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Universal Gravitation

Universal Gravitation is likely the most basic principle you will learn in Gravitation, which, in hindsight, is rather depressing, to say the least. It is based on the following equation, which has cemented itself as one of the most iconic ever in recent years.

What

I have no idea what the above paragraph means but anyways the formula below is very important

\[F_g = G \frac{Mm}{r^2}\]

Let's first get into the definitions of each of the variables in this equations.

The Gravitational Constant, \(G\)

The Gravitational Constant is pretty much just a constant that scientists measured empirically, that has really no relevance to the solution. The only reason it exists is for calculation, and the value we use isn't necessarily accurate.

Empirically determined, \(G\) is defined as follows:

\[G = 6.6743015 \times 10^{-14} \frac{m^3}{kg \cdot s^2}\]

Yes, the units are indeed as cancerous as they seem. We usally use a rounded down value of \(G = 6.67 \times 10^{-14}\) in normal mathematical calculations, since beyond this term, the values approach redundancy due to Significant Figure and Negligibility considerations.

The Masses of the Objects, \(M\) and \(m\)

Note: Don't waste your time

this entire section can summarized as \(M\) is an object with bigger mass than \(m\)

Gravitational Force is dependent of mass, or in this case, the two masses interacting.

Picture two objects that are next to each other

Those 2 masses experience the same \(F_g\), towards each other.

\(M\) is usually for describing much bigger mass than \(m\) such that only the force on \(m\) actually matters.

Often, we use \(M_S\) or \(M_\odot\) to represent the Mass of the Sun, or the Solar Mass. We also use \(M_E\) or \(M_\bigoplus\) to represent the Mass of the Earth.

The following is a list of the variables we use for masses, which are usually represented with \(M\) since they are traditionally huge masses.

Property Value
Mass of the Sun, \(M_S\) or \(M_\odot\) \(1.989 \times 10^{30}\)
Mass of Earth, \(M_E\) or \(M_\bigoplus\) \(5.972 \times 10^{24} \approx 5.97 \times 10^{24}\)
Mass of Moon, \(M_{moon}\) \(7.348 \times 10^{22}\)

Note: You don't actually need to memorize these

There should be a constant sheet given

The distance between the objects, \(r\)

r is the distance between the objects. From the formula you can see that if the distance halves, the force increases by 4. This is the inverse square law, which is useful to know intuitively.

Summary

Warning

All of this is directly copied from the Year 4 Physics Chapter 4 notes

Consider 2 masses:

There are two forces (one on each mass) attracting both masses together. Those two forces are:

  • equal in magnitude
  • opposite in direction
  • acting on different bodies
  • of the same type (gravitational force)
  • an action-reaction pair (Newton's 3rd law)

In general:

  1. Newton's Law of Gravitation is a universal law, i.e. it works well enough everywhere in the Universe and to everything that has mass
  2. Attractive Nature of gravitational force. It always causes particles to be attracted to each other.
  3. The gravitational force is a field force (more about this later) that always exists between 2 point masses. The medium between them doesn't matter.
  4. This law only works for point masses, i.e. the dimension of the objects are very small compared with \(r\) (distance between the objects)

  5. Inverse-square Law

    Newton's Law of Gravitation is an example of an inverse-square law, i.e.: $$ F \propto \frac{1}{r^2} $$ i.e., when separation doubles, the force is \(\frac{1}{4}\) its original value

  6. Although the law is only applicable for point masses, the law also applies to the force between two bodies that have spherically symmetrical mass distributions - in such cases the body is treated as if its whole mass is concentrated at its center of mass. So \(r\) would represent the distance between the centres of mass of the 2 bodies

Question

Consider Henry Cavendish's famous experiment to directly measure the gravitational force between 2 spheres. Suppose that the mass \(m_1\) of one of the small spheres of a Cavendish balance is 10.0g and the mass \(m_2\) of one of the large spheres is 0.500 kg and the separation between the centres of the two masses is 0.0500m. Find the magnitude of the gravitational force \(F_g\) on each sphere. (\(G=6.674\times10^{-11}\))

\(1.33 \times 10^{-10} N\)