[ Posted originally in the Stack Exchange PHYSICS forum and because of the hostile environment there mirrorred at this place. ]

How does a planetery orbit change with varying masses?

Assuming a circular orbit and two point masses is sufficient for our purpose. With force of gravity equals centripetal force according to Newton's laws it reads: $$ F = G\,\frac{M\,m}{r^2} = m\left(\frac{2\pi}{T}\right)^2 r \quad \Longrightarrow \quad T = 2\pi\sqrt{\frac{r^3}{GM}} $$ With $G = $ gravity constant, $M,m = $ masses, $r = $ distance between masses, $T = $ orbital period.
Now suppose - wild hypothesis - that both $m$ and $M$ become heavier as time proceeds. How will the orbit change?
  1. The period $T$ will become smaller and the distance $r$ remains the same
  2. The distance $r$ will become bigger and the period $T$ remains the same
  3. Both the distance $r$ and the period $T$ will change
  4. None of the above
I'm really at lost what to think about it. Please help.

[ Some users suggesting that this is homework. I do not agree: ]

Update. THIS IS NO HOMEWORK!
Take a look at my profile. How can it be? I am a retired person.

[ Question downvoted and put on hold, comments moved to chat ]
[ Nothing seems to help, so I decide to solve the problem myself: ]

EDIT. Here is the (Delphi Pascal) program unit that calculates distance $r$ (graphically) and period $T$ (numerically) as a function of increasing masses of both the earth and the sun:

unit Unit9;
{
  This software has been designed and is CopyLefted
  by Han de Bruijn:
                          (===)
                         @-O^O-@
                          #/_\#
                           ###
}
interface
uses Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms, Dialogs, ExtCtrls, Grafisch;
type TForm1 = class(TForm) Image1: TImage; procedure Scheppen(Sender: TObject); procedure Toetsdruk(Sender: TObject; var Key: Char); private { Private declarations } public { Public declarations } end;
var Form1: TForm1;
implementation
{$R *.dfm}
procedure Tekenen; type vektor = record x,y : double; end; const maal = 7000; G : double = 6.67408E-11; { Gravitation constant } eps : double = 0.00005; var dmv,r,v : vektor; Mz,Ma,dMa,dMz,L,dt,T,jaar,vorig : double; k : integer; rond : boolean; begin Mz := 1.9891E30; { Mass of sun } Ma := 5.972E24; { Mass of earth } r.x := 149.60E9; r.y := 0; { Orbit radius } v.x := 0; v.y := 2*Pi/(3600*24*365.256)*r.x; { Initial velocity calculated from sidereal year } dt := 3600*24/2; { Time step half an hour } { dMa := 0; dMz := 0; } { Test giving known results } dMa := 0.00001*Ma; dMz := 0.00001*Mz; { Assumed increase in mass of earth and sun } { Writeln(dMa,dMz); } vorig := 0; T := 0; Form1.Image1.Canvas.MoveTo(x2i(r.x),y2j(r.y)); for k := 0 to maal-1 do begin L := sqrt(sqr(r.x)+sqr(r.y)); { Length of radius } dmv.x := -G*Mz*Ma/(L*L*L)*r.x*dt; { Momentum- } dmv.y := -G*Mz*Ma/(L*L*L)*r.y*dt; { difference } v.x := (dmv.x-dMa*v.x)/Ma+v.x; { Orbit } v.y := (dmv.y-dMa*v.y)/Ma+v.y; { velocity } r.x := r.x+v.x*dt; { Orbit } r.y := r.y+v.y*dt; { radius } Form1.Image1.Canvas.LineTo(x2i(r.x),y2j(r.y)); Ma := Ma + dMa; { Assumed } Mz := Mz + dMz; { increase } T := T + dt; { Period } rond := (abs(r.x/L-1) < eps); { After a whole period ("year") } if rond then begin jaar := T/(3600*24*365.256); Writeln('T = ',(jaar-vorig):5:2,' year'); vorig := jaar; end; end; end;
procedure TForm1.Scheppen(Sender: TObject); { On Create } begin TV(Form1.Image1); ClearDevice; xmin := -160.0E9; xmax := 160.0E9; ymin := -160.0E9; ymax := 160.0E9; Tekenen; end;
procedure TForm1.Toetsdruk(Sender: TObject; var Key: Char); { On KeyPress } begin Exit; end;
end.
Output numerically (period becoming smaller):
T =  0.98 year
T =  0.95 year
T =  0.92 year
T =  0.89 year
T =  0.86 year
T =  0.84 year
T =  0.81 year
T =  0.79 year
T =  0.77 year
T =  0.75 year
T =  0.73 year
Output graphically (distance becoming smaller):

It is concluded that (3.) is the right answer. However, both $r$ and $T$ are decreasing.