484 CHAPTER 7. ASTROPHYSICS AND COSMOLOGY
In view of (7.6.52), we obtain that
φ′′=−evR+2
rφ′+1
2
(u′−v′)φ′Again by the Schwarzschild approximation, we have
(7.6.53) φ′′=
(
2
r+
δr 0
r^2)
φ′−R.Integrating (7.6.53) and omittinge±δr^0 /r, we derive thatφ′=−r^2[
ε+∫
r−^2 Rdr]
,
whereεis a constant. Thus (7.6.51) can be rewritten as
(7.6.54) F=mMG
[
−
1
r^2+
r
δr 0R+
(
1 +
2 r
δr 0)(
εr+r∫ R
r^2dr)]
.
The solutions of (7.6.34) can be Taylor expanded. Hence by (7.6.33) we see thatu′(r) =1
r^2∞
∑
k= 0ak(r−r 0 )k.By (7.6.46), the gravitational forceFtakes the following form
F=
1
r^2∞
∑
k= 0bkrk, b 0 =−mMG.In view of (7.6.54), it implies thatRcan be expanded as
R=ε 0
r
−ε 1 +O(r),and by Physical Conclusion7.29,ε 0 andε 1 are two to-be-determined free parameters. Insert-
ingRinto (7.6.54) we obtain that
(7.6.55) F=mMG
[
−
1
r^2−
k 0
r+φ(r)]
forr>r 0.wherek 0 =^12 ε 0 , and
φ(r) =ε 1 +k 1 r+O(r), k 1 =ε+
ε 1
δr 0.
The nature of dark matter and dark energy suggests thatk 0 > 0 , k 1 > 0.Based on Theorem7.30,φ(r)→0 asr→∞, and (7.6.55) can be further simplified as in the
form forr 0 <r<r 1 ,
(7.6.56) F=mMG
[
−
1
r^2−
k 0
r+k 1 r