:
K = -Gm1m2   and for electricity:  K = kq1q2 .  Note that for an attractive force, K<0.   The general solution of Newton's Second Law for r(
q) is:

1/r  =  (-mK/L2) + A cos(q-qc)  =  B + A cos(q-qc)                       (Eq. 1)

where

B = (-mK/L2)   ,                     (Eq. 2)

where A and qo are determined from initial conditions, and L is the angular momentum (a constant) where L = mr2q' .

The total energy, E, is just potential plus kinetic:

E = (1/2)m[r'2+(rq')2] + V(r)   =   (1/2)mr'2 + Veff(r)

where the Veff(r) is the effective potential that incorporates the (1/2)mr2q'2 term:

V(r) = K/r;           Veff(r) = K/r + (1/2)mr2q'2 .

Using L=mr2q', or  q'=L/mr2  we get for Veff(r):

Veff(r) = K/r + (1/2)mr2[L2/m2r4]  =  K/r + L2/2mr2 .

At closest approach (when r = rmin) , r' = 0, so

E = Veff(rmin) =  K/rmin + L2/2mrmin2 ,

or

(L2/2m)(1/rmin)2 + K(1/rmin) - E  =  0 .

This can be solved for (1/rmin) by the quadratic formula:

(1/rmin)  =  [-K + {K2-4(L2/2m)(-E)}1/2] / 2(L2/2m)   ,

or

(1/rmin)  =  -mK/L2 + {(mK/L)2 + 2mE/L2}1/2         (Eq. 3)

where we have chosen the + sign in the quadratic formula, since this will give the smallest rmin.  The - sign would give the biggest value for r, which would correspond to rmax if one exists.

But from Eq. (1) above, 1/rmin = (-mK/L2) + A .  Comparing Eq. 1 and 3, we see that

A =  {(mK/L)2 + 2mE/L2}1/2   (Eq. 4)

We now define e as (with B = -mK/L2):

e  =  A /½B½   =  {(mK/L)2 + 2mE/L2}1/2 / ï[-mK/L2] ï,

or

e   =  {1 + 2EL2/mK2}1/2 (Eq. 5)

Recall that for a bound system, the energy E is negative.  For an unbound system, the energy is positive.  This means that for e<1 we have a bound orbit (circular, elliptical).  For e>1 we have an unbound "orbit" (hyperbola).  The case where ε=1 and E=0 is the parabola.  The preceding section on Conic Sections discussed this in detail.

Inverse Square Law Orbits – Review

0.  General:

1/r  =  B + A cos(q-qc)

where

B = (-mK/L2),              L = mr2θ’

and

K = -Gm1m2 for gravity,   and   K = kq1q2 for electricity.

We have defined eccentricity as:

e  =  A /½B½   and we have e   =  {1 + 2EL2/mK2}1/2 ,

where

E = ½m[r'2+(rq')2] + K/r .

We also have from the General Central Force section a relation involving period and area of orbit:

S = (L/2m)T         ,

where T is the period of orbit and S is the area of the orbit..

1.  Circle    ε = 0;     E < 0;     attractive force at center

r = constant = 1/B = -L2/mK       (K<0, so attractive force only)

ε = 0,    so   A = 0  ;    vr = r’ = 0,    so

E  =  ½mr2θ2 + K/r   =   -mK2/2L2   =   K/2r  ,

where the first expression for energy comes from KE + PE = ½mr’2 + ½mr2θ’2 + K/r;

the second expression for energy comes from e   =  {1 + 2EL2/mK2}1/2 with ε = 0; and

the third expression for energy comes from  r = -L2/mK  to get L2 = -mKr and using this in the second expression for energy: E = -mK2/2L2 = -mK2/2{-mKr) = K/2r.  Recall that K is negative (attractive force), so Energy is negative (particle is bound to the central force). Note also that the positive kinetic energy is half of the negative potential energy.

T = 2Sm/L = 2πr2m/L = 2πr2m/mr2θ’ = 2π/θ’

=  2πr2m/L = 2πr2m/(-mKr)½  =  2π (m/-K)½ r3/2

where we have used r = -L2/mK  to get L2 = -mKr to use in the  2πr2m/L  term for T.

2.  Ellipse   0 < ε < 1;    E < 0;    attractive force at either of two foci

1/rmin = B + A   (when θ=θc);       1/rmax = B – A  (when θ=θc+180o)

rmin + rmax  =  2a  =  r1 + r2  ;    or   a = ½(rmin + rmax)  =  1/{B(1-ε2)}

ε  =  A /½B½   =  {1 + 2EL2/mK2}1/2    =   (rmax - rmin) / (rmax + rmin)

x  =  εa  =  ½(rmax - rmin)  =  distance of foci from center

L = mr2θ’ ;    E = ½mr’2 + ½mr2θ’2 + K/r;

B = -mK/L2  =  ½(1/rmax + 1/rmin)

S =  πab,    where   b  =  a[1-ε2]½

T = 2mS/L .

Example:   You are in circular orbit around the earth in a space shuttle at a height of 150 km.  You are to pick up a geosynchronous satellite (one with a period of 24 hours).  What do you need to do?

Basic idea:  We are in a circular orbit.  We need to get into an elliptical orbit with a perigee at the original circular orbit, and an apogee at the new geosynchronous satellite orbit.  We then need to determine the speed in the original circular orbit, the speed in the elliptical orbit at perigee (so we can speed up to that speed), the speed in the elliptical orbit at apogee, and the speed of the satellite in the geosynchronous orbit so we can match that speed.

a)  To determine the elliptical orbit, we need to know the perigee (rmin) and apogee (rmax).

a-1)  The rmin is just the earth’s radius plus the 150 km height:  rmin = 6,378 km plus 150 km = 6,528 km = 6.528 x 106 m  =  rmin .

a-2)  To get rmax, we need to determine the radius for an orbit that has a period of 24 hours (geosynchronous).  Since we have uniform circular motion, we can start with F = ma, where F = GMm/r2 and a = ac = w2r  to get GM/w2 = r3 ,  or

r = [(6.67 x 10-11 Nt*m2/kg2)*(6 x 1024 kg) / (2p/{24 hrs * 60 min/hr * 60 sec/min})2 ]1/3  =  4.2297 x 107 m  =  rmax =  6.63 * Rearth  .

b)  To get the speeds for the circular orbits, we can again use Newton’s Second Law but this time use ac = v2/r  to get:  GMm/r2 = mv2/r ,  or  v = [GM/r]1/2 .

b-1)  For the original shuttle speed in its circular orbit,  vshuttle = [(6.67 x 10-11 Nt*m2/kg2) * (6 x 1024 kg) / (6.528 x 106 m)]1/2  =  7,830 m/s.

b-2)  For the satellite speed in its circular orbit,  vsat  =   [(6.67 x 10-11 Nt*m2/kg2) *

(6 x 1024 kg) / (4.23 x 107 m)]1/2  =  3,076 m/s.

c)  To get the speed for the elliptical orbit, one at perigee and one at apogee, we need to find either E or L, since E = ½ mv2 – GM/r   and  L = mvr  .  (Note that at perigee and apogee, r’ = 0 so r and v are perpendicular.)  From the basic equation for an ellipse:

1/r  =  B + A cos(q-qc)  we can find B by using rmin (r = rmin when q = qc) and rmax (r = rmax when q = qc+180o)  so  [(1/rmin) + (1/rmax)]  =  2B  where B =  -mK/L2 ,  where

K = -GMm .  This gives:  L = [2m2GM / {(1/rmin) + (1/rmax)}]1/2  =  mvr .  Solving for v gives:  v =  [2GM / {(1/rmin) + (1/rmax)}]1/2  / r .  Solving for the v at rperigee and at rapogee give:  vperigee  =  10,306 m/s;   vapogee  =  1,590 m/s.

d)  Thus we need to speed up from the shuttle’s regular orbital speed:  from 7,830 m/s up to 10,306 m/s.  We then need to speed up again from the elliptical speed at apogee of 1,590 up to the orbital speed of the satellite of 3,076 m/s.  This is a LOT of extra energy, and so far the shuttle has not had this as a mission.

e)  Where in its relative orbit should the first burn of the shuttle to pick up speed be made?  We need to know the periods for the elliptical orbit and the geosynchronous satellite orbit.  The circular orbit of the satellite has a period of 24 hours (by definition).  But what about the period for the elliptical orbit?  We start with our expression for the period:  T = 2mS/L  where  S = area = pab ,  where  rmin + rmax  =  2a   and

where   b  =  a[1-ε2]½ ,  where  ε  =   (rmax + rmin) / (rmax - rmin) .  We have already determined L/m to be  [2GM / {(1/rmin) + (1/rmax)}]1/2  .  Calculating the various values gives:  a = 2.44 x 107 m;  e = 0.733;  b = 1.26 x 107 m;  L/m = 6.73 x 1010 m2/s; and finally Tellipse = 7.97 hours.  To go from rmin to rmax will take ½ of a period, or 3.99 hours.  The satellite goes (3.99/24) of an orbit, or 59.8o during this time.  Thus the shuttle needs to start behind the satellite by 59.8o behind the point of meeting the satellite.

Homework Problem #22:  Explorer I satellite had a perigee of 360 km and an apogee of 2,549 km above the earth’s surface.  Find its distance above the earth’s surface when it passed over a point 90o around the earth from its perigee.

Homework Problem #23:  The earth moves in a fairly circular (assume perfectly circular) orbit with radius of 1.49 x 1011 m.  Mars moves in an elliptical orbit with a perihelion distance of 2.06 x 1011 meters and aphelion distance of 2.485 x 1011 meters.

a)  Find the speed of the Earth in its circular orbit about the sun.

b)  Find the speed of Mars at perihelion.

c)  Find an orbit for a spaceship that has its perihelion at the earth distance, and its aphelion at the perihelion distance of Mars.

d)  Find the speed the spaceship should have at its perihelion (at the Earth’s distance) and compare to the Earth’s speed.  (This difference is the additional speed we would need to supply to the spaceship after we lift it off the earth’s surface.)

e)  Find the speed the spaceship should have at its aphelion (at the Mar’s perihelion) and compare to Mar’s speed at that location.   (This is the speed we would need to add to the spaceship to have it match Mars’ speed.)

3.  Parabola         ε = 1;    E = 0;    attractive force at the one focus

1/rmin  =  2B;   B = -mK/L2;     A = B

L = mr2θ’ ;    E = 0  = ½mr’2 + ½mr2θ’2 + K/r .

If you know rmin, you can get B, and then get L, and then q’at rmin;  if you don’t know rmin, you could use any r as long as you had the (q-qc) value for that r, and use these in the
1/r  =  B + A cos(
q-qc)  equation to determine B (since A=B) instead.

4.  Hyperbola         ε > 1;    E > 0;   attractive or repulsive force at one of the two foci

attractive:  K < 0;  B > 0;  1/rmin  =  B + A  =  B(1+ε);

cos(θcriticalθc)  =  - B/A  =  -1/ε

repulsive:  K > 0;  B < 0;  1/rmin  =  B + A  =   ½B½(ε – 1)

cos(θcriticalθc)  =  - B/A  =  +1/ε

rrepulsive - rattractive  =  2a,   where  a = 1/{½B½2 – 1) }

rrepulsive-min + rattractive-min  =  2εa

Θ = 180o – 2(θcriticalo) ,  where Θ is the angle through which the object is deflected

tan(Θ/2)   =  1 / [ε2 – 1]½     and using  e   =  {1 + 2EL2/mK2}1/2   we get

tan(Θ/2)   =  {mK2/2EL2}1/2

L = mr2θ’ ;    E = ½mr’2 + ½mr2θ’2 + K/r;    B = -mK/L2

With the hyperbola, we often have the incoming projectile start far away from the scatting source, and we know the initial speed, vo, and we know the impact parameter, s.  [The impact parameter is the distance the particle will miss the scattering source if there were no attractive or repulsive force between the scattering source and the projectile.]  From this information, we can determine the energy, E = ½mvo2 , since far away the potential energy is zero.  We also know the angular momentum, L = mvos.  By knowing E and L, we can use  e   =  {1 + 2EL2/mK2}1/2   to get the eccentricity, e.  By knowing L we can get B from   B = -mK/L2 .  By knowing e and B, we can get A since e = A/½B½.  By knowing E and L we can get  Θ  by using  tan(Θ/2)   =  {mK2/2EL2}1/2   . We’ll deal more with the hyperbola in the next section on Scattering.