# What is the equation for gravitational force?

January 25, 2016

Worse than electrodynamics, general relativity is non-linear, in the sense that the field from multiple sources is not just the sum of fields from each isolated source. Even the simple case you are asking about, which is the two-body problem in general relativity, has not been solved exactly.

An even simpler case is the limit $m_2 \to 0$. In that case only $m_1$ affects the geometry of spacetime, and $m_2$ follows a geodesic in that spacetime. This has been solved exactly. The linked article gives details.

[Addendum] To directly answer the original question for the limit $m_2 \to 0$:

Of course if $m_2=0$, then the force between the two masses is $0$. But the thing about classical gravity is that the acceleration due to gravity is independent of mass. (This is the equivalence principle and is actually one of the starting points of the theory of general relativity.) So it still makes sense to ask what would be the acceleration of a negligible mass (aka, a test body) due to the gravity of another mass. The classical answer is $a_2 = \frac{Gm_1}{r^2}$.

In general relativity, the acceleration of a test body due to the gravity of a single spherical, homogenous, non-rotating mass is given exactly by the Schwarzschild solution, which link you can consult for details. The result of which is that

$$\ddot{r} = -\frac{Gm_1}{r^2} + r\dot{\theta}^2 - \frac{3Gm_1}{c^2}\dot{\theta}^2$$ $$\ddot{\theta} = -\frac{2}{r}\dot{r}\dot{\theta}$$ [CORRECTED AND SIMPLIFIED Jan 2]

where $r$ and $\theta$ are polar co-ordinates centred on the gravitating mass, the dots represent differentiation by the proper time of the test body.

So the first term is just the classical radial acceleration $-\frac{Gm_1}{r^2}$. The terms $r\dot{\theta}^2$ and $-\frac{2}{r}\dot{r}\dot{\theta}$ are the classical centrifugal and Coriolis acceleration for polar co-ordinates.

What's not classical is the extra term $\frac{3Gm_1}{c^2}\dot{\theta}^2$. Finally, there is the fact that differentiation is with respect to proper time of the test body. Different test bodies will experience time differently. They can be related by:

$$(1 - \frac{r_s}{r})\dot{t}^2 - \frac{\dot{r}^2}{(1 - \frac{r_s}{r}) c^2} - \frac{r^2\dot{\theta}^2}{c^2} = 1$$

Source: physics.stackexchange.com
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##### What is gravity and centripetal force.

Gravity is the name for the attraction of mass to mass.
Centripetal force is the real force of attraction F=gmM/r^2 to the center of gravitational attraction.