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9—Vector Calculus 1 270

The second derivative on the left side of Eq. ( 39 ) has a double spike that does not appear on the right side.
It can’t be there, so my assumption thatV is discontinuous is false andV must be continuous.
Assume next thatV is continuous but its derivative is not. The graphs ofV, ofdV/dr, and ofd^2 V/dr^2
then look like

V dV/dr d^2 V/dr^2

The second derivative on the left side of Eq. ( 39 ) still has a spike in it and there is no such spike in theρ
on the right side. This is impossible, sodV/drtoo must be continuous.

Back to the Problem
Of the four constants that appear in Eqs. ( 40 ) and ( 41 ), one is already known,C′. For the rest,

V(R−) =V(R+) is 4 πGρ 0

R^2

6

+D′=−

C

R

+D

dV

dr

(R−) =

dV

dr

(R+) is 8 πGρ 0

R

6

= +

C

R^2

These two equations determine two of the constants.

C= 4πGρ 0

R^3

3

, then D−D′= 4πGρ 0

R^2

6

+ 4πGρ 0

R^2

3

= 2πGρ 0 R^2

Put this together and you have

V R

V(r) =

{ 2

3 πGρ^0 r

(^2) − 2 πGρ
0 R
(^2) +D (r < R)
−^43 πGρ 0 R^3

/

r+D (r > R)

(43)