Modern Physics/The Law of Gravitation

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Law of Gravitation

• Aubhro Sengupta Spring 2020*

How do we calculate the force of gravity on an object? It seems like a simple enough task.

$F_{g}=g*m$

However, this is only applicable when we want to calculate the force of gravity from earth on an object on earth. What if we want to calculate the force of gravity of an asteroid in space from another asteroid, or the force of gravity on a space craft from the sun. This is where universal gravitation comes in.

Of Newton's accomplishments, the discovery of the universal law of gravitation ranks as one of his greatest. Imagine two masses, M₁ and M₂, separated by a distance r. The gravitational force has the magnitude

$F_{g}=G{\frac {M_{1}M_{2}}{r^{2}}}$

where G is the gravitational constant:

$G\approx 6.67\times 10^{-11}{\frac {m^{3}}{kg\cdot s^{2}}}$

The force is always attractive, and acts along the line joining the center of the two masses.

Lets notice a couple things here. First of all, since gravity is a force, it is a vector. The formula described above only gives us it's magnitude. What is the direction? Well depends on which force of gravity we are talking about. Since there are 2 objects here, M₁ and M₂, there are also two forces, the force of gravity on M₁ by M₂ and the force of gravity on M₂ by M₁. The magnitude of both forces are the same but the directions are opposite one another. This is actually a case of Newton's second law, as both forces are equal and opposite.

Notice that the ``r`` is squared in the bottom of the fraction. This tells us that not only does gravity get weaker as ``r`` increases, but it gets much weaker as ``r`` increases. This is why although technically all objects in the universe attract every other object through gravity, we can often disregard the force as it is negligible.

Vector Notation

Let's say that we have two masses, M and m, separated by a distance r, and a distance vector R. The relationship between R and r is given by:

$|{\vec {\mathbf {R} }}|=r$

We will also change our force into a force vector, acting in the direction of R:

${\vec {F}}_{g}=G{\frac {M_{1}M_{2}}{r^{2}}}\cdot {\frac {\vec {\mathbf {R} }}{r}}$

And this gives us our final vector equation:

${\vec {F}}_{g}=G{\frac {M_{1}M_{2}{\vec {\mathbf {R} }}}{r^{3}}}$

Notice that since the ratio between R and r is normalized, the addition of these terms does not alter the equation, only the direction in which the force is acting.