Gravitation law equation as R approaches zero

AI Thread Summary
As the radius (R) approaches zero in the gravitational force equation F = GMm/R^2, the force becomes undefined, particularly for point particles, as they cannot perfectly overlap. For larger bodies like planets, modifications are necessary; according to Gauss's law, only the mass within the radius of interest affects the gravitational force. As one moves toward the center of a planet, the enclosed mass decreases, causing the gravitational force to approach zero. Discussions also highlight that while theoretical scenarios can lead to infinite gravity, practical physics maintains a finite radius in unified theories. Understanding these concepts is essential for grasping gravitational interactions in different contexts.
pconstantino
Messages
40
Reaction score
0
Hello,

in the equation which describes the force as a function of the radius:

F = GMm/R^2


What happens as R approaches zero? or even when R is less than the radius of the planet.

mass m will be inside the planet so this formula seems to break down because m will be pulled from both sides.

how can this be handled?

thank you !
 
Physics news on Phys.org
That's an excellent question. For a detailed answer, try looking at http://en.wikipedia.org/wiki/Gauss's_law_for_gravity and the references it cites.

If you have two massive point particles (infinitesimally small) which are gravitationally interacting with each-other, the force becomes undefined as the separation approaches zero. This is a non-physical scenario however, as two particles can never perfectly overlap.

If you're talking about a planet (or other large body) you're exactly right---the equation has to be modified. What "Gauss' law" (the link I posted) states, is that only the mass within the radius you're interested in, matters. (This is an oversimplification, but its the basic idea). If you are half way to the center of the earth, you'll only need to consider the mass within a sphere of radius = half the Earth's radius (approximately 1/8 the mass of the earth). Thus, as your distance from the center approaches zero, the mass enclosed approaches zero, and the force goes to zero.

Does this make sense?
We can go into more math/details if you're interested.
 
Two objects cannot match exactly. The distance is calculated between point of gravity of both objects. So in 3d 2 objects cannot overlap exactly on their point of gravity. Then why thinking about it?
 
pconstantino said:
Hello,

in the equation which describes the force as a function of the radius:

F = GMm/R^2


What happens as R approaches zero? or even when R is less than the radius of the planet.

Of course there at objects which are VERY massive and their radius is in fact considered to be 0. What happens then is that gravity gets larger and larger as you approach them, until it goes to infinity.
 
Raama said:
Two objects cannot match exactly. The distance is calculated between point of gravity of both objects. So in 3d 2 objects cannot overlap exactly on their point of gravity. Then why thinking about it?
Because non-physical questions are still often good ones both per se and for elucidating physical ones. Additionally, as 'objects' are actually waves, they can overlap largely.


Lsos said:
Of course there at objects which are VERY massive and their radius is in fact considered to be 0. What happens then is that gravity gets larger and larger as you approach them, until it goes to infinity.
That's not accurate. In pure gravity theories one would simply say the gravitational force becomes undefined. Otherwise, most unifying theories always preserve a finite radius (e.g. string theory).
 
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Scalar and vector potentials in Coulomb gauge Assume Coulomb gauge so that $$\nabla \cdot \mathbf{A}=0.\tag{1}$$ The scalar potential ##\phi## is described by Poisson's equation $$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$ which has the instantaneous general solution given by $$\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{3}$$ In Coulomb gauge the vector potential ##\mathbf{A}## is given by...
Thread 'Griffith, Electrodynamics, 4th Edition, Example 4.8. (First part)'
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8 and stuck at some statements. It's little bit confused. > Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin. Solution : The surface bound charge on the ##xy## plane is of opposite sign to ##q##, so the force will be...
Back
Top