25.7: Worked Examples
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Example 25.1 Elliptic Orbit
A satellite of mass ms is in an elliptical orbit around a planet of mass mp>>ms. The planet is located at one focus of the ellipse. The satellite is at the distance ra when it is furthest from the planet. The distance of closest approach is rp (Figure 25.11). What is (i) the speed vp of the satellite when it is closest to the planet and (ii) the speed va of the satellite when it is furthest from the planet?

Solution: The angular momentum about the origin is constant and because →rO,a⊥→va and →rO,a⊥→vp, the magnitude of the angular momentums satisfies
L≡LO,p=LOa
Because ms<<mp the reduced mass μ≅ms and so the angular momentum condition becomes
L=msrpvp=mrsrva
We can solve for vp in terms of the constants G,mp,ra and rp as follows. Choose zero for the gravitational potential energy, U(r=∞)=0. When the satellite is at the maximum distance from the planet, the mechanical energy is
Ea=Ka+Ua=12msv2a−Gmsmpra
When the satellite is at closest approach the energy is
Ep=12msv2p−Gmsmprp
When the satellite is at closest approach the energy is
Ep=12msv2p−Gmsmprp
Mechanical energy is constant,
E≡Ea=Ep
therefore
E=12msv2p−Gmsmprp=12msv2a−Gmsmpra
From Equation (25.6.2) we know that
va=(rp/ra)vp
Substitute Equation (25.6.7) into Equation (25.6.6) and divide through by ms/2 yields
v2p−2Gmprp=r2pr2av2p−2Gmpra
We can solve this Equation (25.6.8) for vp
v2p(1−r2pr2a)=2Gmp(1rp−1ra)⇒v2p(r2a−r2pr2a)=2Gmp(ra−rprpra)⇒v2p((ra−rp)(ra+rp)r2a)=2Gmp(ra−rprpra)⇒vp=√2Gmpra(ra+rp)rp
We now use Equation (25.6.7) to determine that
va=(rp/ra)vp=√2Gmprp(ra+rp)ra
Example 25.2 The Motion of the Star SO-2 around the Black Hole at the Galactic Center
The UCLA Galactic Center Group, headed by Dr. Andrea Ghez, measured the orbits of many stars within 0.8′′ × 0.8′′ of the galactic center. The orbits of six of those stars are shown in Figure 25.12.

We shall focus on the orbit of the star S0-2 with the following orbit properties given in Table 25.13 . Distances are given in astronomical units, 1au=1.50×1011m, which is the mean distance between the earth and the sun.
Table 25.1 Orbital Properties of S0-2
Star Period (yrs) Eccentricity Semi-major axis (10−3arcsec ) Periapse (au) Apoapse (au) S0-2 15.2(0.68/0.76)0.8763(0.0063)120.7(4.5)119.5(3.9)1812(73)
The period of S0-2 satisfies Kepler’s Third Law, given by
T2=4π2a3G(m1+m2)
where m1 is the mass of S0-2, m2 is the mass of the black hole, and a is the semi-major axis of the elliptic orbit of S0-2. (a) Determine the mass of the black hole that the star S0- 2 is orbiting. What is the ratio of the mass of the black hole to the solar mass? (b) What is the speed of S0-2 at periapse (distance of closest approach to the center of the galaxy) and apoapse (distance of furthest approach to the center of the galaxy)?
Solution: (a) The semi-major axis is given by
a=rp+ra2=119.5au+1812au2=965.8au
In SI units (meters), this is
a=965.8au1.50×1011m1au=1.45×1014m
The mass m1 of the star S0-2 is much less than the mass of the black hole, and m2 Equation (25.6.11) can be simplified to
T2=4π2a3Gm2
Solving for the mass m2 and inserting the numerical values, yields
m2=4π2a3GT2=(4π2)(1.45×1014m)3(6.67×10−11N⋅m2⋅kg−2)((15.2yr)(3.16×107s⋅yr−1))2=7.79×1034kg
The ratio of the mass of the black hole to the solar mass is
m2msun=7.79×1034kg1.99×1030kg=3.91×106
The mass of black hole corresponds to nearly four million solar masses.
(b) We can use our results from Example 25.1 that
vp=√2Gm2ra(ra+rp)rp=√Gm2raarp
va=rpravp=√2Gm2rp(ra+rp)ra=√Gm2rpara
where a=(ra+rb)/2 is the semi-major axis. Inserting numerical values,
vp=√Gm2ararp=√(6.67×10−11N⋅m2⋅kg−2)(7.79×1034kg)(1.45×1014m)(1812119.5)=7.38×106m⋅s−1
The speed va at apoapse is then
va=rpravp=(1812119.5)(7.38×106m⋅s−1)=4.87×105m⋅s−1
Example 25.3 Central Force Proportional to Distance Cubed
A particle of mass m moves in plane about a central point under an attractive central force of magnitude F=br3. The magnitude of the angular momentum about the central point is equal to L . (a) Find the effective potential energy and make sketch of effective potential energy as a function of r . (b) Indicate on a sketch of the effective potential the total energy for circular motion. (c) The radius of the particle’s orbit varies between r0 and 2r0. Find r0.
Solution: a) The potential energy, taking the zero of potential energy to be at r = 0 , is
U(r)=−∫r0(−br′3)dr′=b4r4
The effective potential energy is
Ueff(r)=L22mr2+U(r)=L22mr2+b4r4
A plot is shown in Figure 25.13a, including the potential (yellow, right-most curve), the term L2/2m (green, left-most curve) and the effective potential (blue, center curve). The horizontal scale is in units of r0 (corresponding to radius of the lowest energy circular orbit) and the vertical scale is in units of the minimum effective potential.
b) The minimum effective potential energy is the horizontal line (red) in Figure 25.13a.

c) We are trying to determine the value of r0 such that Ueff(r0)=Ueff(2r0). Thus
L2mr20+b4r40=L2m(2r0)2+b4(2r0)4
Rearranging and combining terms, we can then solve for r0,
38L2m1r20=154br40r60=110L2mb
In the plot in Figure 25.13b, if we could move the red line up until it intersects the blue curve at two point whose value of the radius differ by a factor of 2 , those would be the respective values for r0 and 2r0. A graph, showing the corresponding energy as the horizontal magenta line, is shown in Figure 25.13b.
Example 25.4 Transfer Orbit
A space vehicle is in a circular orbit about the earth. The mass of the vehicle is ms=3.00×103kg and the radius of the orbit is 2Re=1.28×104km. It is desired to transfer the vehicle to a circular orbit of radius 4Re (Figure 24.14). The mass of the earth is Me=5.97×1024kg. (a) What is the minimum energy expenditure required for the transfer? (b) An efficient way to accomplish the transfer is to use an elliptical orbit from point A on the inner circular orbit to a point B on the outer circular orbit (known as a Hohmann transfer orbit). What changes in speed are required at the points of intersection, A and B?

Solution: (a) The mechanical energy is the sum of the kinetic and potential energies,
E=K+U=12msv2−GmsMeRe
For a circular orbit, the orbital speed and orbital radius must be related by Newton’s Second Law,
Fr=mar−GmsMeR2e=−msv2Re⇒12msv2=12GmsMeRe
Substituting the last result in (25.6.22) into Equation (25.6.21) yields
E=12GmsMeRe−GmsMeRe=−12GmsMeRe=12U(Re)
Equation (25.6.23) is one example of what is known as the Virial Theorem, in which the energy is equal to (1/2) the potential energy for the circular orbit. In moving from a circular orbit of radius 2Re to a circular orbit of radius 4Re, the total energy increases, (as the energy becomes less negative). The change in energy is
ΔE=E(r=4Re)−E(r=2Re)=−12GmsMe4Re−(−12GmsMe2Re)=GmsMe8Re
Inserting the numerical values,
ΔE=18GmsMeRe=14GmsMe2Re=14(6.67×10−11m3⋅kg−1⋅s−2)(3.00×103kg)(5.97×1024kg)(1.28×104km)=2.3×1010J
b) The satellite must increase its speed at point A in order to move to the larger orbit radius and increase its speed again at point B to stay in the new circular orbit. Denote the satellite speed at point A while in the circular orbit as vA,i and after the speed increase (a “rocket burn”) as vA,f Similarly, denote the satellite’s speed when it first reaches point B as vB,i Once the satellite reaches point B , it then needs to increase its speed in order to continue in a circular orbit. Denote the speed of the satellite in the circular orbit at point B by vB,f. The speeds vA,i and vB,f are given by Equation (25.6.22). While the satellite is moving from point A to point B in the elliptic orbit (that is, during the transfer, after the first burn and before the second), both mechanical energy and angular momentum are conserved. Conservation of energy relates the speeds and radii by
12ms(vA,f)2−Gmsme2Re=12ms(vB,t)2−Gmsme4Re
Conservation of angular momentum relates the speeds and radii by
msvA,f(2Re)=msvB,i(4Re)⇒vA,f=2vB,i
Substitution of Equation (25.6.27) into Equation (25.6.26) yields, after minor algebra,
vA,f=√2GMe3,vB,i=√16GMeRe
We can now use Equation (25.6.22) to determine that
vA,i=√12GMeRe,vB,f=√14GMeRe
Thus the change in speeds at the respective points is given by
ΔvA=vA,f−vA,i=(√23−√12)√GMeReΔvB=vB,f−vB,i=(√14−√16)√GMeRe
Substitution of numerical values gives
ΔvA=8.6×102m⋅s−2,ΔvB=7.2×102m⋅s−2