Modification of Gravitational force's law and related problem I'm facing a problem which states the following: 

Suppose that it is discovered that Newton's law of gravitation is incorrect and that the force $F$ on test particle of mass $m$ due to a body of mass $M$ has an additional term that does not depend on $M$, as follows:

$$F = - \frac{GmM}{r^2} + \frac{A m r}{3}$$

Where $A$ is a positive constant.
  Assuming that Newton's sphere theorem continues to hold, derive the appropriate form fo the Friedmann equations in this case.

How I proceeded
Consider the motion of a point like particle of mass $m$ located on the surface of a sphere of homogeneous density $\rho$ and radius $R = a r$. 
The acceleration this particle is subjected is
$$\ddot{R} = r \ddot{a} = \frac{-GM}{r^2} + \frac{AR}{3} = -\frac{G}{R^2}\left(\frac{4}{3}\pi R^3 \rho\right) + \frac{AR}{3} = -\frac{4}{3}\pi G\rho R + \frac{AR}{3}$$
Hence
$$\ddot{R} = r\ddot{a} =  \frac{1}{3}\left[A - 4\pi\rho G\right]ra$$
$$3\frac{\ddot{a}}{\dot{a}} = A - 4\pi\rho G$$
Adding the pressure term and arranging:
$$3\frac{\ddot{a}}{\dot{a}} = A - 4\pi\rho G(\rho + 3P)$$
Is this correct? 
Thank you so much!
 A: If we want to derive the Friedmann Equation by using Newtonian Mechanics we can always use the Energy of the system.
$$F=ma=-k/R^2+AmR/3$$ 
And the potential energy related to this force would be,
$$U=-\int Fdr$$
$$U=-\int(\frac{-MmG}{R^2}+\frac{AmR} {3})dr$$
$$U=-\frac {MmG} {R}-\frac {AmR^2} {6}$$
Total energy of the particle of mass m will be,
$$E=T+U$$
$$E=\frac {mV^2} {2}-\frac {MmG} {R}-\frac {AmR^2} {6}$$
Multiplying both side with $\frac {2} {mR^2}$
$$\frac {2E} {mR^2} = \frac {V^2} {R^2}-\frac {2MG} {R^3}-\frac {A} {3}$$
but $V=\dot{R}=\dot{a}(t)r$ and we know that $R=a(t)r$ hence
$$\frac {2E} {mR^2} = \frac {\dot{a}(t)^2} {a(t)^2}-\frac {2MG} {R^3}-\frac {A} {3}$$
finally writing 
$$\frac {\dot{a}(t)^2} {a(t)^2}=\frac {8\pi G\rho} {3}+\frac {A} {3}+\frac {2E} {ma(t)^2r^2}$$
if we wanted to write in relativistic form we would have,$2E/mr^2=-\kappa c^2/R^2$ and energy density $\epsilon=\rho c^2$ or $\rho=\epsilon /c^2$ where $\rho$ is the matter density.
so we have,
$$\frac {\dot{a}(t)^2} {a(t)^2}=\frac {8\pi G} {3c^2}\epsilon(t)+\frac {A} {3}-\frac {\kappa c^2} {a(t)^2R^2}$$
For the acceleration, from
$$F=ma=m\ddot{R}=-k/R^2+AmR/3$$
$$\ddot{R}=-MG/R^2+AR/3$$
$$\frac {\ddot{R}} {R} =-MG/R^3+A/3$$
or $$\frac {\ddot{a}(t)} {a(t)}= -\frac {4\pi G\rho} {3} + \frac {A} {3}$$
In general the acceleration equation is written as $$\frac {\ddot{a}(t)} {a(t)}= -\frac {4\pi G}{3c^2}(\epsilon+3P)$$
But the equation of state for matter is $P=w\epsilon=0$ since $w=0$ for matter hence we have 
$$\frac {\ddot{a}(t)} {a(t)}= -\frac {4\pi G}{3c^2}(\epsilon) + \frac {A} {3}$$
(Since we have an additional term, our equation is a bit different)
or in terms of matter density,
$$\frac {\ddot{a}(t)} {a(t)}= -\frac {4\pi G\rho} {3} + \frac {A} {3}$$
Which is exactly what we derived using the $F=ma$
