Where 'd' is density of the two bodies experiencing gravitational force instead of mass 'm' only.
In support .....
A hot air balloon rises upwards defying gravity without any thrust .
It rises only due to the concept of hot air rising upwards due to decrease in its density.
Universally. ...
If a body with same mass that of earth but with significantly greater volume is placed in earth's orbit, it will wobble and chances are that it may leave the orbit without acknowledging the sun's gravitational pull.
This can be more clear with the following explanation:
Earthy analogy....
Let us rotate a metal ball of mass 'm' and volume 'v1' around a fixed axis with radius of it's orbit 'r' and with the same mass 'm' and radius of orbit 'r' but with greater volume 'v2' let us take another object (say a balloon ).
We may observe that on revolution of both the objects the metal ball stays around its orbit acknowledging the gravity whereas the ballon wobbles and tries to escape.
Nevertheless an air bubble when suspended in an atmosphere of dry ice stays at it's place without falling down.
Here the air bubble stays because it has lower density than that of solid CO2 . Here the gravitational pull is counter balanced by the lower density of it than the atmosphere in which it is subjected to.
We can also consider a cloud. The clouds never get to touch the surface of the earth as their density is less than the stratosphere to experience the gravitational pull of the earth.
Gravity changes the shape of space around an object . When another body is suspended around in the changed space due to gravity it must have a significant density to experience the gravitational pull of the former.
Similarly if the former object has no significant density it's gravity will fail to change the space around it and hence exert no gravitational pull towards any other body.
This must give us the idea that to experience gravitational force not only the masses of the bodies is required but also the volumes and hence density.
~RV
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